13 CALCULUS OF VECTOR-VALUED FUNCTIONS

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1 CALCULUS OF VECTOR-VALUED FUNCTIONS. Vector-Valued Functions LT Section 4.) Preliminar Questions. Which one of the following does not parametrize a line? a) r t) 8 t,t,t b) r t) t i 7t j + t k c) r t) 8 4t, + 5t, 9t a) This is a parametrization of the line passing through the point 8,, ) in the direction parallel to the vector,,, since: 8 t,t,t 8,, + t,, b) Using the parameter s t we get: t, 7t,t s, 7s, s s, 7, This is a parametrization of the line through the origin, with the direction vector v, 7,. c) The parametrization 8 4t, + 5t, 9t does not parametrize a line. In particular, the points 8,, ) at t ), 4, 7, 9) at t ), and 4,, 7) at t ) are not collinear.. What is the projection of rt) ti + t 4 j + e t k onto the z-plane? The projection of the path onto the z-plane is the curve traced b ti + e t k t,,e t. This is the curve z e in the z-plane.. Which projection of cos t,cos t,sin t is a circle? The parametric equations are cos t, cos t, z sin t The projection onto the z-plane is cos t,, sin t. Since + z cos t + sin t, the projection is a circle in the z-plane. The projection onto the -plane is traced b the curve cos t,cos t,. Therefore, cos t and cos t. We epress in terms of : cos t cos t The projection onto the -plane is a parabola. The projection onto the z-plane is the curve, cos t,sin t. Hence cos t and z sin t. We find as a function of z: cos t sin t z The projection onto the z-plane is again a parabola. 4. What is the center of the circle with parametrization The parametric equations are rt) + cos t)i + j + sin t)k? + cos t,, z sin t Therefore, the curve is contained in the plane, and the following holds: + ) + z ) cos t + sin t We conclude that the curve rt) is the circle of radius in the plane centered at the point,, ). 5 Ma 6,

2 SECTION. Vector-Valued Functions LT SECTION 4.) 5 5. How do the paths r t) cos t,sin t and r t) sin t,cos t around the unit circle differ? The two paths describe the unit circle. However, as t increases from to π, the point on the path sin ti + cos tj moves in a clockwise direction, whereas the point on the path cos ti + sin tj moves in a counterclockwise direction. 6. Which three of the following vector-valued functions parametrize the same space curve? a) + cos t)i + 9j + sin t)k b) + cos t)i 9j + sin t)k c) + cos t)i + 9j + sin t)k d) cos t)i + 9j + + sin t)k e) + cos t)i + 9j + + sin t)k All the curves ecept for b) lie in the vertical plane 9. We identif each one of the curves a), c), d) and e). a) The parametric equations are: Hence, + cos t, 9, z sin t + ) + z ) cos t) + sin t) This is the circle of radius in the plane 9, centered at, 9, ). c) The parametric equations are: Hence, + cos t, 9, z sin t + ) + z ) cos t) + sin t) This is the circle of radius in the plane 9, centered at, 9, ). d) In this curve we have: Hence, cos t, 9, z + sin t + ) + z ) cos t) + sin t) Again, the circle of radius in the plane 9, centered at, 9, ). e) In this parametrization we have: Hence, + cos t, 9, z + sin t ) + z ) cos t) + sin t) This is the circle of radius in the plane 9, centered at, 9, ). We conclude that a), c) and d) parametrize the same circle whereas b) and e) are different curves. Eercises. What is the domain of rt) e t i + t j + t + ) k? rt) is defined for t and t, hence the domain of rt) is: D {t R : t,t }. Evaluate What is r) the and domain r ) of for rs) rt) e s i sin + π sj t,t + cos,t sk? + ). Since rt) sin π t,t,t + ), then r) sin π, 4, 5, 4, 5 and r ) sin π,,,, Ma 6,

3 5 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) 5. Find a vector parametrization of the line through P, 5, 7) in the direction Does either of P 4,, ) or Q, 6, 6) lie on the path rt) v,, + t, + t.,t 4? We use the vector parametrization of the line to obtain: rt) OP + tv, 5, 7 + t,, + t, 5, 7 + t or in the form: rt) + t)i 5j t)k, <t< 7. Match the space curves in Figure 8 with their projections onto the -plane in Figure 9. Find a direction vector for the line with parametrization rt) 4 t)i + + 5t)j + tk. z z z A) B) C) FIGURE 8 i) ii) FIGURE 9 iii) The projection of curve C) onto the -plane is neither a segment nor a periodic wave. Hence, the correct projection is iii), rather than the two other graphs. The projection of curve A) onto the -plane is a vertical line, hence the corresponding projection is ii). The projection of curve B) onto the -plane is a periodic wave as illustrated in i). 9. Match the vector-valued functions a) f) with the space curves i) vi) in Figure. Match the space curves in Figure 8 with the following vector-valued functions: a) rt) t + 5,e.8t cos t,e.8t sin t b) rt) cos t,sin t,sin t a) r t) cos t,cos t,sin t 5t c) rt) b) r t) t,t, t,cos t,sin t + t c) r t),t,t d) rt) cos t,sin t,sin t e) rt) t,t, t f) rt) cos t,sin t,cos t sin t z z z i) ii) iii) z z z iv) v) vi) FIGURE a) v) b) i) c) ii) d) vi) e) iv) f) iii) Ma 6,

4 SECTION. Vector-Valued Functions LT SECTION 4.) 5. Match Which the of space the curves following A) C) curves in have Figure the same with projection their projections onto the i) iii) -plane? onto the -plane. a) r t) t,t,e t b) r t) e t,t,t c) r t) t,t, cos t z z z A) B) C) z z z i) ii) FIGURE iii) Observing the curves and the projections onto the -plane we conclude that: Projection i) corresponds to curve C); Projection ii) corresponds to curve A); Projection iii) corresponds to curve B). In Eercises Describe 6, the the projections function of rt) thetraces circlea rt) circle. sin Determine t,, 4 + the cos radius, t ontocenter, the coordinate and planeplanes. containing the circle.. rt) 9 cos t)i + 9 sin t)j Since t) 9 cos t, t) 9 sin t we have: + 8 cos t + 8 sin t 8cos t + sin t) 8 This is the equation of a circle with radius 9 centered at the origin. The circle lies in the -plane. 5. rt) sin t,, 4 + cos t rt) 7i + cos t)j + sin t)k t) sin t, zt) 4 + cos t, hence: + z 4) sin t + cos t is the equation of the z-plane. We conclude that the function traces the circle of radius, centered at the point,, 4), and contained in the z-plane. 7. Let C be the curve rt) t cos t, t sin t, t. rt) 6 + sin t,9, 4 + cos t a) Show that C lies on the cone + z. b) Sketch the cone and make a rough sketch of C on the cone. t cos t, t sin t and z t, hence: + t cos t + t sin t t cos t + sin t ) t z. + z is the equation of a circular cone, hence the curve lies on a circular cone. As the height z t increases linearl with time, the and coordinates trace out points on the circles of increasing radius. We obtain the following curve: z rt) t cos t,t sin t,t Ma 6,

5 54 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) In Eercises 9 and, let Use a computer algebra sstem to plot the projections onto the - and z-planes of the curve rt) t cos t,t sin t,t in Eercise 7. rt) sin t,cos t,sin t cos t as shown in Figure. z FIGURE 9. Find the points where rt) intersects the -plane. The curve intersects the -plane at the points where z. That is, sin t cos t and so either sin t or cos t. The s are, thus: t πk or t π 4 + πk, k, ±, ±,... ) The values t πk ield the points: sin πk,cos πk,), ) k,. The values t π 4 + πk ield the points: k : sin π 4, cos π ) ) 4,,, k : sin π4, cos π4 ),, ), k : sin 5π4, cos 5π4 ),, ), k : sin 7π4, cos 7π4 ), ),, Other values of k do not provide) new points). We) conclude that the curve ) intersects the ) -plane at the following points:,, ),,, ),,,,,,,,,,,,. Parametrize the intersection of the surfaces Show that the projection of rt) onto the z-plane is the curve z z, + z 9 for using t as the parameter two vector functions are needed as in Eample ). We solve for z and in terms of. From the equation + z 9 we have z 9 or z ± 9. From the second equation we have: z + 9 ) + 7 Taking t as a parameter, we have z ± 9 t, t 7, ielding the following vector parametrization: rt) t 7,t,± 9 t, for t.. Viviani s Curve C is the intersection of the surfaces Figure ) Find a parametrization of the curve in Eercise using trigonometric functions. + z, z a) Parametrize each of the two parts of C corresponding to and, taking t z as parameter. b) Describe the projection of C onto the -plane. c) Show that C lies on the sphere of radius with center,, ). This curve looks like a figure eight ling on a sphere [Figure B)]. Ma 6,

6 SECTION. Vector-Valued Functions LT SECTION 4.) 55 z + z Viviani's curve A) z B) Viviani s curve viewed from the negative -ais. FIGURE Viviani s curve is the intersection of the surfaces + z and z. a) We must solve for and in terms of z which is a parameter). We get: z z ± z ± z z 4 Here, the ± from ± z z 4 represents the two parts of the parametrization: + for, and for. Substituting the parameter z t we get: t, ± t t 4 ±t t. We obtain the following parametrization: rt) ±t t,t,t for t ) b) The projection of the curve onto the -plane is the curve on the -plane obtained b setting the z-coordinate of rt) equal to zero. We obtain the following curve: ±t t,t,, t We also note that since ±t t, then t t ), but also t, so that gives us the equation ) for the projection onto the plane. We rewrite this as follows. ) /4 /4 + /) /) We can now identif this projection as a circle in the plane, with radius /, centered at the point, /). c) The equation of the sphere of radius with center,, ) is: + ) + z ) To show that C lies on this sphere, we show that the coordinates of the points on C given in )) satisf the equation of the sphere. Substituting the coordinates from ) into the left side of ) gives: + ) + z ±t t ) + t ) + t t t ) + t ) + t t )t t ) + t We conclude that the curve C lies on the sphere of radius with center,, ). 5. Use sine and cosine to parametrize Show that an point on + the z intersection of the clinders + and + z use two vector-valued functions). Then describe the projections can be written of this in curve the form onto z the cos three θ,zsin coordinate θ,z) for planes. some θ. Use this to find a parametrization of Viviani s curve Eercise ) with θ as parameter. The circle + z inthez-plane is parametrized b cos t, z sin t, and the circle + in the -plane is parametrized b cos s, sin s. Hence, the points on the clinders can be written in the form: + z : cos t,,sin t, t π + : cos s, sin s, z, t π Ma 6,

7 56 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) The points,,z) on the intersection of the two clinders must satisf the following equations: cos t cos s sin s z sin t The first equation implies that s ±t + πk. Substituting in the second equation gives sin ±t + πk) sin ±t) ± sin t. Hence, cos t, ±sin t, z sin t. We obtain the following vector parametrization of the intersection: rt) cos t,± sin t,sin t The projection of the curve on the -plane is traced b cos t,± sin t, which is the unit circle in this plane. The projection of the curve on the z-plane is traced b cos t,, sin t which is the unit circle in the z-plane. The projection of the curve on the z-plane is traced b, ± sin t,sin t which is the two segments z and z for. 7. Use sine and cosine to parametrize the intersection of the surfaces + Use hperbolic functions to parametrize the intersection of the surfaces and z 4 Figure 4). 4 and z. z FIGURE 4 Intersection of the surfaces + and z 4. The points on the clinder + and on the parabolic clinder z 4 can be written in the form: + : z 4 : cos t,sin t,z,,4 The points,,z)on the intersection curve must satisf the following equations: We obtain the vector parametrization: cos t sin t cos t, sin t, z 4 cos t z 4 rt) cos t,sin t,4 cos t, t π Using the CAS we obtain the following curve: z 4 rt) cos t,sin t,4 cos t Ma 6,

8 SECTION. Vector-Valued Functions LT SECTION 4.) 57 In Eercises 8, two paths r t) and r t) intersect if there is a point P ling on both curves. We sa that r t) and r t) collide if r t ) r t ) at some time t. 9. Determine whether r Which of the following and r statements collide or intersect: are true? a) If r and r intersect, then the collide. r t) t +,t +, 6t b) If r and r collide, then the intersect. r t) 4t,t,t 7 c) Intersection depends onl on the underling curves traced b r and r, but collision depends on the actual parametrizations. To determine if the paths collide, we must eamine whether the following equations have a : We simplif to obtain: t + 4t t + t 6 t t 7 t 4t + t )t ) t t 7t 6 The of the second equation is t. This is also a of the first and the third equations. It follows that r ) r ) so the curves collide. The curves also intersect at the point where the collide. We now check if there are other points of intersection b solving the following equation: Equating coordinates we get: t +,t +, 6 t r t) r s) 4s, s,s 7 t + 4s t + s 6 t s 7 B the second equation, t s. Substituting into the first equation ields: s ) + 4s 4s s s Substituting s and s into the second equation gives: The s of the first two equations are: s 4s + s, s t + t t + t t, s ; t, s We check if these s satisf the third equation: 6 6 t 6, s s t t, s t s 7 We conclude that the paths intersect at the endpoints of the vectors r ) and r ) or equivalentl r ) and r )). That is, at the points 4,, 6) and, 4, ). In Eercises Determine 4, whether find a parametrization r and r collide ofor the intersect: curve.. The vertical line passing through the point r t), t,t,,t ), r t) 4t + 6, 4t, 7 t The points of the vertical line passing through the point,, ) can be written as,,z). Using z t as parameter we get the following parametrization: rt),,t, <t< Ma 6,

9 58 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4). The line through the origin whose projection on the -plane is a line of slope and whose projection on the z-plane The line passing through,, 4) and 4,, ) is a line of slope 5 i.e., z/ 5) We denote b,,z)the points on the line. The projection of the line on the -plane is the line through the origin having slope, that is the line in the -plane. The projection of the line on the z-plane is the line through the origin with slope 5, that is the line z 5. Thus, the points on the desired line satisf the following equalities: z 5, z 5 5 We conclude that the points on the line are all the points in the form,,5). Using t as parameter we obtain the following parametrization: rt) t,t,5t, <t<. 5. The circle of radius with center,, 5) in a plane parallel to the z-plane The horizontal circle of radius with center,, 4) The circle is parallel to the z-plane and centered at,, 5), hence the -coordinates of the points on the circle are. The projection of the circle on the z-plane is a circle of radius centered at, 5). This circle is parametrized b: + cos t, z 5 + sin t We conclude that the points on the required circle can be written as, + cos t,5 + sin t). This gives the following parametrization: rt), + cos t,5 + sin t, t π. 7. The intersection of) the plane ) with the sphere + + z The ellipse + inthe-plane, translated to have center 9, 4, ) Substituting in the equation of the sphere gives: ) + + z + z 4 This circle in the horizontal plane has the parametrization cos t, z sin t. Therefore, the points on the intersection of the plane and the sphere + + z, can be written in the form cos t, ), sin t, ielding the following parametrization: rt) cos t,, sin t, t π. ) z ) 9. The The ellipse intersection + of the surfaces inthez-plane, translated to have center,, 5) [Figure 5A)] z and z + z z A) B) FIGURE 5 The ellipses described in Eercises 9 and 4. The translated ellipse is in the vertical plane, hence the -coordinate of the points on this ellipse is. The and z coordinates satisf the equation of the ellipse: ) ) z 5 +. This ellipse is parametrized b the following equations: + cos t, z 5 + sin t. Ma 6,

10 SECTION. Vector-Valued Functions LT SECTION 4.) 59 Therefore, the points on the translated ellipse can be written as + cos t,, 5 + sin t). This gives the following parametrization: rt) + cos t,, 5 + sin t, t π. Further Insights and Challenges ) z ) The ellipse +, translated to have center,, 5) [Figure 5B)] 4. Sketch the curveparametrized b rt) t +t, t t. We have: t +t { { t t t > ; t t t t t> As t increases from to, the -coordinate is zero and the -coordinate is positive and decreasing to zero. As t increases from to +, the -coordinate is zero and the -coordinate is positive and increasing to +. We obtain the following curve: rt) t + t, t t 4. Find Let the maimum C be the curve height obtained above the b -plane intersecting of a point clinder on rt) of radius e t r, sin andt,t4 a plane. t) Insert. two spheres of radius r into the clinder above and below the plane, and let F and F be the points where the plane is tangent to the spheres [Figure 6A)]. Let K be the vertical distance between the equators of the two spheres. Rediscover Archimedes s proof that C is an ellipse b showing that ever point P on C satisfies PF + PF K Hint: If two lines through a point P are tangent to a sphere and intersect the sphere at Q and Q as in Figure 6B), then the segments PQ and PQ have equal length. Use this to show that PF PR and PF PR. R F P F K Q Q R P A) FIGURE 6 B) To show that C is an ellipse, we show that ever point P on C satisfies: F P + F P K We denote the points of intersection of the vertical line through P with the equators of the two spheres b R and R see figure). Ma 6,

11 6 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) R F P F K R We denote b O and O the centers of the spheres. O r F P Since F is the tangenc point, the radius O F is perpendicular to the plane of the curve C, and therefore it is orthogonal to the segment PF on this plane. Hence, O F P is a right triangle and b Pthagoras Theorem we have: O F + PF O P r + PF O P PF O P r ) O r R P O R P is also a right triangle, hence b Pthagoras Theorem we have: O R + R P O P r + R P O P PR O P r ) Combining ) and ) we get: PF PR ) Similarl we have: PF PR 4) We now combine ), 4) and the equalit PR + PR K to obtain: F P + F P PR + PR K Thus, the sum of the distances of the points P on C to the two fied points F and F is a constant K>, hence C is an ellipse. 45. Now reprove the result of Eercise 4 using vector Assume that the clinder in Figure 6 has equation geometr.assume + r that the clinder has equation + and the plane has equation z a + b. Find r and the plane has equation z a + b. a vector parametrization rt) of the curve of intersection using the trigonometric functions cos t and sin t. a) Show that the upper and lower spheres in Figure 6 have centers ) C,,r a + b + ) C,, r a + b + Ma 6,

12 SECTION. Calculus of Vector-Valued Functions LT SECTION 4.) 6 b) Show that the points where the plane is tangent to the sphere are r F a,b,a + b ) a + b + r F a,b,a + b ) a + b + Hint: Show that C F and C F have length r and are orthogonal to the plane. c) Verif, with the aid of a computer algebra sstem, that Eq. ) holds with K r a + b + To simplif the algebra, observe that since a and b are arbitrar, it suffices to verif Eq. ) for the point P r,,ar). a) and b) Since F is the tangenc point of the sphere and the plane, the radius to F is orthogonal to the plane. Therefore to show that the center of the sphere is at C and the tangenc point is the given point we must show that: C F r ) C F is orthogonal to the plane. ) We compute the vector C F : ra C F a + b +, rb a + b +, ra + b ) a a + b + r + b r + a,b, a + b + Hence, C F r r a,b, a + b + a + b + ) r a + b + We, thus, proved that ) is satisfied. To show ) we must show that C F is parallel to the normal vector a,b, to the plane z a + b i.e., a + b z ). The two vectors are parallel since b ) C F is a constant multiple of a,b,. In a similar manner one can show ) and ) for the vector C F. c) This is an etremel challenging problem. As suggested in the book, we use P r,,ar), and we also use the epressions for F and F as given above. This gives us: PF + a + b a + a + b ) r PF + a + b + a + a + b ) r Their sum is not ver inspiring: PF + PF + a + b a + a + b ) r + + a + b + a + a + b ) r Let us look, instead, at P F + PF ), and show that this is equal to K. Since everthing is positive, this will impl that PF + PF K, as desired. P F + PF ) r + 4 a r + b r + r 4 + b r 4 + b 4 r 4 r + 4 a r + b r + + b )r 4r + a + b ) K. Calculus of Vector-Valued Functions LT Section 4.) Preliminar Questions. State the three forms of the Product Rule for vector-valued functions. The Product Rule for scalar multiple ft)of a vector-valued function rt) states that: d dt ft)rt) ft)r t) + f t)rt) The Product Rule for dot products states that: d dt r t) r t) r t) r t) + r t) r t) Ma 6,

13 6 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Finall, the Product Rule for cross product is d dt r t) r t) r t) r t) + r t) r t). In Questions 6, indicate whether the statement is true or false, and if it is false, provide a correct statement.. The derivative of a vector-valued function is defined as the limit of the difference quotient, just as in the scalar-valued case. The statement is true. The derivative of a vector-valued function rt) is defined a limit of the difference quotient: r r t + h) rt) t) lim t h in the same wa as in the scalar-valued case.. There are two Chain Rules for vector-valued functions: one for the composite of two vector-valued functions and one for the composite of a vector-valued and a scalar-valued function. This statement is false. A vector-valued function rt) is a function whose domain is a set of real numbers and whose range consists of position vectors. Therefore, if r t) and r t) are vector-valued functions, the composition r r )t) r r t)) has no meaning since r t) is a vector and not a real number. However, for a scalar-valued function ft), the composition rf t)) has a meaning, and there is a Chain Rule for differentiabilit of this vector-valued function. 4. The terms velocit vector and tangent vector for a path rt) mean one and the same thing. This statement is true. 5. The derivative of a vector-valued function is the slope of the tangent line, just as in the scalar case. The statement is false. The derivative of a vector-valued function is again a vector-valued function, hence it cannot be the slope of the tangent line which is a scalar). However, the derivative, r t ) is the direction vector of the tangent line to the curve traced b rt), atrt ). 6. The derivative of the cross product is the cross product of the derivatives. The statement is false, since usuall, d dt r t) r t) r t) r t) The correct statement is the Product Rule for Cross Products. That is, d dt r t) r t) r t) r t) + r t) r t) 7. State whether the following derivatives of vector-valued functions r t) and r t) are scalars or vectors: d a) dt r d t) b) r t) r t) ) d c) r t) r t) ) dt dt a) vector, b) scalar, c) vector. Eercises In Eercises 6, evaluate the limit.. lim t, 4t, t t B the theorem on vector-valued limits we have: lim t, 4t, lim t t t t, lim 4t, lim 9,,. t t t. lim lim e t i sin + lnt ti + t cos )j tj + + 4k tan 4tk t π Computing the limit of each component, we obtain: ) ) ) ) lim e t i + ln t + ) j + 4k lim t t et i + lim lnt + ) j + lim 4 k e i + ln )j + 4k i + 4k t t Ma 6,

14 SECTION. Calculus of Vector-Valued Functions LT SECTION 4.) 6 rt + h) rt) 5. Evaluate lim lim t t +, et for rt) t, sin t,4. h h, 4t t This limit is the derivative dr dt. Using componentwise differentiation ields: r t + h) rt) lim dr d t h h dt ), d dt dt sin t), d dt 4), cos t,. t In Eercises 7, compute rt) the derivative. Evaluate lim for rt) sin t, cos t, t. 7. rt) t,t,t t t Using componentwise differentiation we get: dr d dt dt t), d dt t ), d dt t ), t,t 9. rs) e s rt),e s,s 4 7 t,4 t,8 Using componentwise differentiation we get: dr d ds ds es ), d ds e s ), d ds s4 ) e s, e s, 4s. ct) t i bt) e t 4 e t k,e 6 t,t + ) Using componentwise differentiation we get: c t) t ) i e t ) k t i e t k. Calculate r t) and r t) for aθ) cos θ)i + sin rt) t,t,t. θ)j + tan θ)k We perform the differentiation componentwise to obtain: r t) t),t ),t ), t,t We now differentiate the derivative vector to find the second derivative: r t) d, t,t,, 6t. dt 5. Sketch the curve r t) t,t Sketch the curve rt) together t,t with its tangent vector at t. Then do the same for r t) t,t 6. for t. Compute the tangent vector at t and add it to the sketch. Note that r t), t and so r ),. The graph of r t) satisfies. Likewise, r t) t, 6t 5 and so r ), 6. The graph of r t) also satisfies. Both graphs and tangent vectors are given here. r t) r t) In Eercises 7, evaluate the derivative Sketch the ccloid rt) b using the t sin t, cos t appropriate Product Rule, where together with its tangent vectors at t π and π 4. r t) t,t,t, r t) e t,e t,e t d 7. r t) r t) ) dt d dt r t) r t)) r t) r t) + r t) r t) t,t,t e t, e t,e t + t,t, e t,e t,e t t e t + t e t + te t + te t + t e t + e t t + t)e t + t + t )e t + t + )e t Ma 6,

15 64 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) 9. d r d t) t 4 r t) dt r t) ) ) dt d dt r t) r t)) r t) r t) + r t) r t) t,t,t e t, e t,e t + t,t, e t,e t,e t i j k t t t e t e t e t + i j k t t e t e t e t t e t te t )i + te t t e t )j + t e t t e t )k + t e t e t )i + e t te t )j + te t t e t )k [t + t )e t t + )e t ]i +[t + )e t t + t)e t ]j +[t + t)e t t + t )e t ]k In Eercises d and, rt) r t) ) let t, assuming that dt r t) t,, t, r t),,e t r),,, r ), 4,. Compute d dt r t) r t) in two was: t a) Calculate r t) r t) and differentiate. b) Use the Product Rule. a) First we will calculate r t) r t): r t) r t) t,, t,,e t t + + te t And then differentiating we get: d dt r t) r t)) d dt t + + te t ) t + te t + e t d dt r t) r t)) + e + e + 4e t b) First we differentiate: r t) t,, t, r t) t,, r t),,e t, r t),,e t Using the Product Rule we see: d dt r t) r t)) r t) r t) + r t) r t) t,, t,,e t + t,,,,e t te t + t + e t d dt r t) r t)) e + + e + 4e t Compute d In Eercises 6, evaluate d dt r t) r dt t) rgt)) in using two was: the Chain Rule. t. rt) a) Calculate t, r t t), gt) r t) eand t differentiate. b) Use We the Product first differentiate Rule. the two functions: r t) d dt t, t t, g t) d dt et ) e t Ma 6,

16 SECTION. Calculus of Vector-Valued Functions LT SECTION 4.) 65 Using the Chain Rule we get: d dt rgt)) g t)r gt)) e t e t, e t, e t 5. rt) rt) et,e t t,t, 4,, gt) gt) sin 4t + t 9 We first differentiate the two functions: r t) d e t,e t, 4 e t, e t, dt g t) d 4t + 9) 4 dt Using the Chain Rule we get: d dt r gt)) g t)r gt)) 4 e 4t+9, e 4t+9), 4e 4t+9, 8e 8t+8, 7. Let rt) t, t,4t. Calculate the rt) 4 sin t,6 cos t, gt) t derivative of rt) at) at t, assuming that a),, and a ), 4,. B the Product Rule for dot products we have At t we have d dt d dt rt) at) rt) a t) + r t) at) rt) at) r) a ) + r ) a) ) t We compute the derivative r ): r t) d t, t,4t t,, 4 dt r ) 4,, 4 ) Also, r),, 4 4,, 8. Substituting the vectors in the equation above, we obtain: d rt) at) dt 4,, 8, 4, + 4,, 4,, ) ) t The derivative of rt) at) at t is. In Eercises 9 4, find Let vs) s a parametrization i + sj + 9s of the tangent k. Evaluate d ds vgs)) line at at s the 4, point assuming indicated. that g4) and g 4) rt) t,t 4, t The tangent line has the following parametrization: lt) r ) + tr ) ) We compute the vectors r ) and r ): r ) ), ) 4 4, 6 r t) d t,t 4 t,4t r ) 4, dt Substituting in ) gives: The parametrization for the tangent line is, thus, lt) 4, 6 + t 4, 4 4t,6 t 4 4t, 6 t, <t<. To find a direct relation between and, we epress t in terms of and substitute in 6 t. This gives: 4 4t t 4 4. Hence, 6 t ) The equation of the tangent line is 8 6. Ma 6,

17 66 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4). rt) rt) t, 5t,t, cos t,sin t t, t π 4 The tangent line is parametrized b: lt) r) + tr ) ) We compute the vectors in the above parametrization: r), 5,,, 6 r t) d t, 5t,t t,5, 6t dt r ) 4, 5, 4 Substituting the vectors in ) we obtain the following parametrization:. rs) 4s rt) i 8 4t,5t,9t s k, s, t 4 The tangent line is parametrized b: lt),, 6 + t 4, 5, 4 4t, + 5t,6 + 4t We compute the vectors in the above parametrization: r) 4) i 8 ) k i k ls) r) + sr ) ) 4s i 8 ) s k 4s i + 8s 4 k r ) i + k r s) d ds Substituting the vectors in ) we obtain the following parametrization: lt) i ) k + s i + k ) s)i + s ) k 5. Users) Eample ln 4s)i to+ calculate s d j + 9sk, dt r s r ), where rt) t,t,e t. In Eample 4 it is proved that: d dt r r r r ) We compute the derivatives r t) and r t): r t) d t,t,e t, t,e t dt r t) d, t,e t,,e t dt Using ) we get d dt r r r r t,t,e t,,e t i j k t t e t e t t e t e t ) i te t ) j + t ) k t ) e t i + te t j + tk t t ) e t,te t, t 7. Show that the derivative of the norm is not equal to the norm of the derivative b verifing that rt) r t) for Let rt) cos t,5 sin t,4 cos t. Show that rt) is constant and conclude, using Eample 7, that rt) and rt) r t,,. t) are orthogonal. Then compute r t) and verif directl that r t) is orthogonal to rt). First let us compute rt) for rt) t,, : rt) d dt t + ) t t + Now, first let us compute the derivative, r t): r t),, and then computing the norm: r t),, It is clear in this eample, that rt) r t). Ma 6,

18 SECTION. Calculus of Vector-Valued Functions LT SECTION 4.) 67 In Eercises 9 46, d evaluate the integrals. Show that dt a r) a r for an constant vector a. 9. 8t t,6t + t dt Vector-valued integration is defined via componentwise integration. Thus, we first compute the integral of each component. 8t tdt 8 t t 7 9 ) 8 ) 6t + tdt t4 + t ) + ) 4 Therefore, 8t t,6t + t dt 8t tdt, 6t + tdt, 4 4. u i + u 5 ) j du + s, s + s ds The vector-valued integration is defined via componentwise integration. Thus, we first compute the integral of each component. u du u u 5 du u Therefore, ) u i + u 5 ) ) j du u du i + u 5 du j i + j 4. t,4t, te t cos t dt ) i + t lnt + )j dt The vector valued integration is defined via componentwise integration. Therefore, t,4t, cos t dt t dt, 4t dt, cos t dt t, t sin t,,, sin 4 / u, u 4, 45. t i + 4 t j 8t / ) k dt u 5 du We perform the integration componentwise. Computing the integral of each component we get: 4 t 4 dt ln t ln 4 ln ln tdt 4 4 t/ 8 ) 4 / t / dt t5/ 6 ) 4 5/ Hence, 4 t i + 4 ) tj 8t / k dt ln 4) i + 56 j k t si + 6s ) j + 9k ds Ma 6,

19 68 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) In Eercises 47 54, find both the general of the differential equation and the with the given initial condition. dr 47. t,4t, r), dt We first find the general b integrating dr rt) Since r),, we have: Substituting in ) gives the : t,4t dt dt : t) dt, r), + c, c, rt) t t, t +, t + t +, t r t) r t i + 5tj + k, r) j + k t) i j, r) i + k We first find the general b integrating r t): ) ) rt) t i + 5tj + k) dt t dt i + 5t dt j + The which satisfies the initial condition must satisf: ) ) 5 r) i + j + k + c j + k That is, c i j + k 4t dt t t, t + c ) ) ) ) 5 dt k t i + t j + tk + c ) Substituting in ) gives the following : ) ) 5 rt) t i + t j + tk i j + k t ) 5t ) i + j + t + ) k 5. r t) 6k, r),,, r t) sin t,sin t,t, r r ),, π ) To find the general we first, 4, find π r t) b integrating r t): 4 r t) r t) dt 6k dt 6t) k + c ) We now integrate r t) to find the general rt): rt) r t) dt 6t) k + c ) dt ) 6t) dt k + c t + c 8t )k + c t + c ) We substitute the initial conditions in ) and ). This gives: r ) c,, j Combining with ) we obtain the following : r) k + c + c,, c,, i rt) 8t )k + tj + i i + tj + 8t )k 5. r t) r,, t) e t, r),t,,, r ),,,, r),,, r ),, To find the general we first find r t) b integrating r t): r t) r t) dt,, dt, t, + c ) We now integrate r t) to find the general rt): rt) r t) dt, t, + c ) dt,t, + c t + c ) Ma 6,

20 SECTION. Calculus of Vector-Valued Functions LT SECTION 4.) 69 We substitute the initial conditions in ) and ). This gives: r ), 6, + c,, c, 6, r), 9, + c ) + c,,, 9, +, 8, + c,, c,, Combining with ) we obtain the following : rt),t, + t, 6, +,,,t 6t +,t 55. Find r the location t) e t at t, sin t,cos t of a particle whose path Figure, r),,, r 8) satisfies ),, dr dt t t + ), t 4, r), 8 5, 8) t t FIGURE 8 Particle path. To determine the position of the particle in general, we perform integration componentwise on r t) to obtain: rt) r t) dt Using the initial condition, observe the following: t t + ), t 4 dt t + t +,t 4t + c r), + c, 8 c, 8 Therefore, rt) t + t +,t 4t +, 8 t + t + +,t 4t + 8 and thus, the location of the particle at t isr) 45/4, 5.5, A fighter plane, which can shoot a laser beam straight ahead, travels along the path rt) 5 t, t Find the location and velocit at t 4 of a particle whose path satisfies, t /7. Show that there is precisel one time t at which the pilot can hit a target located at the origin. dr B the given information the laser dt t beam /, travels 6, 8t, in ther) direction 4, of 9, r t). The pilot hits a target located at the origin at the time t when r t) points towards the origin, that is, when rt) and r t) are parallel and point to opposite directions. rt) r t) Ma 6,

21 7 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) We find r t): r t) d 5 t, t, t, t, t dt 7 9 We first find t such that rt) and r t) are parallel, that is, we find t such that the cross product of the two vectors is zero. We obtain: i j k r t) rt) t t 9 5 t t 7 t ) ) ) ) t t + t 7 9 t ) i t + t9 7 5 t) j + t ) + t5 t) ) k t 4 ) ) 7 + 7t 6t i t 7 + 5t 9 j + t + t ) k Equating each component to zero we obtain the following equations: t t 6t t 7 + 5t 9 t + t t 7)t ) The third equation implies that t ort 7. Onl t satisfies the other two equations as well. We now must verif that r) and r ) point in opposite directions. We find these vectors: r) 5,,,, 7 r ),,, 6, 9 Since r) r ), the vectors point in opposite direction. We conclude that onl at time t can the pilot hit a target located at the origin. 59. Find all s to r t) v with initial condition r) w, where The fighter plane of Eercise 57 travels along the path rt) v t t and w are, t constant, t vectors in R.. Show that the pilot cannot We denote the components of the constant vector v b v v hit an target on the -ais.,v,v and integrate to find the general. This gives: rt) v dt v,v,v dt v dt, v dt, v dt We let c c,c,c and obtain: v t + c,v t + c,v t + c t v,v,v + c,c,c rt) tv + c c + tv Notice that the s are the vector parametrizations of all the lines with direction vector v. We are also given the initial condition that r) w, using this information we can determine: r) )v + c w Therefore c w v and we get: rt) w v) + tv t )v + w 6. Find all s to r t) rt) Let u be a constant vector in R where rt) is a vector-valued function. Find the of the equation r in three-space. t) sin t)u satisfing r ). We denote the components of rt) b rt) t), t), zt). Then, r t) t), t), z t). Substituting in the differential equation we get: t), t), z t) t), t), zt) Equating corresponding components gives: t) t) t) t) z t) zt) t) c e t t) c e t zt) c e t Ma 6,

22 SECTION. Calculus of Vector-Valued Functions LT SECTION 4.) 7 We denote the constant vector b c c,c,c and obtain the following s: rt) c e t,c e t,c e t e t c,c,c e t c 6. Prove that the Bernoulli spiral Figure 9) with parametrization rt) e t cos 4t,e t sin 4t Show that wt) sint + 4), sint ), cos t satisfies the differential equation w has the propert that the angle ψ between the position vector and the tangent vector is constant. Find the angle ψ in degrees. t) 9wt). ψ t t π ψ FIGURE 9 Bernoulli spiral. First, let us compute the tangent vector, r t): rt) e t cos 4t,e t sin 4t, r t) 4e t sin 4t + e t cos 4t,4e t cos 4t + e t sin 4t Then recall the identit that a b a b cos θ, where θ is the angle between a and b, so then, rt) r t) e t cos 4t,e t sin 4t 4e t sin 4t + e t cos 4t,4e t cos 4t + e t sin 4t ψ 4e t sin 4t cos 4t + e t cos 4t + 4e t sin 4t cos 4t + e t sin 4t e t cos 4t + sin 4t) e t Then, computing norms, we get: rt) e t cos 4t + e t sin 4t e t cos 4t + sin 4t) e t r t) 4e t sin 4t + e t cos 4t) + 4e t cos 4t + e t sin 4t) 6e t sin 4t 4e t sin 4t cos 4t + e t cos 4t + 6e t cos 4t + 4e t sin 4t cos 4t + e t sin 4t 6e t sin 4t + cos 4t) + e t cos 4t + sin 4t) 6e t + e t 7e t Then using the dot product relation listed above we get: Hence e t e t 7e t ) cos θ 7e t cos θ cos θ 7, θ Therefore, the angle between the position vector and the tangent vector is constant. 65. A curve Prove in polar that if form rt) r takes fθ)has on a parametrization local minimum or maimum value at t, then rt ) is orthogonal to r t ). Eplain how this result is related to Figure. Hint: Observe that if rt ) is a minimum, then rt) is tangent at t to the sphere of radius rt ) centered at the origin. rθ) fθ)cos θ,sin θ z Let ψ be the angle between the radial and tangent vectors Figure ). Prove that r t tan ψ r dr/dθ fθ) ) f θ) Hint: Compute rθ) r θ) and rθ) r θ). rt ) rt) FIGURE Ma 6,

23 7 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Suppose that rt) takes on a minimum or maimum value at t t. Hence, rt) also takes on a minimum or maimum value at t t, therefore dt d rt) tt. Using the Product Rule for dot products we get d dt rt) d tt dt tt rt) rt) rt ) r t ) + r t ) rt ) rt ) r t ) Thus rt ) r t ), which implies the orthogonalit of rt ) and r t ). In Figure, rt ) is a minimum and the path intersects the sphere of radius rt ) at a single point. Therefore, the point of intersection is a tangenc point which implies that r t ) is tangent to the sphere at t. We conclude that rt ) and r t ) are orthogonal. Further Newton s Insights Second andlaw Challenges of Motion in vector form states that F dp where F is the force acting on an object of dt 67. mass Let rt) m and t), p mr t) t) trace is thea object s plane curve momentum. C. Assume The that analogs t ) of force. Show andthat momentum the slopefor ofrotational the tangent motion vectorare r t the ) is equal torque the τ slope r d/d F and angular of the curve momentum at rt ). J rt) pt) a) B the Chain Rule we have Use the Second Law to prove that τ dj dt. Hence, at the points where d dt we have: d dt d d d d d dt d dt d dt t) t) b) The line lt) a,b + tr t ) passes through a, b) at t. It holds that: l) a,b + r t ) a,b That is, a, b) is the terminal point of the vector l), hence the line passes through a, b). The line has the direction vector r t ) t ), t ), therefore the slope of the line is t ) d t ) which is equal to d b part a). tt 69. Verif the Sum d and Product Rules for derivatives of vector-valued functions. Prove We that first dt r verif r the r Sum )) Rule r r stating: r ). r t) + r t)) r t) + r t) Let r t) t), t), z t) and r t) t), t), z t). Then, r t) + r t)) d dt t) + t), t) + t), z t) + z t) t) + t)), t) + t)), z t) + z t)) t) + t), t) + t), z t) + z t) t), t), z t) + t), t), z t) r t) + r t) The Product Rule states that for an differentiable scalar-valued function ft)and differentiable vector-valued function rt), it holds that: d dt ft)rt) ft)r t) + f t)rt) To verif this rule, we denote rt) t), t), zt). Then, d df ft)rt) d f t)t), f t)t), f t)zt) dt Appling the Product Rule for scalar functions for each component we get: d dt ft)rt) ft) t) + f t)t), f t) t) + f t)t), f t)z t) + f t)zt) ft) t), f t) t), f t)z t) + f t)t), f t)t), f t)zt) ft) t), t), z t) + f t) t), t), zt) ft)r t) + f t)rt) Verif the Chain Rule for vector-valued functions. Ma 6,

24 SECTION. Arc Length and Speed LT SECTION 4.) 7 7. Verif the Product Rule for cross products [Eq. 5)]. Let r t) a t), a t), a t) and r t) b t), b t), b t). Then we omit the independent variable t for simplicit): i j k r t) r t) a a a b b b a b a b ) i a b a b ) j + a b a b ) k Differentiating this vector componentwise we get: d dt r r a b + a b a b a b ) i a b + a b a b a b ) j + a b + a b a b a b ) k a b a b ) i a b a b ) j + a b a b ) ) k + a b a b ) i a b a b ) j + a b a b ) ) k Notice that the vectors in each of the two brackets can be written as the following formal determinants: d dt r i j k r a a a b b b + i j k a a a b b b a, a,a b,b,b + a,a,a b, b,b r r + r r 7. Prove the Substitution Rule where gt) is a differentiable scalar function): Verif the linearit properties b g rgt))g crt) dt b) t) dt c rt) dt ru) c du an constant) a g a) r t) + r t) ) Note that an earl edition of the tetbook dt hadrthe t) integral dt + limits r t) dt as ga) and gb); the should actuall be g a) and g b).) We denote the components of the vector-valued function b rt) dt t), t), zt). Using componentwise integration we have: b b b b rt) dt t)dt, t)dt, zt) dt a a a a b b g Write t)dt as s)ds. Let s gt), sods g b) t) dt. The substitution gives us gt))g t) dt.a a a g a) similar procedure for the other two integrals gives us: b g b) g rt) dt gt)) g b) g t) dt, gt)) g b) t) dt, z gt)) g t) dt a g a) g a) g a) g b) gt)) g t), gt)) g t), z gt)) g t) dt g a) g b) g gt)),gt)),zgt)) g b) t) dt r gt)) g t) dt g a) g a) Prove that if rt) K for t [a,b], then b rt) dt Kb a) a. Arc Length and Speed LT Section 4.) Preliminar Questions. At a given instant, a car on a roller coaster has velocit vector r 5, 5, in miles per hour). What would the velocit vector be if the speed were doubled? What would it be if the car s direction were reversed but its speed remained unchanged? The speed is doubled but the direction is unchanged, hence the new velocit vector has the form: λr λ 5, 5, for λ> We use λ, and so the new velocit vector is 5, 7,. If the direction is reversed but the speed is unchanged, the new velocit vector is: r 5, 5,. Ma 6,

25 74 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4). Two cars travel in the same direction along the same roller coaster at different times). Which of the following statements about their velocit vectors at a given point P on the roller coaster is/are true? a) The velocit vectors are identical. b) The velocit vectors point in the same direction but ma have different lengths. c) The velocit vectors ma point in opposite directions. a) The length of the velocit vector is the speed of the particle. Therefore, if the speeds of the cars are different the velocities are not identical. The statement is false. b) The velocit vector is tangent to the curve. Since the cars travel in the same direction, their velocit vectors point in the same direction. The statement is true. c) Since the cars travel in the same direction, the velocit vectors point in the same direction. The statement is false.. A mosquito flies along a parabola with speed vt) t. Let st) be the total distance traveled at time t. a) How fast is st) changing at t? b) Is st) equal to the mosquito s distance from the origin? a) B the Arc Length Formula, we have: t t st) r t) dt vt) dt t t Therefore, s t) vt) To find the rate of change of st) at t we compute the derivative of st) at t, that is, s ) v) 4 b) st) is the distance along the path traveled b the mosquito. This distance is usuall different from the mosquito s distance from the origin, which is the length of rt). Distance Lt) t rt) t Distance from the origin 4. What is the length of the path traced b rt) for 4 t if rt) is an arc length parametrization? Since rt) is an arc length parametrization, the length of the path for 4 t is equal to the length of the time interval 4 t, which is 6. Eercises In Eercises 6, compute the length of the curve over the given interval.. rt) t,4t, 6t +, t We have t) t, t) 4t, zt) 6t + hence t), t) 4, z t) 6. We use the Arc Length Formula to obtain: s r t) dt t) + t) + z t) dt dt 6. rt) rt) t,ln ti t,t tk,, t t 4 5 The derivative of rt) is r t), t, t. We use the Arc Length Formula to obtain: 4 4 ) s r 4 t) dt + + t) t dt 4t t dt t + ) dt t 4 t + ) dt t 4 + ln t t 6 + ln 4) + ln ) 5 + ln 4 Ma 6,

26 SECTION. Arc Length and Speed LT SECTION 4.) rt) t cos rt) t,t t sin t,t, +, t,t t π, t The derivative of rt) is r t) cos t t sin t,sin t + t cos t,. The length of r t) is, thus, r t) cos t t sin t) + sin t + t cos t) + 9 cos t t cos t sin t + t sin t + sin t + t sin t cos t + t cos t + 9 ) ) cos t + sin t + t sin t + cos t + 9 t + Using the Arc Length Formula and the integration formula given in Eercise 6, we obtain: π π s r t) dt t + dt t t + + ) ln t + t π + ) π 4π + + 5ln π + 4π + 5ln π 4π + + 5ln π + 4π + 9. t In Eercises rt) 7 and ti + 8, tj compute + t the )k, arc length t function. Usest) the formula: r u) du for the given value of a. t + a dt a t t + a + a ln 7. rt) t t + t, t,t, a + a ) The derivative of rt) is r t) t,4t,t. Hence, r t) t) + 4t) + t ) 4t + 6t + 9t 4 t + 9t Hence, t t st) r u) du u + 9u du We compute the integral using the substitution v + 9u, dv 8udu. This gives: st) 8 +9t v / dv 8 +9t v/ + 9t ) / /). 7 In Eercises 9, rt) 4t / find the speed, ln t,t at the given value of t., a 9. rt) t +, 4t, 5 t, t 4 The speed is the magnitude of the derivative r t), 4,. That is, vt) r t) ) rt) sin rt) t,cos e t 4t,cos,, t 5t, t π, t The velocit vector is r t) cos t, 4 sin 4t, 5 sin 5t. Att π the velocit vector is r π ) cos π, 4 sin π, 5 sin 5π,, 5. The speed is the magnitude of the velocit vector: π ) v,, What is the velocit vector of a particle traveling to the right along the hperbola rt) cosh t,sinh t,t, t with constant speed 5 cm/s when the particle s location is, )? The position of the particle is given as rt) t. The magnitude of the velocit vector r t) is the speed of the particle. Hence, r t) 5 ) The velocit vector points in the direction of motion, hence it is parallel to the tangent line to the curve and points to the right. We find the slope of the tangent line at : m d d d d ) 4 Ma 6,

27 76 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) We conclude that the vector, 4 is a direction vector of the tangent line at, and for some λ> we have at the given instance: r λ, 4 ), ) 4, 4 4 To satisf ) we must have: r λ + ) 7 λ 5 λ ) Substituting in ) we obtain the following velocit vector at, : r,, Let A bee with velocit vector r t) starts out at the origin at t and flies around for T seconds. Where is the bee ) T ) πnt πnt T located at time T if rt) r u) du R cos? What does,rsin the quantit,t h r, h u) du represent? t h a) Show that rt) parametrizes a heli of radius R and height h making N complete turns. b) Guess which of the two springs in Figure 5 uses more wire. c) Compute the lengths of the two springs and compare. 4 cm cm turns, radius 7 cm A) 5 turns, radius 4 cm B) We first verif that the projection pt) point moving around the circle of radius R. We have: t) + t) R cos πnt h FIGURE 5 Which spring uses more wire? ) R cos πnt h,rsin ) + R sin πnt h πnt h ), onto the -plane describes a ) ) )) R cos πnt + sin πnt R h h This is the equation of the circle of radius R in the -plane. As t changes in the interval t h the argument πnt h changes from to πn, that is, it covers N periods of the cos and sin functions. It follows that the projection onto the -plane describes a point moving around the circle of radius R, making N complete turns. The height of the heli is the maimum value of the z-component, which is t h. a) The second wire seems to use more wire than the first one. b) Setting R 7, h 4 and N in the parametrization in Eercise 5 gives: r t) 7 cos π t, 7 sin 4 π t,t 4 Setting R 4, h and N 5 in this parametrization we get: π 5t π 5t r t) 4 cos, 4 sin,t 7 cos πt πt, 7 sin,t, t 4 4 cos πt πt, 4 sin,t, t Ma 6,

28 SECTION. Arc Length and Speed LT SECTION 4.) 77 We find the derivatives of the two vectors and their lengths: r t) π πt sin, π πt cos, r t) 4π πt sin, 4π πt cos, Using the Arc Length Formula we obtain the following lengths: r t) 44π 4 r t) 6π π + 4 6π s 44π + 4 dt 44π + 4 s 6π + 9 dt 6π We see that the first spring uses more wire than the second one. 7. The ccloid generated b the unit circle has parametrization Use Eercise 5 to find a general formula for the length of a heli of radius R and height h that makes N complete turns. rt) t sin t, cos t a) Find the value of t in [, π] where the speed is at a maimum. b) Show that one arch of the ccloid has length 8. Recall the identit sin t/) cos t)/. One arch of the ccloid is traced as t π. B the Arc Length Formula we have: We compute the derivative and its length: π s r t) dt ) r t) cos t,sin t r t) cos t) + sin t) cos t + cos t + sin t cos t cos t) sin t sin t. For t π, we have t π, sosin t. Therefore we ma omit the absolute value sign and write: r t) sin t Substituting in ) and computing the integral using the substitution u t, du dt, gives: π s sin t π π dt sin u du) 4 sin udu π 4 cos u) 4 cos π cos )) 4 + ) 8 The length of one arc of the ccloid is s 8. The speed is given b the function: vt) r t) sin t, t π To find the value of t in [, π] where the speed is at maimum, we first find the critical point in this interval: v t) cos t cos t cos t t π t π Since v t) sin t, we have v π) sin π <, hence the speed vt) has a maimum value at t π. 9. Let rt) t +, 4t 5, t. Which of the following is an arc length t parametrization of a circle of radius 4 centered at the origin? a) Evaluate a) r t) the 4 arcsin length t,4 cos integral t st) r u) du. b) Find b) rthe t) inverse 4 sin 4t,4 cos 4t c) r t) gs) of st). 4 sin 4 t, 4 cos 4 t c) Verif that r s) rgs)) is an arc length parametrization. Ma 6,

29 78 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) a) We differentiate rt) componentwise and then compute the norm of the derivative vector. This gives: We compute st): r t), 4, r t) t t st) r t u) du 9 du 9 u 9t b) We find the inverse gs) ts) b solving s 9t for t. We obtain: s 9t t gs) s 9 We obtain the following arc length parametrization: r s) r s 9 ) s 9 +, 4s 9 5, s 9 To verif that r s) is an arc length parametrization we must show that r s). We compute r s): r s) d ds s 9 +, 4s 9 5, s, ,, 4, 9 9 Thus, r s) 9, 4, Let rt) w + tv be the parametrization of a line. Find an arc length parametrization of the t line a) Show that the arc length function st) r u) du is given b st) t v. This shows that rt) is an arc length parametrizaton if and onl if v is a unit vector. b) Find an arc length parametrization of the line with w,, and v, 4, 5. a) Since rt) w + tv, then r t) v and r t) v. Then computing st) we get: If we consider st), t t st) r u) du v du t v st) t if and onl if v b) Since v, 4, 5, then from part a) we get: st) t v t t 5, t gs) s 5 Therefore, since we are given rt) w + tv, the arc length parametrization is: r s),, + s, 4, 5 + s, s 5, + 5s 5. Find a path that traces the circle in the plane with radius 4 and center,, ) with constant speed 8. Find an arc length parametrization of the circle in the plane z 9 with radius 4 and center, 4, 9). We start with the following parametrization of the circle: rt),, + 4 cos t,, sin t + 4 cos t,, + 4 sin t We need to reparametrize the curve b making a substitution t gs), so that the new parametrization r s) r gs)) satisfies r s) 8 for all s. We find r s) using the Chain Rule: r s) d ds r gs)) g s)r gs)) ) Ma 6,

30 SECTION. Arc Length and Speed LT SECTION 4.) 79 Net, we differentiate rt) and then replace t b gs): r t) 4 sin t,, 4 cos t r gs)) 4 sin gs),, 4 cos gs) Substituting in ) we get: r s) g s) 4 sin gs),, 4 cos gs) 4g s) sin gs),, cos gs) Hence, r s) 4 g s) sin gs)) + cos gs)) 4 g s) To satisf r s) 8 for all s, we choose g s). We ma take the antiderivative gs) s, and obtain the following parametrization: r s) r gs)) rs) + 4 coss),, + 4 sins). This is a parametrization of the given circle, with constant speed Find an arc length parametrization of rt) t,t Find an arc length parametrization of rt) e t. sin t,e t cos t,e t. We follow two steps. Step. Find the inverse of the arc length function. The arc length function is the following function: t st) r u) du ) In our case r t) t,t hence r t) 4t + 9t t t. We substitute in ) and compute the resulting integral using the substitution v 4 + 9u, dv 8udu. This gives: t st) 4 + 9u udu 4+9t v / dv t v/ 4 + 9t ) / 4 /) t ) ) / 8 7 We find the inverse of t st) b solving for t in terms of s. This function is invertible for t and for t. s ) 4 + 9t ) / 8 7 7s t ) / 7s + 8) / 4 9t t ) 7s + 8) / s + 8)/ 4 9 t ± 7s + 8) / 4 ) Step. Reparametrize the curve. The arc length parametrization is obtained b replacing t b ) in rt): r s) 9 7s + 8)/ 4 9, ± ) 7s + 8) / / Find an arc length parametrization of the line m for an arbitrar slope m. Find an arc length parametrization of the ccloid with parametrization rt) t sin t, cos t. Step. Find the inverse of the arc length function. We are given the line m and a parametrization of this line is rt) t,mt, thus r t),m and r t) + m We then compute st): Solving s t + m for t we get: t st) + m du t + m t s + m Ma 6,

31 8 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Step. Reparametrize the curve using the t we just found. r s) s + m, sm + m 9. The curve known as the Bernoulli spiral Epress the arc length s of Figure 6) has parametrization rt) e t cos 4t,e t sin 4t. for 8 as an integral in two was, using the parametrizations r t) t,t t a) Evaluate st) and r t) r u) t,t du. 9. It Do is not convenient evaluatetothe take integrals, lower limit but use substitution because r ) to show that, the. ield the same result. b) Use a) to find an arc length parametrization of rt). t π t FIGURE 6 Bernoulli spiral. a) We differentiate rt) and compute the norm of the derivative vector. This gives: r t) e t cos 4t 4e t sin 4t,e t sin 4t + 4e t cos 4t e t cos 4t 4 sin 4t,sin 4t + 4 cos 4t r t) e t cos 4t 4 sin 4t) + sin 4t + 4 cos 4t) e t cos 4t 8 cos 4t sin 4t + 6 sin 4t + sin 4t + 8 sin 4t cos 4t + 6 cos 4t ) / e t cos 4t + sin 4t + 6 sin 4t + cos 4t ) e t + 6 7e t We now evaluate the improper integral: t t st) r u) du lim 7e u du lim 7e u t lim 7e t e R ) R R R R R 7e t ) 7e t b) An arc length parametrization of rt) is r s) r gs)) where t gs) is the inverse function of st). We find t gs) b solving s 7e t for t: s 7e t e t s s t gs) ln 7 7 An arc length parametrization of rt) is: r s) r gs)) e lns/ 7)) cos 4ln s cos 4ln 7 ) s, sin 4ln 7 ) s,e lns/ 7)) sin 4ln 7 ) s 7 ) s 7 ) Further Insights and Challenges. The unit circle with the point, ) removed has parametrization see Eercise 7 in Section.) Prove that the length of a curve as computed using the arc length integral does not depend on its parametrization. More precisel, let C be the curve traced b rt) tfor a t b. Let fs)be a differentiable function such that f s) > and that fc) a and fd) b. rt) Then r s) rf s)) parametrizes C for c s d. Verif that + t, t + t, <t< b d Use this parametrization to compute the length of the r t) unit dt circle as an r improper s) ds integral. Hint: The epression for r t) a c simplifies. Ma 6,

32 SECTION. Arc Length and Speed LT SECTION 4.) 8 We have t) t +t, t) t +t. Hence, t) + t ) ) t t) + t + + t t + t 4 + 4t + t ) + t + t 4 + t ) + t ) + t ) It follows that the path rt) lies on the unit circle. We now show that the entire circle is indeed parametrized b rt) as t moves from to. First, note that t) can be written as [ t + t ) t t ) ] / + t ) which is 4t/ + t ). So, for t negative, t) is an increasing function, t) is negative, and since lim t) and t lim t), we conclude that rt) does indeed parametrize the lower half of the circle for negative t. A similar t argument proves that we get the upper half of the circle for positive t. We now compute r t) and its length: r t + t ) t t ) t) + t ), + t ) t t + t ) 4t + t ), t + t ) + t ) 4t, t ) r t) + t ) 6t + 4 t ) + t ) t 4 + t + + t ) t + ) t + ) + t ) + t That is, r t) + t We now use the Arc Length Formula to compute the length of the circle: ) s r dt t) dt + t lim R tan R lim R tan R π π )) π. The curve rt) t tanh t,sech t is called a tractri see Eercise 9 in Section.). The involute of a t circle Figure 7), traced b a point at the end of a thread unwinding from a circular spool of a) Show radiusthat R, has st) parametrization r u) du seeiseercise equal to 6 st) in Section lncosh.) t). b) Show that t gs) lne s + e s rθ) ) is Rcos an inverse θ + θ of sinst) θ),rsin and verif θ θthat cos θ) Find an arc length parametrization r s) of the tanh involute. e s) e s,e s is an arc length parametrization of the tractri. a) We compute the derivative vector and its length: r t) sech t, sech t tanh t r t) sech t) + sech t tanh t sech t + sech 4 t + sech t tanh t sech t tanh t) + + sech 4 t We use the identit tanh t sech t to write: r t) sech t + sech t) + + sech 4 t sech t tanh t tanh t sech t sech 4 t + + sech 4 t For t, tanh t hence, r t) tanh t. We now appl the Arc Length Formula to obtain: t t st) r t u) du tanh u) du lncosh u) lncosh t) lncosh ) lncosh t) ln lncosh t) That is: st) lncosh t) Ma 6,

33 8 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) b) We show that the function t gs) ln e s + ) e s is an inverse of st). First we note that s t) tanh t, hence s t) > for t>, which implies that st) has an inverse function for t. Therefore, it suffices to verif that gst)) t. We have: gst)) ln e lncosh t) ) ) + e lncosh t) ln cosh t + cosh t Since cosh t sinh t we obtain for t ): ) g st)) ln cosh t + sinh e t + e t t ln cosh t + sinh t) ln + et e t ) ln e t ) t We thus proved that t gs) is an inverse of st). Therefore, the arc length parametrization is obtained b substituting t gs) in rt) t tanh t,sech t. We compute t, tanh t and sech t in terms of s. We have: s ln cosh t) e s cosh t sech t e s Also: tanh t sech t e s tanh t e s t tanh e s Substituting in rt) gives: r s) t tanh t,sech t c) The tractri is shown in the following figure: tanh e s e s,e s 4.4 Curvature LT Section 4.4) Preliminar Questions. What is the unit tangent vector of a line with direction vector v,,? A line with direction vector v has the parametrization: rt) OP + tv hence, since OP and v are constant vectors, we have: Therefore, since v, the unit tangent vector is: Tt) r t) v r t) r t) v /, /, / v. What is the curvature of a circle of radius 4? The curvature of a circle of radius R is R, hence the curvature of a circle of radius 4 is 4.. Which has larger curvature, a circle of radius or a circle of radius 4? The curvature of a circle of radius is, and it is larger than the curvature of a circle of radius 4, which is What is the curvature of rt) + t,7t,5 t? rt) parametrizes the line,, 5 + t, 7,, and a line has zero curvature. Ma 6,

34 SECTION.4 Curvature LT SECTION 4.4) 8 5. What is the curvature at a point where T s),, in an arc length parametrization rs)? The curvature is given b the formula: κt) T t) r t) In an arc length parametrization, r t) for all t, hence the curvature is κt) T t). Using the given information we obtain the following curvature: κ,, What is the radius of curvature of a circle of radius 4? The definition of the osculating circle implies that the osculating circles at the points of a circle, is the circle itself. Therefore, the radius of curvature is the radius of the circle, that is, What is the radius of curvature at P if κ P 9? The radius of curvature is the reciprocal of the curvature, hence the radius of curvature at P is: R κ P 9 Eercises In Eercises 6, calculate r t) and Tt), and evaluate T).. rt) 4t, 9t We differentiate rt) to obtain: We now find the unit tangent vector: r t) 8t,9 r t) Tt) 8t) t + 8 r t) r t) 8t,9 64t + 8 For t we obtain the vector: Tt) 8, , rt) + rt) 4t, e t,t 5t,9t We first find the vector r t) and its length: The unit tangent vector is therefore: r t) 4, 5, 9 r t) Tt) r t) r t) 4, 5, ) + 9 We see that the unit tangent vector is constant, since the curve is a straight line. 5. rt) cos rt) πt,sin πt,t + t,t, t We compute the derivative vector and its length: r t) π sin πt,π cos πt, r t) The unit tangent vector is thus: π sin πt) + π cos πt) + Tt) 4, 5 9, π sin πt + cos πt)+ π + r t) r t) π sin πt,π cos πt, π + For t we get: π T) π sin π, π cos π,, π,, π + π + π +,. π + Ma 6,

35 84 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) In Eercises 7, rt) e t use,e t Eq.,t ) to calculate the curvature function κt). 7. rt),e t,t We compute the first and the second derivatives of rt): r t),e t,, r t),e t,. Net, we find the cross product r t) r t): r t) r i j k t) e t e t et e t i j + e t e t k et i e t,, We need to find the lengths of the following vectors: r t) r t) e t,, e t r t) + e t ) + + e t We now use the formula for curvature to calculate κt): κt) r t) r t) e t r t) ) e t + e t + e t ) / 9. rt) rt) 4t +, 4t, t 4 cos t,t,4 sin t B Formula ) we have: We compute r t) and r t): κt) r t) r t) r t) r t) 4, 4,, r t),, Thus r t) r t),,, r t) r t), and κt), as epected. In Eercises 4, rt) t use,,t Eq. ) to evaluate the curvature at the given point.. rt) /t,/t,t, t B the formula for curvature we know: κt) r t) r t) r t) We now find r t), r t) and the cross product. These give: r t) t, t, t, r ),, r t) t, 6t 4,, r ), 6, r ) r ) 6, 6, Now finding the norms, we get: r ) ) + + ) r ) r ) ) Therefore, κ ) r ) r ) r ) rt) cos rt) t,sin t,t t,e t 4, t, 8t t π, t 4 B the formula for curvature we know: κt) r t) r t) r t) Ma 6,

36 SECTION.4 Curvature LT SECTION 4.4) 85 We now find r t), r t) and the cross product. These give: r t) sin t,cos t,t r π/),,π r t) cos t, sin t, r π/),, r π/) r π/) π,, Now finding norms we get: r π/) r π/) r π/) + π ) + + π π + ) + π + 5 Therefore, κπ/) r π/) r π/) r π/) π π ) π π.8 ) / In Eercises 5 8, rt) find the curvature cosh t,sinh t,t of the plane curve at the point indicated., t 5. e t, t We use the curvature of a graph in the plane: κt) f t) + f t) ) / In our case ft) e t, hence f t) f t) e t and we obtain: κt) e t + e t ) / κ) e + e 6 ) /.5 7. t 4, cos t, B the curvature of a graph in the plane, we have: κt) f t) + f t) ) / In this case ft) t 4, f t) 4t, f t) t. Hence, κt) t + 4t ) ) / t + 6t 6 ) / At t we obtain the following curvature: κ) ) / ) / 9. Find the curvature t n of rt) sin t,cos t,t at t π, t and t π Figure 6). z t π FIGURE 6 The curve rt) sin t,cos t,t. Ma 6,

37 86 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) B the formula for curvature we have: κt) r t) r t) r t) ) We compute the first and second derivatives: r t) cos t, sin t,, r t) sin t, 9 cos t, At the points t π and t π we have: r π ) cos π, sin π, cos π, sin π,,, r π ) sin π, 9 cos π,, 9, r π ) cos π, sin π,,, r π ) sin π, 9 cos π,,, We compute the cross products required to use ): r π ) r π ) i j k 9 9 i j + 9 k 9i j + 9k r π ) r π ) i j k i j + k j + 6k Hence, At t π we have: r π r π ) r π ) 9) + r π ) r π r π ) + + ) ) ) ) + + Substituting the values for t π and t π in ) we obtain the following curvatures: π ) κ ) 4.54 π ) κ ).. Show that the tractri rt) t tanh t,sech t has the curvature function κt) sech t. Find the curvature function κ) for sin. Use a computer algebra sstem to plot κ) for π. Prove that Writing the curvature rt) t), takes its t), maimum we haveatt) π t and tanh π. t Hint: and t) As a shortcut sech t. We to finding compute the the ma, firstobserve and second that derivatives the maimum of theseofunctions. numerator We use and tanh the minimum t sech of the t denominator obtain: of κ) occur at the same points. t) sech t tanh t t) sech t sech t tanh t) sech t tanh t t) sech t tanh t t) sech t tanh t + sech t ) sech t tanh t sech t ) sech t sech t ) Ma 6,

38 SECTION.4 Curvature LT SECTION 4.4) 87 We compute the cross product r r : We compute the length of r : Hence Substituting, we obtain t) t) t) t) tanh t sech t sech t) + sech t tanh t [ ] tanh t sech t sech t + sech t tanh t sech t t) + t) tanh 4 t + sech t tanh t tanh ttanh t + sech t) tanh t r tanh t ) / tanh t κt) sech t tanh t tanh t sech t tanh t tanh t sech t tanh t. Find the value of α such that the curvature of e α at is as large as possible. Show that curvature at an inflection point of a plane curve f)is zero. Using the curvature of a graph in the plane we have: κ) ) + ) ) / ) In our case ) αe α, ) α e α. Substituting in ) we obtain κ) α e α + α e α) / The curvature at the origin is thus κ) α e α + α e α ) / α + α ) / Since κ) and κ ) have their maimum values at the same values of α, we ma maimize the function: We find the stationar points: gα) κ α 4 ) + α ) g α) 4α + α ) α 4 ) + α ) α + α ) 6 α + α ) α ) + α ) 6 The stationar points are the s of the following equation: α + α ) α ) α or α α α ± Since gα) and g), α is a minimum point. Also, g α) is positive immediatel to the left of and negative to the right. Hence, α is a maimum point. Since gα) is an even function, α is a maimum point as well. Conclusion: κ) takes its maimum value at the origin when α ±. 5. Show that the curvature function of the parametrization Find the point of maimum curvature on e rt) a cos t,bsin t of the ellipse ) ). + is a b κt) ab b cos t + a sin t) / 9 The curvature is the following function: κt) r t) r t) r t) ) Ma 6,

39 88 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) We compute the derivatives and their cross product: Thus, r t) a sin t,bcos t, r t) a cos t, b sin t r t) r t) a sin ti + b cos tj) a cos ti b sin tj) ab sin tk + ab cos tk ab sin t + cos t ) k abk r t) r t) abk ab r t) Substituting in ) we obtain the following curvature: a sin t) + b cos t) a sin t + b cos t ab ab κt) ) a sin t + b cos t a sin t + b cos t ) / 7. In the notation of Eercise 5, assume that a b. Show that b/a Use a sketch to predict where the points of minimal and maimal κt) a/b curvature occur for all t. on an ellipse. Then use Eq. 9) to confirminor Eercise refute our 5 we prediction. showed that the curvature of the ellipse rt) a cos t,bsin t is the following function: κt) ab b cos t + a sin t ) / Since a b> the quotient becomes greater if we replace a b b in the denominator, and it becomes smaller if we replace b b a in the denominator. We use the identit cos t + sin t to obtain: ab a cos t + a sin t ) / κt) ab b cos t + b sin t ) / ab a cos t + sin t )) / κt) ab b cos t + sin t )) / ab a ab ab ab κt) a / ) b / ) b b a κt) a b In Eercises Use Eq. 9, ) to use prove Eq. ) that for to compute a plane curve the curvature rt) t), at thet), given point. 9. t,t, t κt) t) t) t) t) For the given parametrization, t) t, t) t) t +, hence t) ) / At the point t we have Substituting in Eq. ) we get t) t t) t) t t) 6t ) 4, ), ), ) κ) ) ) ) ) ) + ) ) /. t cos t,sin t, t π cosh s, s, s We have t) t cos t and t) sin t, hence: t) cos t t sin t π) cos π π sin π ) / 4. 6/ t) sin t sin t + t cos t) sin t t cos t π) sin π π cos π π t) cos t π) cos π t) sin t π) sin π Ma 6,

40 SECTION.4 Curvature LT SECTION 4.4) 89 Substituting in Eq. ) gives the following curvature: κπ) π) π) π) π) π) + π) ) / π ) ) + ) ) / π. t. Let sin st) s, sin 4s r, u) s duπ for the Bernoulli spiral rt) e t cos 4t,e t sin 4t see Eercise 9 in Section.). Show that the radius of curvature is proportional to st). The radius of curvature is the reciprocal of the curvature: Rt) κt) We compute the curvature using the equalit given in Eercise 9 in Section : κt) t) t) t) t) t) + t) ) / ) In our case, t) e t cos 4t and t) e t sin 4t. Hence: t) e t cos 4t 4e t sin 4t e t cos 4t 4 sin 4t) t) e t cos 4t 4 sin 4t) + e t 4 sin 4t 6 cos 4t) e t 5 cos 4t + 8 sin 4t) t) e t sin 4t + 4e t cos 4t e t sin 4t + 4 cos 4t) t) e t sin 4t + 4 cos 4t) + e t 4 cos 4t 6 sin 4t) e t 8 cos 4t 5 sin 4t) We compute the numerator in ): t) t) t) t) e t cos 4t 4 sin 4t) 8 cos 4t 5 sin 4t) +e t 5 cos 4t + 8 sin 4t) sin 4t + 4 cos 4t) e t 68 cos 4t + 68 sin 4t ) 68e t We compute the denominator in ): t) + t) e t cos 4t 4 sin 4t) + e t sin 4t + 4 cos 4t) e t cos 4t 8 cos 4t sin 4t + 6 sin 4t + sin 4t + 8 sin 4t cos 4t + 6 cos 4t ) e t cos 4t + sin 4t + 6 sin 4t + cos 4t )) e t + 6 ) 7e t ) Hence t) + t) ) / 7 / e t Substituting in ) we have κt) 68et 7 / e t 4 e t 7 7 R 4 et ) On the other hand, b the Fundamental Theorem and ) we have s t) r t) t) + t) 7e t 7e t We integrate to obtain 7 st) e t dt 7 e t + C 4) t Since st) r u) du, we have lim st), hence b 4): t ) lim 7e t + C + C C. t Ma 6,

41 9 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Substituting C in 4) we get: Combining ) and 5) gives: st) 7e t 5) Rt) 4 st) which means that the radius of curvature is proportional to st). 5. The Cornu Plot and spiral compute is the the plane curvature curve rt) κt) of t), the clothoid t), where rt) t), t), where t t t) sin u t t) sin u t du, du, t) t) cos u cos u du du Verif that Weκt) use the t. following Since the formula curvature for the increases curvature linearl, giventhe earlier): Cornu spiral is used in highwa design to create transitions between straight and curved road segments Figure 7). κt) t) t) t) t) t) + t) ) ) / We compute the first and second derivatives of t) and t). Using the Fundamental Theorem and the Chain Rule we get: t) sin t t) t cos t t cos t t) cos t ) t) t sin t t sin t Substituting in ) gives the following curvature function: ) sin t t sin t t cos t cos t ) t sin t + cos t κt) ) ) ) / / t sin t + cos t That is, κt) t. Here is a plot of the curvature as a function of t: κ κt) t 7. Find the unit normal vector Nt) to rt) 4, sin t,cos t. Find the unit normal vector Nθ) to rθ) R cos θ,sin θ, the circle of radius R. Does Nθ) point inside or outside the Wecircle? first find Draw thenθ) unit tangent at θ vector: π 4 with R 4. t We have Tt) r t) r t) r t) d 4, sin t,cos t, cos t, sin t, cos t, sin t dt r t) + cos t + sin t) + ) Ma 6,

42 SECTION.4 Curvature LT SECTION 4.4) 9 Substituting in ) gives:, cos t, sin t Tt) The normal vector is the following vector:, cos t, sin t Nt) T t) T t) We compute the derivative of the unit tangent vector and its length: ) Substituting in ) we obtain: T t) d, cos t, sin t, sin t, cos t, sin t,cos t dt T t) + sin t + cos t + Nt), sin t,cos t, sin t, cos t 9. Find the normal vectors to rt) Sketch the graph of rt) t,cos t,t t at. Since r t π t), t, the unit normal Nt) points in one of the two directions ± t, 4 and t π 4. The. Which normal sign vector is correct to rt) at t t,cos? Which t is T is t), correct whereattt) t? r t) r t) is the unit tangent vector. We have r t), sin t r t) + sin t) + sin t Hence, Tt) + sin, sin t t We compute the derivative of Tt) to find the normal vector.we use the Product Rule and the Chain Rule to obtain: ) T d t) + sin, sin t + t dt + sin, sin t t At t π 4 we obtain the normal vector:, At t π 4 T π ) 4 T π 4 + we obtain: ) + + sin, cos t t + sin t sin t cos t + sin, sin t t + sin, cos t t, + + sin t + sin t ) /, sin t ) /,,,, ) /,, +,, 4. Find the unit normal to the clothoid Eercise 5) at t π Find the unit normal to the Cornu spiral Eercise 4) at /. t π. The Clothoid is the plane curve rt) t), t) with t t) sin u t du, t) cos u du The unit normal is the following vector: Nt) T t) T t) ) Ma 6,

43 9 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) We first find the unit tangent vector Tt) r t) r t). B the Fundamental Theorem we have r t) sin t, cos t r t) sin t + cos t Hence, Tt) sin t, cos t We now differentiate Tt) using the Chain Rule to obtain: T t t) cos t, t sin t t cos t, sin t Hence, ) T t) t cos t + sin t t Substituting in ) we obtain the following unit normal: Nt) cos t, sin t At the point T π / the unit normal is Nπ / ) cos π/ ), sin π/ ) cos π, sin π, In Eercises 4 48, use Eq. ) to find N at the point Method for Computing N Let vt) r indicated. t). Show that 4. t,t, t Nt) vt)r t) v t)r t) We use the equalit vt)r t) v t)r t) Hint: N is the unit vector in the direction T t). Differentiate Tt) r t)/vt) to show that vt)r t) v t)r t) is a positive multiple of T t). Nt) vt)r t) v t)r t) vt)r t) v t)r t) For rt) t,t we have r t) t,t r t), 6t vt) r t) t) + t ) 4t + 9t 4 At the point t weget v t) 8t + 6t 4t + 8t 4t + 9t4 4t + 9t 4 r ), 6, v ) , and also r ),, v) Hence, v)r ) v )r ), ,, 8, v)r ) v )r ) 8, 8) Ma 6,

44 SECTION.4 Curvature LT SECTION 4.4) 9 Substituting in ) gives the following unit normal: N) 8, 6, 45. t /,t /,t, t t sin t, cos t, t π We use the following equalit: Nt) vt)r t) v t)r t) vt)r t) v t)r t) We compute the vectors in the equalit above. For rt) t /,t /,t we get: r t) t,t, r t), t, vt) r t) t + t 4 + At the point t we get: v t) t + t 4 + ) / 4t 4t + t + t) t + t 4 + r ),, r ),, v ) 6 v) Hence, v)r ) v )r ),,,,,, v)r ) v )r ) + ) + ) 6 We now substitute these values in ) to obtain the following unit normal: N) v)r ) v )r ),, v)r ) v )r ) t,e t,t t, t,t,t, t We use the equalit For rt) t,e t,t we have At the point t we have Nt) r t),e t, r t),e t, vt) r t) v t) vt)r t) v t)r t) vt)r t) v t)r t), + e t ) + e t + e t e t + et e t +, r,,, r ),,, v), v ), Ma 6,

45 94 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Hence, v)r ) v )r ),,,,,,,, v)r ) v )r ) Substituting in ) we obtain the following unit normal: N),, 6,, cosh t,sinh t,t ) 49. Let f). Show that the center of the osculating circle at, t, 4 ) is given b, +. vector We parametrize the curve b r),. The center Q of the osculating circle at has the position OQ r ) + κ ) N ) ) We first find the curvature, using the formula for the curvature of a graph in the plane. We have f ) and f ), hence, κ) f ) + f ) ) / + 4 ) / κ ) + 4 )/ To find the unit normal vector N ) we use the following considerations: The tangent vector is r ),, hence the vector, is orthogonal to r ) since their dot product is zero). Hence N ) is one of the two unit vectors ±,. +4 The graph of f) shows that the unit normal vector points in the positive -direction, hence, the appropriate choice is: N ), ) + 4 f) We now substitute ), ), and r ), in ) to obtain OQ, ) / + 4, + 4, + 4 ),, ), 4, + The center of the osculating circle is the terminal point of OQ, that is, Q 4, ) + Use Eq. 8) to find the center of curvature to rt) t,t at t. Ma 6,

46 SECTION.4 Curvature LT SECTION 4.4) 95 In Eercises 5 58, find a parametrization of the osculating circle at the point indicated. 5. rt) cos t,sin t, t π 4 The curve rt) cos t,sin t is the unit circle. B the definition of the osculating circle, it follows that the osculating circle at each point of the circle is the circle itself. Therefore the osculating circle to the unit circle at t π 4 is the unit circle itself. 5., rt) sin t,cos t, t Let f). We use the parametrization r), and proceed b the following steps. Step. Find κ and N. We compute κ using the curvature of a graph in the plane: κ) f ) + f ) ) / We have f ), f ), therefore, κ) + ) ) / + 4 ) / κ) 5 / ) To find N) we notice that the tangent vector is r ), hence, is orthogonal to r ) their dot product is zero). Therefore, N) is the unit vector in the direction of, or, that points to the inside of the curve. As shown in the figure, the unit normal vector points in the positive -direction, hence: N), 4 + N) 5, ) Step. Find the center of the osculating circle. The center Q at r) has the position vector Substituting ), ) and r), we get: OQ, + 5/ OQ r) + κ) N) 5 /,, + 5, 4, 7 Step. Parametrize the osculating circle. The osculating circle has radius R κ) 5/ and it is centered at the point ) 4, 7, therefore it has the following parametrization: 55. t sin t, cos t, sin, π t π Step. Find κ and N. In Eercise 44 we found that: To find κ we use the formula for curvature: ct) 4, 7 + 5/ cos t,sin t Nπ), ) κπ) r π) r π) r π) ) Ma 6,

47 96 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) For rt) t sin t, cos t we have: r t) cos t,sin t r π) cos π, sin π, r t) sin t,cos t r π) sin π, cos π, Hence, r π) r π) i j) k r π) r π) k and r π), Substituting in ) we get: κ π) 4 ) Step. Find the center of the osculating circle. The center Q of the osculating circle at r π) π, has position vector Substituting ), ) and r π) π, we get: OQ r π) + κπ) N π) ) OQ π, +, π, +, 4 π, 4 Step. Parametrize the osculating circle. The osculating circle has radius R κπ) and it is centered at π, ), hence it has the following parametrization: ct) π, + 4 cos t,sin t 57. rt) cos rt) t,sin t t,t /,t, /,t t, t The curvature is the following quotient: We compute the vectors r t) and r t): We now compute the following cross product: r t) r i j k t) sin t cos t cos t sin t κt) r t) r t) r t) ) r t) d cos t,sin t,t sin t,cos t, ) dt r t) d sin t,cos t, cos t, sin t, dt cos t sin t i sin t cos t j + sin t cos t cos t sin t k sin t)i cos t)j + k ) We calculate the norms of the vectors in ). B ) and ) we have: r t) r t) sin t + cos t) + + r t) sin t) + cos t + + 4) Substituting 4) in ) ields the following curvature: κt) ) κ) 5) The unit normal vector is the following vector: Nt) T t) T t) 6) Ma 6,

48 SECTION.4 Curvature LT SECTION 4.4) 97 B ) and 4) we have: Tt) Combining 6) and 7) gives: r t) r t) sin t,cos t, T t) cos t, sin t, 7) T t) cos t) + sin t) + Nt) cos t, sin t, N),, 8) The center of curvature at t is: OQ r) + κ) N) B 5), 8) and r),, we get: OQ,, +,,,, +,,,, Finall, we find a parametrization of the osculating circle at t. The osculating circle has radius R κ) and center,,, hence it has the following parametrization: ct),, +N) cos t + T) sin t,, +,, cos t +,, sin t ct) cos t, sin t, sin t 59. Figure 8 shows rt) the graph cosh t,sinh t,t of the half-ellipse ± r p, t, where r and p are positive constants. Show that the radius of curvature at the origin is equal to r. Hint: One wa of proceeding is to write the ellipse in the form of Eercise 5 and appl Eq. 9). r r FIGURE 8 The curve ± r p and the osculating circle at the origin. The radius of curvature is the reciprocal of the curvature. We thus must find the curvature at the origin. We use the following simple variant of the formula for the curvature of a graph in the plane: κ) ) + ) ) / ) The traditional formula of κ) ) + ) ) / is inappropriate for this problem, as ) is undefined at.) We find in terms of : r p p r + r p We solve for and obtain: ± p r p + r p,. Ma 6,

49 98 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) We find and : p ± p r p ± r p At the origin we get: r p p r ± p r p ± r p + p r p ) / ± r r p ) / ), ) ±r ± r / ) r Substituting in ) gives the following curvature at the origin: We conclude that the radius of curvature at the origin is ) κ) + ) ) / ± r + ) / r r R κ) r 6. The angle of inclination at a point P on a plane curve is the angle θ between the unit tangent vector T and the -ais In a recent stud of laser ee surger b Gatinel, Hoang-Xuan, and Azar, a vertical cross section of the cornea is Figure ). Assume that rs) is a arc length parametrization, and let θ θs) be the angle of inclination at rs). Prove modeled b the half-ellipse of Eercise 59. Show that the half-ellipse can be written in the form f), where that f) p r r p ). During surger, tissue is removed to a depth t)at height for S S, where t) is given b Munnerln s equation for some R>r): κs) dθ ds t) R S R r S + r Hint: Observe that Ts) cos θs), sin θs). After surger, the cross section of the cornea has the shape f)+ t)figure 9). Show that after surger, the radius of curvature at the point P where ) is R. P θ T cos θ, sin θ FIGURE The curvature at P is the quantit dθ/ds. Since Tt) is a unit vector that makes an angle θt) with the positive -ais, we have Differentiating this vector using the Chain Rule gives: Tt) cos θt),sin θt). T t) θ t) sin θt),θ t) cos θt) θ t) sin θt),cos θt) We compute the norm of the vector T t): T t) θ t) sin θt),cos θt) θ t) sin θt)) + cos θt)) θ t) θ t) When rs) is a parametrization b arc length we have: κs) dt ds dt dt dθ dt dθ ds θ t) θ dθ t) ds dθ ds as desired. 6. Let θ) be the angle of inclination at a point A particle moves along the path on the graph f)see Eercise 6). with unit speed. How fast is the tangent turning i.e., how fast is the angle a) Use of inclination the relation changing) f ) when tan θ the to prove particle that passes dθ through the point, 8)? d f ) + f ) ). b) Use the arc length integral to show that ds d + f ). c) Now give a proof of Eq. 5) using Eq. ). Ma 6,

50 SECTION.4 Curvature LT SECTION 4.4) 99 a) B the relation f ) tan θ we have θ tan f ). Differentiating using the Chain Rule we get: dθ d d tan f ) d ) f ) f ) ) d + f ) d + f ) b) We use the parametrization r),f). Hence, r ),f ) and we obtain the following arc length function: S) r u) du,f u) du + f u) du Differentiating using the Fundamental Theorem gives: ds d d ) + f d u) du + f ) c) B Eq. ), κs) dθ ds ) Using the Chain Rule and the equalities in part a) and part b), we obtain: dθ ds dθ d d ds dθ d ds d f ) + f ) + f ) Combining with ) we obtain the curvature as the following function of : κ) f ) + f ) ) / f ) + f ) ) / which proves Eq. 5). In Eercises Use the 65 67, parametrization use Eq. ) rθ) to find the fθ)cos curvature θ,fθ)sin of the curve θ togiven showinthat polar a curve form. r fθ)in polar coordinates has curvature 65. fθ) cos θ B Eq. ):, κθ) fθ) + f θ) fθ)f θ) fθ) + f θ) ) / κθ) fθ) + f θ) fθ)f θ) fθ) + f θ) ) / We compute the derivatives f θ) and f θ) and evaluate the numerator of κθ). This gives: f θ) sin θ f θ) cos θ fθ) + f θ) fθ)f θ) 4 cos θ + 4 sin θ cos θ cos θ) 8 cos θ + 8 sin θ 8 We compute the denominator of κθ): fθ) + f θ) ) / 4 cos θ + 4 sin ) / θ 4 / 8 Hence, κθ) fθ) e fθ) θ θ B Eq. ) we have the following curvature: κθ) fθ) + f θ) fθ)f θ) fθ) + f θ) ) / Ma 6,

51 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Since fθ) e θ also f θ) f θ) e θ. We compute the numerator and denominator of κθ): fθ) + f θ) fθ)f θ) e θ + e θ e θ e θ e θ fθ) + f θ) ) / e θ + e θ ) / e θ ) / e θ Substituting in the formula for κθ) we obtain: κθ) eθ e θ e θ 69. Show that both r t) and r t) lie in the osculating plane for a vector function rt). Use Eq. ) to find the curvature of the general Bernoulli spiral r ae bθ Hint: Differentiate r t) vt)tt). in polar form a and b are constants). The osculating plane at P is the plane through P determined b the unit tangent T and the unit normal N at P. Since Tt) r t) r t) we have r t) vt)tt) where vt) r t). That is, r t) is a scalar multiple of Tt), hence it lies in ever plane containing Tt), in particular in the osculating plane. We now differentiate r t) vt)tt) using the Product Rule: B Nt) r t) v t)tt) + vt)t t) ) T t) T t) we have T t) bt)nt) for bt) T t). Substituting in ) gives: r t) v t)tt) + vt)bt)nt) We see that r t) is a linear combination of Tt) and Nt), hence r t) lies in the plane determined b these two vectors, that is, r t) lies in the osculating plane. 7. Two vector-valued functions r Show that s) and r s) are said to agree to order at s if r s ) r s ), r γs) rt ) + s κ N ) + r s ), r s ) r s ) ) sin κs)t cos κs)n κ Let rs) be an arc length parametrization of a path C, and let P be the terminal point of r). Let γs)be the arc length parametrization is an arc length of the parametrization osculating circle of the given osculating Eercise circle 7. atshow rt ). that rs) and γs)agree to order at s in fact, the osculating circle is the unique circle that approimates C to order at P ). The arc length parametrization of the osculating circle at P, described in the -coordinate sstem with P at the origin and the and aes in the directions of T and N respectivel, is given in Eercise 7 b: γs) κ N + ) sin κs)t cos κs)n κ Hence We get: γ) κ N + ) sin )T cos )N κ κ N + κ N) κ N κ N r) OP γ) r) ) Differentiating γs)gives notice that N, T, and κ are fied): γ s) ) κ cos κs)t + κ sin κs)n cos κs)t + sin κs)n κ Hence: γ ) cos κ ) T + sin κ ) N T + N T N P T Ma 6,

52 SECTION.4 Curvature LT SECTION 4.4) Also, since rs) is the arc length parametrization, r s), hence: We conclude that: We differentiate γ s) to obtain: Hence: For the arc length parametrization rs) we have: Hence: We conclude that: T T) r ) r ) r ) γ ) r ) ) γ s) κ sin κs) T + κ cos κs) N γ ) κ sin ) T + κ cos ) N T + κn κn r s) T s) T s) Ns) r s) κs)ns) κs)ns) r ) κ)n) κn ), ), and ) show that rs) and γs)agree to order two at s. γ ) r ) ) Let rt) t), t), zt) be a path with curvature κt)and define the scaled path r t) λt), λt), λzt), Further Insights and Challenges where λ is a constant. Prove that curvature varies inversel with the scale factor. That is, prove that the curvature 7. Show κ t) of that r the t) is curvature κ t) of λ Viviani s κt). This curve, eplains givenwh b rt) the curvature + cos oft, asin circle t, of sint/), radius Risis proportional to /R in fact, it is equal to /R). Hint: Use Eq. ). + cos t κt) + cos t) / We use the formula for curvature: κt) r t) r t) r t) ) Differentiating rt) gives r t) sin t,cos t, cos t r t) cos t, sin t, sin t sin t,cos t,cos t We compute the cross product in ): r t) r i j k t) sin t cos t cos t cos t sin t sin t cos t sin t + sin t cos t ) i sin t sin t + cos t cos t ) j + k We find the length of the cross product: r t) r t) cos t sin t + sin t cos t ) + sin t sin t + cos t cos t ) + t ) 4 sin cos t + sin t + cos t ) sin t + cos t + 4 sin t + cos t + Ma 6,

53 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) We use the identities sin t + cos t and cos t + cos t to write: r t) r t) 4 sin t + cos t ) cos t + 8 cos t + 8 sin t + t ) cos + t 4 cos + Hence: r t) r t) 8 + cos t ) We compute the length of r t): Hence, r t) sin t) + cos t + cos t + cos t + + cos t ) + cos t Substituting ) and ) in ) gives: κt) + cos t 8 ) + cos t r t) + cos t ) 8 + cos t + cos t) / + cos t + cos t) / In Eercises Let rs) 75 8, be an letarc B denote lengththe parametrization binormal vector of a at closed a point curve on ac space of length curve L. C, We defined call Cb anb oval T if dθ/ds N. > see Eercise 6). Observe that N points to the outside of C. For k>, the curve C defined b r s) rs) kn is 75. Show that B is a unit vector. called the epansion of cs) in the normal direction. a) ShowT that and r N are s) orthogonal r s) +kκs). unit vectors, therefore the length of their cross product is: b) As P moves around the oval counterclockwise, θ increases b π [Figure A)]. Use this and a change of L B T N T N sin π variables to prove that κs)ds π. Therefore B is a unit vector. c) Show that C has length L + πk. 77. Show that if C is contained in a plane P, then B is a unit vector normal to P. Conclude that τ for a plane curve. Follow steps a) c) to prove that there is a number τ lowercase Greek tau ) called the torsion such that If C is contained in a plane P, then the unit normal N and the unit tangent T are in P. The cross product B T N is orthogonal to T and N which are in the db plane, ds τn hence B is normal to the plane. Thus, B is a unit vector normal to the plane. There are onl two different unit normal vectors to a plane, one pointing up and the other pointing down. Thus, we can assume due to continuit) that B is a constant vector, therefore a) Show that db ds T dn and concludedb that db/ds is orthogonal to T. ds or τ. b) Differentiate B B with respect to s to ds show that db/ds is orthogonal to B. c) Conclude that db/ds is a multiple of N. 79. Use the identit Torsion means twisting. Is this an appropriate name for τ? Eplain b interpreting τ geometricall. a b c) a c)b a b)c to prove N B T, B T N 5 We use the given equalit and the definition B T N to write: N B N T N) N N) T N T) N ) The unit normal N and the unit tangent T are orthogonal unit vectors, hence N N N and N T. Therefore, ) gives: N B T N T To prove the second equalit, we substitute T N B and then use the given equalit. We obtain: B T B N B) B B) N B N) B ) Ma 6,

54 SECTION.5 Motion in Three-Space LT SECTION 4.5) Now, B is a unit vector, hence B B B. Also, since B T N, B is orthogonal to N which implies that B N. Substituting in ) we get: B T N B N. 8. Show that r r is a multiple of B. Conclude that Follow steps a) b) to prove B dn r r r κt + τb ds r 7 a) Show B that the dn/ds definition is orthogonal of the binormal to N. vector, Conclude B that T dn/ds N. We lies denote in the at) plane r spanned t) and b write: T and B, and hence, dn/ds at + bb for some scalars a,b. b) Use N T to show that T dn dt N Tt) r t) and compute a. Compute b similarl. Equations 4) and 6) ds ds r t) at)r t) ) together with dt/dt κn are called the Frenet formulas and were discovered b the French geometer Jean Frenet We differentiate Tt) using the Product Rule: 86 9). T t) at)r t) + a t)r t) We denote bt) T t) and obtain: For c at) bt) and c a t) bt) we have: Nt) T t) T t) at) bt) r t) + a t) bt) r t) We now find B as the cross product of Tt) in ) and Nt) in ). This gives: Nt) c t)r t) + c t)r t) ) Bt) at)r t) c t)r t) + c t)r t) ) at)c t)r t) r t) + at)c t)r t) r t) at)c t)r t) r t) + at)c t)r t) r t) We see that B is parallel to r r. Since B is a unit vector we have: B r r r r. The vector N can be computed using N B T [Eq. 5)] with B, as in Eq. 7). Use this method to find N in.5 the following Motioncases: in Three-Space LT Section 4.5) a) rt) cos t,t,t at t Preliminar b) rt) Questions t,t,t at t. If a particle travels with constant speed, must its acceleration vector be zero? Eplain. If the speed of the particle is constant, the tangential component, a T t) v t), of the acceleration is zero. However, the normal component, a N t) κt)vt) is not necessaril zero, since the particle ma change its direction.. For a particle in uniform circular motion around a circle, which of the vectors vt) or at) alwas points toward the center of the circle? For a particle in uniform circular motion around a circle, the acceleration vector at) points towards the center of the circle, whereas vt) is tangent to the circle.. Two objects travel to the right along the parabola with nonzero speed. Which of the following statements must be true? a) Their velocit vectors point in the same direction. b) Their velocit vectors have the same length. c) Their acceleration vectors point in the same direction. a) The velocit vector points in the direction of motion, hence the velocities of the two objects point in the same direction. b) The length of the velocit vector is the speed. Since the speeds are not necessaril equal, the velocit vectors ma have different lengths. c) The acceleration is determined b the tangential component v t) and the normal component κt)vt). Since v and v ma be different for the two objects, the acceleration vectors ma have different directions. Ma 6,

55 4 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) 4. Use the decomposition of acceleration into tangential and normal components to eplain the following statement: If the speed is constant, then the acceleration and velocit vectors are orthogonal. If the speed is constant, v t). Therefore, the acceleration vector has onl the normal component: at) a N t)nt) The velocit vector alwas points in the direction of motion. Since the vector Nt) is orthogonal to the direction of motion, the vectors at) and vt) are orthogonal. 5. If a particle travels along a straight line, then the acceleration and velocit vectors are choose the correct description): a) Orthogonal b) Parallel Since a line has zero curvature, the normal component of the acceleration is zero, hence at) has onl the tangential component. The velocit vector is alwas in the direction of motion, hence the acceleration and the velocit vectors are parallel to the line. We conclude that b) is the correct statement. 6. What is the length of the acceleration vector of a particle traveling around a circle of radius cm with constant velocit 4 cm/s? The acceleration vector is given b the following decomposition: at) v t)tt) + κt)vt) Nt) ) In our case vt) 4 is constant hence v t). In addition, the curvature of a circle of radius is κt). Substituting vt) 4, v t) and κt) in ) gives: at) 4 Nt) 8Nt) The length of the acceleration vector is, thus, at) 8cm/s 7. Two cars are racing around a circular track. If, at a certain moment, both of their speedometers read mph. then the two cars have the same choose one): a) a T b) a N The tangential acceleration a T and the normal acceleration a N are the following values: a T t) v t); a N t) κt)vt) At the moment where both speedometers read mph, the speeds of the two cars are v mph. Since the track is circular, the curvature κt) is constant, hence the normal accelerations of the two cars are equal at this moment. Statement b) is correct. Eercises. Use the table below to calculate the difference quotients the velocit and speed at t. r + h) r) h for h.,.,.,.. Then estimate r.8) r.9) r) r.) r.).557,.459, ,.64,.74.54,.84, ,.78,.5.45,.4,.48 r.) r). r.) r). h.) r.8) r)..7,.8,.57. h.) r.9) r)..9,.7, ,.459,.97.54,.84, ,.9, ,.64,.74.54,.84,.44..9,.7,.97 Ma 6,

56 SECTION.5 Motion in Three-Space LT SECTION 4.5) 5 r +.) r). r +.) r). The velocit vector is defined b: h.) h.) We ma estimate the velocit at t b: r.) r)..4,.7,.48. r.) r)..5,.5, ,.78,.5.54,.84,.44..4,.7, ,.4,.48.54,.84, ,.55, 5.75 vt) r rt + h) rt) t) lim h h v).,.,.5 and the speed b: v) v) In Eercises 6, calculate the velocit and acceleration r + h) vectors r) and the speed at the time indicated. Draw the vectors r + h) r) and for h.5 for the path in Figure. Draw v) using a. h rough rt) estimate t, for t,4t its, length). t In this case rt) t, t,4t hence: vt) r t) t,, 8t v),, 8 at) r t) 6t,, 8 a) 6,, 8 The speed is the magnitude of the velocit vector, that is, v) v) + ) rθ) sin rt) e t θ,cos θ,cos θ, θ π j cost)k, t Differentiating rθ) sin θ, cos θ, cos θ gives: vθ) r θ) cos θ, sin θ, sin θ π ) v cos π, sin π, sin π,, aθ) r θ) sin θ, cos θ, 9 cos θ π ) a sin π, cos π, 9 cos π,, 9 The speed is the magnitude of the velocit vector, that is: π ) v π ) ) ) v Find at) for a particle moving around a circle of radius 8 cm at a constant speed of v 4 cm/s see Eample 4). Draw the rs) path and acceleration + s, s vector + s, s at t π 4. The position vector is: rt) 8 cos ωt, sin ωt Ma 6,

57 6 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Hence, vt) r t) 8 ω sin ωt, ω cos ωt 8ω sin ωt, cos ωt ) We are given that the speed of the particle is v 4cm/s. The speed is the magnitude of the velocit vector, hence: v 8ω sin ωt) + cos ωt 8ω 4 ω rad/s Substituting in ) we get: vt) 4 sin t, cos t We now find at) b differentiating the velocit vector. This gives at) v t) 4 cos t, sin t cos t, sin t The path of the particle is rt) 8 cos t, sin t and the acceleration vector at t π 4 is: π ) a cos π 4 8, sin π.85,.77 8 The path rt) and the acceleration vector at t π 4 are shown in the following figure: 8 rt) 8 cos t, sin t 9. Sketch the path rt) t,t Sketch the path rt) together t with, t the velocit and acceleration vectors at t. for t, indicating the direction of motion. Draw the velocit and acceleration We compute vectors atthe t velocit and t and. acceleration vectors at t : vt) r t) t,t v), at) v t), 6t a), 6 The following figure shows the path rt) t,t and the vectors v) and a): a) v) t rt) t, t ) In Eercises 4, find vt) The paths rt) given t,t at) and the initial and r t) t 4,t 6 velocit. trace the same curve, and r ) r). Do ou epect either. the at) velocit t,4, vectors v) or the acceleration vectors of these paths at t to point in the same direction? Compute these, vectors and draw them on a single plot of the curve. We find vt) b integrating at): t t t t vt) au)du u, 4 du u, 4u + v, 4t + v Ma 6,

58 SECTION.5 Motion in Three-Space LT SECTION 4.5) 7 The initial condition gives: Hence, v), + v, t vt), 4t +, v, t +, 4t 6. at) k, at) v) e t i,,t +, v),, We compute vt) b integrating the acceleration vector: t t t vt) au) du k du ku Substituting the initial condition gives: Combining with ) we obtain: v) k + v i v i vt) i + tk + v tk + v ) In Eercises at) 5 8, t k, find v) rt) and i j vt) given at) and the initial velocit and position. 5. at) t,4, v),, r), We first integrate at) to find the velocit vector: t u t vt) u, 4 du, 4u t + v, 4t + v ) The initial condition v), gives: v), + v, v, Substituting in ) we get: t vt), 4t t +, +, 4t We now integrate the velocit vector to find rt): t u u t rt) +, 4u du 6 + u, u u + r The initial condition r), gives: r), + r, r, t 6 + t,t t + r Hence, rt) t 6 + t,t t 7. at) tk, at) e t v) i,, t,t + r) j, v),,, r),, Integrating the acceleration vector gives: t vt) uk du u k t + v t k + v ) The initial condition for vt) gives: v) k + v i v i Ma 6,

59 8 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) We substitute in ): We now integrate vt) to find rt): t rt) The initial condition for rt) gives: Combining with ) gives the position vector: vt) t k + i i + t k ) i + u k du ui + u 6 k t + r ti + t 6 k + r ) r) i + k + r j r j rt) ti + j + t 6 k In Eercises at) 9 4, cos tk, recall v) that g i 9.8 j, m/s r) is the i acceleration due to gravit on the earth s surface. 9. A bullet is fired from the ground at an angle of 45. What initial speed must the bullet have in order to hit the top of a -m tower located 8 m awa? We place the gun at the origin and let rt) be the bullet s position vector. Step. Use Newton s Law. The net force vector acting on the bullet is the force of gravit F, gm m, g. B Newton s Second Law, F mr t), hence: m, g mr t) r t), g We compute the position vector b integrating twice: t t r t) r u) du, g du, gt + v t t rt) r u) du, gu + v ) du, g t + v t + r That is, rt), g t + v t + r ) Since the gun is at the origin, r. The bullet is fired at an angle of 45, hence the initial velocit v points in the direction of the unit vector cos 45, sin 45, therefore, v v,. Substituting these initial values in ) gives: rt), g t + tv, Step. Solve for v. The position vector of the top of the tower is 8,, hence at the moment of hitting the tower we have, rt), g t + tv, 8, tv, g t + tv 8, Equating components, we get the equations: tv 8 g t + tv Ma 6,

60 SECTION.5 Motion in Three-Space LT SECTION 4.5) 9 The first equation implies that t 6 v. We substitute in the second equation and solve for v we use g 9.8m/s ): 9.8 ) 6 ) 6 + v v v ) v 6 v ) 49 6 v 49 v m/s The initial speed of the bullet must be v 4 m/s m/s.. Show that a projectile fired at an angle θ with initial speed v Find the initial velocit vector v of a projectile released with travels a total distance v initial speed m/s that /g) sin θ before hitting the reaches a maimum height ground. Conclude that the maimum distance for a given v of m. ) is attained for θ 45. We place the gun at the origin and let rt) be the projectile s position vector. The net force acting on the projectile is F, mg m, g. B Newton s Second Law, F mr t), hence: Integrating twice we get: m, g mr t) r t), g t t r t) r u) du, g du, gt + v t t rt) r u) du, g u + v ) du, g t + v t + r ) Since the gun is at the origin, r. The firing was at an angle θ, hence the initial velocit points in the direction of the unit vector cos θ,sin θ. Hence, v v cos θ,sin θ. We substitute the initial vectors in ) to obtain: rt), g t + v t cos θ,sin θ ) The total distance is obtained when the -component of rt) is zero besides the original moment, that is, g t + v sin θ) t t g ) t + v sin θ t or t v sin θ g The appropriate choice is t v sin θ g. We now find the total distance T b substituting this value of t in the -component of rt) in ). We obtain: t) v t cos θ T v cos θ v sin θ g v cos θ sin θ g v sin θ g The maimum distance is attained when sin θ, that is θ 9 or θ 45.. A bullet is fired at an angle θ π One plaer throws a baseball to 4 at another a tower plaer located standing d 6 5 m awa awa, with with initial initial speed speed 8 v m/s. Find the m/s. Use the result of height H at which the bullet hits the tower. Eercise to find two angles θ at which the ball can be released. Which angle gets the ball there faster? We place the gun at the origin and let rt) be the bullet s position vector. Step. Use Newton s Law. The net force vector acting on the bullet is the force of gravit F, gm m, g. B Newton s Second Law, F mr t), hence: m, g mr t) r t), g We compute the position vector b integrating twice: t t r t) r u) du, g du, gt + v t t rt) r u) du, gu + v ) du, g t + v t + r Ma 6,

61 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) That is, rt), g t + v t + r ) Since the gun is at the origin, r. The bullet is fired at an angle of π/4 radians, hence the initial velocit v points in the direction of the unit vector cos π/4, sin π/4, therefore, v v,. Substituting these initial values in ) gives: rt), g t + tv, Step. Solve for H. The position vector for the point at which the bullet hits the tower, 6 meters awa, is 6,H, hence at the moment of hitting the tower we have, Therefore, for v : and, g t + tv, 6,H tv 6 t 6 5 gt + tv 9.85) + 5 )) H Hence, H 55 meters. The bullet hits the tower at 55 meters high. 5. A constant force F 5, in newtons) acts on a -kg mass. Find the position of the mass at t s if it is located Show that a bullet fired at an angle θ will hit the top of an h-meter tower located d meters awa if its initial speed at the origin at t and has initial velocit v is, in meters per second). We know that F ma and thus 5, a so then a.5,.. Using integration we know g/ d sec θ v vt) at) d dt tan θ ta + h c and we know v), c. Therefore, vt) ta + v t.5,. +,.5t +,.t Again, integrating, rt) vt) dt ta + v dt t a + tv + c t.5,. + t,.5t + t,.t t + r Using the initial condition r), c, we conclude rt).5t + t,.t t and hence the position of the mass at t is r) 45,. 7. A particle A forcefollows F 4t,6 a path rt) 8t for innewtons) t T acts, beginning on a 4-kgatmass. the origin Find O. thethe position vector of v the mass T at t r t) dt s ifisitcalled is located the at, ) at t and has zero initial velocit. T average velocit vector. Suppose that v. Answer and eplain the following: a) Where is the particle located at time T if v? b) Is the particle s average speed necessaril equal to zero? a) If the average velocit is, then the particle must be back at its original position at time t T. This is perhaps best seen b noting that v T r T t) dt rt) T. Ma 6,

62 SECTION.5 Motion in Three-Space LT SECTION 4.5) b) The average speed need not be zero! Consider a particle moving at constant speed around a circle, with position vector rt) cos t,sin t. Fromtoπ, this has average velocit of, but constant average speed of. 9. At a certain moment, a particle moving along a path has velocit v,, and acceleration a,,. At a certain moment, a moving particle has velocit v,, and a, 4,. Find T, N, and the Is the particle speeding up or slowing down? decomposition of a into tangential and normal components. We are asked if the particle is speeding up or slowing down, that is if v or v is increasing or decreasing. We check v ) : v ) v v) v v a v,,,, 4 + 6) < So the speed is decreasing. In Eercises, use Eq. ) to find the coefficients a T and a N as a function of t or at the specified value of t).. rt) t,cos rt) t t,sin,t t We find a T and a N using the following equalities: We compute v and a b differentiating r twice: The unit tangent vector T is, thus: a T a T,a N vt) r t), sin t,cos t vt) at) r t), cos t, sin t Tt) a v. v vt) vt), sin t,cos t + sin t) + cos t Since the speed is constant v vt) ), the tangential component of the acceleration is zero, that is: a T To find a N we first compute the following cross product: i j k a v cos t sin t sin t cos t cos t sin t cos t + sin t sin t cos t i sin t cos t j + cos t sin t ) i sin tj + cos tk i sin tj + cos tk, sin t,cos t k Hence, a N a v v ) + sin t) + cos t.. rt) e t rt),t,e t t,, ln t,t t, t We will use the following equalities: a v a T a T, a N. v We first find a and v b twice differentiating r. We get: vt) r t) e t,, e t at) r t) 4e t,,e t Then evaluating at t we get: v),,, v) a) 4,, + + ) 6 Ma 6,

63 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Hence, T v v,, and we obtain: 6 a T a T 4,, 6,, ) 7 6 To find a N we first compute the following cross product: i j k a v 4, 6, 4 Therefore, a N a v v ) In Eercise 4 4, find the decomposition of at) into tangential and normal components at the point indicated, as in Eample rt) rt) t e, t, t t, t First note here that: vt) r t) t, At t we have: Thus, at) r t) t, v r ) 4, a r ) 4, a v 4, 4, 6 v Recall that we have: T v 4, 4 v 5 5, 5 a T a v v 6 5 Net, we compute a N and N: This vector has length: a N N a a T T 4, a N a N N 4 5, 6 5 5, ) + 48 ) and thus, N a NN a N 6 5, 48 5 /5 5, 4 5 Finall we obtain the decomposition, where T 45, 5 and N 5, 4 5. a 4, 6 5 T + 5 N rt) t, t, 6 t, t Ma 6,

64 7. rt) t, t, 6 t, t 4 First note here that: SECTION.5 Motion in Three-Space LT SECTION 4.5) vt) r t),t, t at) r t),,t At t 4 we have: v r 4), 4, 8 a r 4),, 4 Thus, Recall that we have: Net, we compute a N and N: This vector has length: and thus, Finall we obtain the decomposition, a v,, 4, 4, 8 6 v T v, 4, 8 v 9 9, 4 9, 8 9 a T a v v a N N a a T T,, 4 4 9, 4 9, , 7 9, 4 9 a N a N N N a NN a N where T 9, 4 9, 8 9 and N 4 9, 7 9, rt) t,e rt) t,te t, t 4 t,t +,t, t First note here that: At t we have: Thus, , 7 9, 4 9 a,, 4 4T + )N 4 9, 7 9, 4 9 vt) r t),e t,t + )e t at) r t),e t,t + )e t v r ),, a r ),, a v,,,, v + + Recall that we have: T v v,, a T a v v Ma 6,

65 4 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Net, we compute a N and N: This vector has length: and thus, Finall we obtain the decomposition, a N N a a T T,,,,,, a N a N N + N a NN,,,, a N a,, T + N where T,, and N,,. 4. rt) t,cos t,t sin t, t π rθ) cos θ,sin θ,θ, θ First note here that: vt) r t), sin t,t cos t + sin t, at) r t), cos t, t sin t + cos t At t π we have: v r π/),, a r ),, π Thus, a v,, π,, π v + + Recall that we have: Net, we compute a N and N: T v v,, a T a v v π/ π a N N a a T T,, π + π,, π + π,, 6 π 6, π 6, π π,, 6,, This vector has length: and thus, a N a N N π,, 6 π 6 N a NN a N π 6 6 π 6,, π 6 6,, π 6 Finall we obtain the decomposition, where T,, and N 6,,. a,, π π T + π N 6 Ma 6,

66 SECTION.5 Motion in Three-Space LT SECTION 4.5) 5 4. Find the components Let rt) t a T, 4t and a N of the acceleration vector of a particle moving along a circular path of radius R cm with constant velocit. FindvTt) 5and cm/s. Nt), and show that the decomposition of at) into tangential and normal components is Since the particle moves with constant speed, we have v ) t), hence: ) t 4 at) a T v t) T + N t + 4 t + 4 The normal component of the acceleration is a N κt)vt). The curvature of a circular path of radius R is κt) R, and the velocit is the constant value vt) v 5. Hence, a N R v 5.5 cm/s 45. Suppose that the Ferris wheel in Eample 5 is rotating clockwise and that the point P at angle 45 has acceleration In the notation of Eample 5, find the acceleration vector for a person seated in a car at a) the highest point of vector a, 5 m/min pointing down, as in Figure. Determine the speed and tangential acceleration of the Ferris the Ferris wheel and b) the two points level with the center of the wheel. wheel. Ferris wheel 45 FIGURE The normal and tangential accelerations are both 5/ 5 m/min. The normal acceleration is v /R v / 5, so the speed is v 58). 47. A space shuttle orbits the earth at an altitude 4 km above the earth s surface, with constant speed v 8, km/h. At time t, a moving particle has velocit vector v i and acceleration vector a i + 8k. Determine the Find the magnitude of the shuttle s acceleration in km/h ), assuming that the radius of the earth is 678 km Figure ). curvature κt ) of the particle s path at time t. FIGURE Space shuttle orbit. The shuttle is in a uniform circular motion, therefore the tangential component of its acceleration is zero, and the acceleration can be written as: a κv N ) The radius of motion is km hence the curvature is κ Also b the given information the constant speed is v 8 km/h. Substituting these values in ) we get: ) a N km/h )N The magnitude of the shuttle s acceleration is thus: a km/h In units of m/s we obtain a m/s Ma 6,

67 6 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) 49. A runner runs along the heli rt) cos t,sin t,t. When he is at position r π ) A car proceeds along a circular path of radius R m centered at the origin. Starting, his speed is m/s and he is at rest, its speed increases accelerating at a rate at of a t rate m/s of. Find m/s the. acceleration Find his acceleration vector a at vector time t a at this s and moment. determine Note: its The decomposition runner s acceleration into normal vector and does tangential not coincide components. with the acceleration vector of rt). We have r t) sin t,cos t,, r t) sin t) + cos t +, T sin t,cos t, B definition, N is the unit vector in the direction of dt dt Therefore N cos t, sin t,.att π/, we have cos t, sin t, N cos t, sin t, T,,, N,, The acceleration vector is a v T + κv N We need to find the curvature, which happens to be constant: κ dt ds dt dt cos t, sin t, r Now we have a v T + κv N, 9, ) T + ) )N ) ),, + 9,, 5. Eplain Figurewh 4 shows the vector acceleration w in Figure vectors cannot of a particle be the acceleration moving clockwise vector ofaround a particle a circle. moving In along each case, the circle. state whether Hint: the Consider particlethe is speeding sign of w up, N. slowing down, or momentaril at constant speed. Eplain. A) B) C) FIGURE 4 In A) and B) the acceleration vector has a nonzero tangential and normal components; these are both possible acceleration vectors. In C) the normal component of the acceleration toward the inside of the curve is zero, that is, a is parallel to T,soκ vt), so either κ meaning our curve is not a circle) or vt) meaning our particle isn t moving). Either wa, C) is not a possible acceleration vector. 5. Suppose that r rt) a lies v on a sphere of radius R for all t. Let J r r. Show that r J r)/ r. Hint: ObserveProve that rthat andar N are perpendicular.. v a) Solution. Since r rt) lies on the sphere, the vectors r rt) and r r t) are orthogonal, therefore: We use the following well-known equalit: r r ) a b c) a c) b a b) c Ma 6,

68 SECTION.5 Motion in Three-Space LT SECTION 4.5) 7 Using this equalit and ) we obtain: Divided b the scalar r we obtain: J r r r ) r r r r ) r r ) r r r) r ) r r ) r + r r r + r r r r r J r r b) Solution. The cross product J r r is orthogonal to r and r. Also, r and r are orthogonal, hence the vectors r, r and J are mutuall orthogonal. Now, since r is orthogonal to r and J, the right-hand rule implies that r points in the direction of J r. Therefore, for some α> we have: r αj r r J r J r ) B properties of the cross product and since J, r, and r are mutuall orthogonal we have: J r J r r r r r r r r r Substituting in ) we get: r r J r r r J r r Further Insights and Challenges In Eercises 55 59, we consider an automobile of mass m traveling along a curved but level road. To avoid skidding, The orbit of a planet is an ellipse with the sun at one focus. The sun s gravitational force acts along the radial the road must suppl a frictional force F ma, where a is the car s acceleration vector. The maimum magnitude of the line from the planet to the sun the dashed lines in Figure 5), and b Newton s Second Law, the acceleration vector frictional force is μmg, where μ is the coefficient of friction and g 9.8 m/s points in the same direction. Assuming that the orbit has positive eccentricit. Let v be the car s speed in meters per the orbit is not a circle), eplain wh second. the planet must slow down in the upper half of the orbit as it moves awa from the sun) and speed up in the lower 55. Show half. Kepler s that the car Second will not Law, skid discussed if the curvature in the net κ of section, the road is is a such precise that version with R of this /κ) qualitative conclusion. Hint: Consider the decomposition of a into normal and tangential components. v ) v ) + μg) 5 R Note that braking v < ) and speeding up v > ) contribute equall to skidding. To avoid skidding, the frictional force the road must suppl is: F ma where a is the acceleration of the car. We consider the decomposition of the acceleration a into normal and tangential directions: at) v t)tt) + κv t)nt) Since N and T are orthogonal unit vectors, T N and T T N N. Thus: ) ) a v T + κv N v T + κv N v T T + κv v N T + κ v 4 N N Therefore: v + κ v 4 v + v 4 R a v ) + v 4 R Since the maimal fractional force is μmg we obtain that to avoid skidding the curvature must satisf: m v ) v 4 + R mμg. Ma 6,

69 8 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Hence, v ) v 4 + R μg), which becomes: v ) v ) + μg) R 57. Beginning Suppose at that rest, the an maimum automobile radius drives of around curvature a circular along a track curved of radius highwa R is R m, 8 accelerating m. How fast at can a rate an of automobile. m/s. After travel how man at constant seconds speed) will the along car the begin highwa to skid without if the coefficient skidding if of the friction coefficient is μ of.6? friction is μ.5? B Eercise 55 the car will begin to skid when: We are given that v. and v. Integrating gives: v ) + v4 R μ g ) t t v v dt. dt.t + v.t We substitute v t, v., R, μ.6 and g 9.8 in ) and solve for t. This gives:.) +.4 t After 9.9 s or. minutes, the car will begin to skid. t ). 4 8,6, t 9.9 s 59. What is the smallest radius R about which an automobile can turn without skidding at km/h if μ.75 a tpical You want to reverse our direction in the shortest possible time b driving around a semicircular bend Figure value)? 6). If ou travel at the maimum possible constant speed v that will not cause skidding, is it faster to hug the inside curve radius In Eercise r) or the 55outside we showed curb radius that ther)? car will Hint: not Use skid Eq. if5) thetofollowing show that inequalit at maimum holds: speed, the time required to drive around the semicircle is proportional to the square root of the radius. v ) v 4 + R <μ g In case of constant speed, v, so the inequalit becomes: Solving for R we get: v 4 R <μ g v 4 <μ g R v 4 μ g <R R> v μg The smallest radius R in which skidding does not occur is, thus, R v μg We substitute v km/h, μ.75, and g 7,8 km/h to obtain: R.5 km..75 7,8 Ma 6,

70 SECTION.6 Planetar Motion According to Kepler and Newton LT SECTION 4.6) 9.6 Planetar Motion According to Kepler and Newton LT Section 4.6) Preliminar Questions. Describe the relation between the vector J r r and the rate at which the radial vector sweeps out area. The rate at which the radial vector sweeps out area equals half the magnitude of the vector J. This relation is epressed in the formula: da dt J.. Equation ) shows that r is proportional to r. Eplain how this fact is used to prove Kepler s Second Law. In the proof of Kepler s Second Law it is shown that the rate at which area is swept out is da dt J, where J rt) r t) To show that J is constant, show that J is constant. This is done using the proportionalit of r and r which implies that rt) r t). Using this we get: dj dt d dt r r ) r r + r r + J const. How is the period T affected if the semimajor ais a is increased four-fold? Kepler s Third Law states that the period T of the orbit is given b: T 4π ) a GM or T π GM a / If a is increased four-fold the period becomes: That is, the period is increased eight-fold. π 4a) / π 8 a / GM GM Eercises. Kepler s Third Law states that T /a has the same value for each planetar orbit. Do the data in the following table support this conclusion? Estimate the length of Jupiter s period, assuming that a 77.8 m. Planet Mercur Venus Earth Mars a m) T ears) Using the given data we obtain the following values of T /a, where a, as alwas, is measured not in meters but in m: Planet Mercur Venus Earth Mars T /a The data on the planets supports Kepler s prediction. We estimate Jupiter s period using the given a) ast a 4.9 ears.. Ganmede, one of Jupiter s moons discovered b Galileo, has an orbital period of 7.54 das and a semimajor ais of.7 Finding 9 m. the UseMass Eercise of a Star to estimate Using Kepler s the massthird of Jupiter. Law, show that if a planet revolves around a star with period 4π ) a ) T and semimajor B Eercise ais, a, the then mass theof mass Jupiter of the can star beis computed M using the G T following. equalit: M 4π a G T Ma 6,

71 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) We substitute the given data T ,5.6 a.7 9 m and G 6.67 m kg s, to obtain: 4π.7 9) M ,5.6) kg. 5. Mass of the Milk Wa The sun revolves around the center of mass of the Milk Wa gala in an orbit that An astronomer observes planet orbiting a star with a period of 9.5 ears and a semimajor ais of 8 is approimatel circular, of radius a.8 7 km and velocit v 5 km/s. Use the result of Eercise to estimate km. the mass Find of the the mass portion of the of star the using Milk Eercise Wa inside. the sun s orbit place all of this mass at the center of the orbit). Write a.8 m and v 5 m/s. The circumference of the sun s orbit which is assumed circular) is πa m; since the sun s speed is a constant v m/s, its period is T πa s. B Eercise, the mass of the v portion of the Milk Wa inside the sun s orbit is 4π ) a ) M G T Substituting the values of a and T from above, G 6.67 m kg s gives M 4π a G av 4π a ) G.8 5 ) kg. v The mass of the sun is.989 kg, hence M is. times the mass of the sun billions times the mass of the sun). 7. Show that a planet in a circular orbit travels at constant speed. Hint: Use that J is constant and that rt) is orthogonal A satellite orbiting above the equator of the earth is geosnchronous if the period is T 4 hours in this case, the to r t) for a circular orbit. satellite stas over a fied point on the equator). Use Kepler s Third Law to show that in a circular geosnchronous orbit, theit distance shown from in the theproof center of of Kepler s the earth Second is R Law 4,46 that the km. vector Then compute J rt) the r altitude t) is constant, h of thehence orbit above its length the is constant: earth s surface. The earth has mass M kg and radius R 67 km. J rt) r t) const ) We consider the orbit as a circle of radius R, therefore, rt) and r t) are orthogonal and rt) R. B ) and using properties of the cross product we obtain: rt) r t) rt) r t) sin π R r t) const We conclude that r t) is constant, that is the speed v r t) of the planet is constant. 9. Prove that if a planetar orbit is circular of radius R, then vt πr, where Verif that the circular orbit v is the planet s speed constant b k Eercise 7) and T is the period. Then use Kepler s Third Law to prove that v rt) R cos ωt, R sin ωt R. B the Arc Length Formula and since the speed v r t) is constant, the length L of the circular satisfies the differential equation, Eq. ), provided that ω kr.then deduce Kepler sthird Law T 4π orbit ) can be computed b the following integral: R k T T for this orbit. L r T t) dt vdt vt vt On the other hand, the length of a circular orbit of radius R is πr, so we obtain: vt πr T πr ) v In a circular orbit of radius R, a R, hence b Kepler s Third Law we have: T 4π GM R ) We now substitute ) in ) and solve for v. This gives: ) πr 4π R v GM 4π R v 4π R GM v R GM GM v R Find the velocit of a satellite in geosnchronous orbit about the earth. Hint: Use Eercises 6 and 9. Ma 6,

72 SECTION.6 Planetar Motion According to Kepler and Newton LT SECTION 4.6). A communications satellite orbiting the earth has initial position r 9,,,, in km) and initial velocit r,, in km/s), where the origin is the earth s center. Find the equation of the plane containing the satellite s orbit. Hint: This plane is orthogonal to J. The vectors rt) and r t) lie in the plane containing the satellite s orbit, in particular the initial position r 9,,,, and the initial velocit r,,. Therefore, the cross product J r r is perpendicular to the plane. We compute J: J r r i j k 9,,, i 9, j + 9,, k,i 9,j + 9k,, 9,, 9 We now use the vector form of the equation of the plane with n J,, 9,, 9 and,,z r 9,,,,, to obtain the following equation: 9,,,, 9,, z 9,,,, 9 9,,,, 9 9,, 9,,z 9,, 9 9,,,, z 84, 4, + 8, 9 + 9z 5, The plane containing the satellite s orbit is, thus: P {,,z) : 9 + 9z 5, } Eercises 9: The perihelion and aphelion are the points on the orbit closest Assume that the earth s orbit is circular of radius R 5 6 to and farthest from the sun, respectivel km it is nearl circular with eccentricit Figure 8). The distance from the sun at the perihelion is denoted r e.7). Find the rate at which the earth s radial vector sweeps per and the speed at this point out area in units of km is denoted v /s. What is the per. Similarl, magnitude we write r of the ap and v vector J ap for the r r distance and speed at the aphelion. for the earth in units of km The semimajor ais is denoted a. per second)? Aphelion r v per F O F v ap Perihelion a Semimajor ais FIGURE 8 r and v r are perpendicular at the perihelion and aphelion.. Use the polar equation of an ellipse p r + e cos θ to show that r per a e) and r ap a + e). Hint: Use the fact that r per + r ap a. We use the polar equation of the elliptic orbit: r p + e cos θ ) Apogee F r ap r F r per Perigee At the perigee, θ and at the apogee θ π. Substituting these values in ) gives the distances r per and r ap respectivel. That is, p r per + e cos θ p + e p r ap + e cos π p e To obtain the s in terms of a rather than p, we notice that: r per + r ap a ) ) Ma 6,

73 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Hence: ielding a p + e + p p e) + p + e) p e + e) e) + e) e) p a + e) e) Substituting in ) and ) we obtain: a + e) e) r per a e) + e a + e) e) r ap a + e) e 5. Use the fact that J r r is constant to prove Use the result of Eercise to prove the formulas v e r per e) v ap r ap + e) per, p r apr per Hint: r is perpendicular to r at the perihelion and r ap aphelion. + r per r ap + r per Since the vector Jt) rt) r t) is constant, it is the same vector at the perigee and at the apogee, hence we ma equate the length of Jt) at these two points. Since at the perigee and at the apogee rt) and r t) are orthogonal we have b properties of the cross product: Equating the two values gives: r ap r ap r ap r ap r apv ap r per r per r per r per r perv per r ap v ap r per v per ) In Eercise we showed that r per a e) and r ap a + e). Substituting in ) we obtain: a + e)v ap a e)v per + e)v ap e)v per 7. Conservation of Energ The total mechanical energ kinetic energ plus potential energ) of a planet of mass m Compute r per and r ap for the orbit of Mercur, which has eccentricit e.44 see the table in Eercise for orbiting a sun of mass M with position r and speed v r is the semimajor ais). E mv GMm 8 r a) Prove the equations d dt mv v ma), d dt GMm r v GMm ) r r b) Then use Newton s Law to show that E is conserved that is, de. dt We start b observing that since r r r, we have using Eq. 4) in Theorem, Section.) Equating these two epressions gives d dt r r d dt r, and d dt r d r r r r dt d r r r dt r a) Appling ) to r, we have d dt mv d dt m r m r d dt r m r r r r r mr ) v ma) proving half of formula. For the other half, note that again b ), d GMm dt r GMm d dt r GMm r d dt r GMm r r r r r GMm ) r r v GMm ) r r ) Ma 6,

74 SECTION.6 Planetar Motion According to Kepler and Newton LT SECTION 4.6) b) We have b part a) de dt d dt ) mv d dt GMm r ) ) GMm v ma) + v r r v ma + GMm ) r r ) B Newton s Law, formula ) in the tet, r GM r e r GM r r ) Substituting ) into ), and noting that v r and a r gives de dt r mr + GMm ) r r r GMm r r + GMm ) r r ) GM + e 9. Prove Show that that v the total energ [Eq. 8)] of a planet in a circular orbit of radius R is E GMm per as follows: a e. Hint: Use Eercise R a) Use 9. Conservation of Energ Eercise 7) to show that vper v ap GM rper ) r ap b) Show that rper r ap e a e using Eercise. ) c) Show that vper v ap 4 e + e) v per using Eercise 5. Then solve for v per using a) and b). a) The total mechanical energ of a planet is constant. That is, E mv GMm r const. Therefore, E has equal values at the perigee and apogee. Hence, mv per GMm r per mv ap GMm r ap ) m vper v ap GMm r per r ap vper v ap GM rper r ap b) In Eercise we showed that r per a e) and r ap a + e). Therefore, c) In Eercise 5 we showed that Hence, We compute the following difference, rper r ap a e) + e e) a + e) a e) + e) e a e ) v per e) v ap + e) v ap e + e v per ) vper e ) ) v ap v per + e v per vper e + e vper + e) e) + e) vper + e + e e + e ) e + e) 4 + e) v per We combine this equalit with the equalit in part a) to write 4e ) + e) v per GM rper r ap ) ) Ma 6,

75 4 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Replacing the difference in the right-hand side b e a e ) from part b)) and solving for v per we obtain: 4e + e) v per GM e a e ) v per 4GMe + e) GM + e) a e) + e) 4e a e) or, GM + e v per a e. Prove that v GM Show that a planet inran ) at an point on an elliptical orbit, where r r, elliptical orbit has total mechanical energ E GMm v is the velocit, and a is the a, where a is the semimajor ais. semimajor a Hint: Use ais Eercise of the orbit. 9 to compute the total energ at the perihelion. The total energ E mv GMm r is conserved, and in Eercise we showed that its constant value is GMm a. We obtain the following equalit: mv GMm GMm r a Algebraic manipulations ield: v GM r GM a GM r ) a Two space shuttles A and B orbit the earth along the solid trajector in Figure 9. Hoping to catch up to B, the Further Insights and Challenges pilot of A applies a forward thrust to increase her shuttle s kinetic energ. Use Eercise to show that shuttle A will Eercises move off and into 4 a larger prove Kepler s orbit as shown Third Law. in the Figure figure. Then shows use an Kepler s elliptical Third orbit Law with to polar show equation that A s orbital period T will increase and she will fall farther and farther r behind B)! p + e cos θ where p J /k. The origin of the polar coordinates is at F. Let a and b be the semimajor and semiminor aes, respectivel. B a a Semiminor ais b F C F A a Semimajor ais FIGURE. This eercise shows that b pa. a) Show that CF ae. Hint: r per a e) b Eercise. b) Show that a p e. c) Show that F A + F A a. Conclude that F B + F B a and hence F B F B a. d) Use the Pthagorean Theorem to prove that b pa. a) Since CF AF, we have: Therefore, F A CA CF a F A F A + F A a ) B C F F A The ellipse is the set of all points such that the sum of the distances to the two foci F and F is constant. Therefore, F A + F A F B + F B ) Ma 6,

76 SECTION.6 Planetar Motion According to Kepler and Newton LT SECTION 4.6) 5 Combining ) and ), we obtain: The triangle F BF is isosceles, hence F B F B and so we conclude that F B + F B a ) F B F B a b) The polar equation of the ellipse, where the focus F is at the origin is r p + e cos θ B r C F F A The point A corresponds to θ, hence, The point C corresponds to θ π hence, F A p + e cos p + e p F C + e cos π p e We now find F A. Using the equalit CF AF we get: That is, Combining ), 4), and 5) we obtain: Hence, a F A F F + F A F F + F C F C F A p e p + e + p e a p e p + e + p ) p e) + p + e) p e + e) e) e ) p e 4) 5) c) We use Pthagoras Theorem for the triangle OBF : OB + OF BF 6) B b a F A Using 4) we have Also OB b and BF a, hence 6) gives: We solve for b: OF a F A a p + e b + a p ) a + e b + a ap + e + p + e) a b ap + e + p + e) Ma 6,

77 6 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) In part b) we showed that a b p e. We substitute to obtain: p + e p e + p + e) b p + e) e) p + e) p p e) + e) e) p + e) + e) e) p e Hence, b p e Since e p a we also have b p p a ap 5. The area According A of the toellipse Eq. 7) isthe A velocit πab. vector of a planet as a function of the angle θ is a) Prove, using Kepler s First Law, that A JT, where T is the period of the orbit. b) Use Eercise to show that A π p)avθ) /. k J e θ + c Use c) this Deduce to eplain Kepler s the following Third Law: statement: T 4π As a GM a planet. revolves around the sun, its velocit vector traces out a circle of radius k/j with center c Figure ). This beautiful but hidden propert of orbits was discovered b William Rowan Hamilton in 847. B vθ) A vθ) θ A C θ B c C D D Planetar orbit Velocit circle FIGURE The velocit vector traces out a circle as the planet travels along its orbit. Recall that e θ sin θ,cos θ, so that vθ) k J sin θ,cos θ+c k J sin θ),cos θ)+c The first term is obviousl a clockwise due to having θ instead of θ) parametrization of a circle of radius k/j centered at the origin. It follows that vθ) is a clockwise parametrization of a circle of radius k/j and center c. CHAPTER REVIEW EXERCISES. Determine the domains of the vector-valued functions. a) r t) t,t + ), sin t b) r t) 8 t, ln t,e t a) We find the domain of r t) t,t + ), sin t. The function t is defined for t. t + ) is defined for t and sin t is defined for t. Hence, the domain of r t) is defined b the following inequalities: t t <t< t or <t Ma 6,

78 Chapter Review Eercises 7 b) We find the domain of r t) 8 t, ln t,e t. The domain of 8 t is 8 t. The domain of ln t is t> and e t is defined for t. Hence, the domain of r t) is defined b the following inequalities: 8 t t> t t 8 t> <t. Find a vector parametrization of the intersection of the surfaces Sketch the paths r θ) θ,cos θ and r θ) cos θ,θ in the + -plane. 4 + z 6 and in R. We need to find a vector parametrization rt) t), t), zt) for the intersection curve. Using t as a parameter, we have t and t. We substitute in the equation of the surface z 6 and solve for z in terms of t. This gives: t 4 + t 4 + z 6 t 4 + z 6 z t 4 z t 4 We obtain the following parametrization of the intersection curve: rt) t,t, t 4. In Eercises Find 5, a vector calculate parametrization the derivative usingindicated. trigonometric functions of the intersection of the plane + + z and ) z ) 5. the r t), elliptical rt) clinder t,t, ln + t inr. 8 We use the Theorem on Componentwise Differentiation to compute the derivative r t). We get r t) t),t ),ln t), t, t 7. r ), r rt) e t), rt) t,e t 4t, 4t,e, 7t 6t We differentiate rt) componentwise to find r t): r t) e t ),e 4t ),e 6t ) e t, 8te 4t, 6e 6t The derivative r ) is obtained b setting t inr t). This gives r ) e, 8 e 4, 6e 6,, 6 d 9. r ), rt) t,t + ),t t dt et,t,t Using the Product Rule for differentiation gives d dt et,t,t e t d,t,t + e t ),t,t e t,, t + e t,t,t dt e t,, t +,t,t ) e t, + t,t + t In Eercises d dθ rcos 4, θ), calculate rs) the derivative s, s, s at t, assuming that r ),,, r ),, r ),,, r ),, 4 d. dt 6r t) 4 r t)) Using Differentiation Rules we obtain: d dt 6r t) 4r t)) 6r ) 4r ) 6,, 4,, 4 t,, 6, 8, 6, 8,. d r d t) e t r t) dt r t) ) ) dt Using Product Rule for Dot Products we obtain: d dt r t) r t) r t) r t) + r t) r t) Ma 6,

79 8 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Setting t gives: d dt r t) r t) r ) r ) + r ) r ),,,, 4 +,,,, + t 5. Calculate d 4t +,t r t) r t) ), 4t dt. dt B the definition of vector-valued integration, we have 4t +,t, 4t dt 4t + )dt, t dt, 4t dt ) We compute the integrals on the right-hand side: 4t + )dt t + t t dt t 9 4t dt t Substituting in ) gives the following integral: 4t +,t, 4t dt 7, 9, 8 7. A particle located π at,, ) at time t follows a path whose velocit vector is vt),t,t. Find the particle s locationcalculate t. sin θ,θ,cos θ dθ. We first find the path rt) b integrating the velocit vector vt): rt),t,t dt dt, tdt, t dt t + c, t + c, t + c Denoting b c c,c,c the constant vector, we obtain: rt) t, t, t + c ) To find the constant vector c, we use the given information on the initial position of the particle. At time t itisat the point,, ). That is, b ): or, r),, + c,, c,, We substitute in ) to obtain: rt) t, t, t +,, t +, t +, t Finall, we substitute t to obtain the particle s location at t : r) +, +,,, 6 At time t the particle is located at the point,, 6 ) Ma 6,

80 Chapter Review Eercises 9 9. Calculate rt) assuming that Find the vector-valued function rt) t), t) in R satisfing r t) rt) with initial conditions r),. r t) 4 6t,t t, r ),, r), Using componentwise integration we get: r t) 4 6t,t t dt 4 6t dt, t tdt 4t 8t, 4t t + c Then using the initial condition r ), we get: r ), c so then r t) 4t 8t, 4t t +, 4t 8t +, 4t t Then integrating componentwise once more we get: rt) 4t 8t +, 4t t 4t 8t + dt, dt 4t t dt t 8 t + t,t 4 t + c 6 Using the initial condition r), we have: r), c Therefore, rt) t 8 t + t,t 4 t +, t 8 6 t + t,t 4 t6 +. Compute the length of the path Solve r t) t,t +,t subject to the initial conditions r),, and r ),, rt) sin t,cos t,t for t We use the formula for the arc length: s r t) dt ) We compute the derivative vector r t) and its length: r t) cos t, sin t, r t) cos t) + sin t) + 4 ) cos t + sin t We substitute in ) and compute the integral to obtain the following length: s dt t. 4 cos t + 4 sin t + 9 Epress the length of the path rt) ln t,t,e t for t as a definite integral, and use a computer algebra sstem to find its value to two decimal places. Ma 6,

81 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4). Find an arc length parametrization of a heli of height cm that makes four full rotations over a circle of radius 5 cm. Since the radius is 5 cm and the height is cm, the heli is traced b a parametrization of the form: Since the heli makes eactl 4 full rotations, we have: rt) 5 cos at,5 sin at,t, t The parametrization of the heli is, thus: rt) The heli is shown in the following figure: a 4 π a π 5 5 cos πt πt, 5 sin 5 5,t, t To find the arc length parametrization for the heli, we use: t st) r u) du ) We find r t) and its length: r t) 5 π πt sin 5 5, 5 π πt cos 5 5, π sin πt πt, π cos 5 5, r t) 4π sin πt 5 + 4π cos πt 5 + 4π sin πt πt + cos 5 5 ) + + 4π Substituting in ) we get: t st) + 4π du t + 4π Therefore, we let s t + 4π and thus, t s + 4π gs) Thus, we can write rs) 5 cos sa + 4π, 5 sin sa + 4π, s, s + 4π π 5. A projectile fired at an angle of 6 lands 4 m awa. What was Find the minimum speed of a particle with trajector rt) its t,e t initial,e 4 t speed?. Place the projectile at the origin, and let rt) be the position vector of the projectile. Step. Use Newton s Law Gravit eerts a downward force of magnitude mg, where m is the mass of the bullet and g 9.8 m/s. In vector form, F, mg m, g Ma 6,

82 Chapter Review Eercises Newton s Second Law F mr t) ields m, g mr t) or r t), g. We determine rt) b integrating twice: t t r t) r u) du, g du, gt + v t rt) r u) du t, gu + v )du, gt + tv + r Step. Use the initial conditions B our choice of coordinates, r. The initial velocit v has unknown magnitude v, but we know that it points in the direction of the unit vector cos 6, sin 6. Therefore, v v cos 6, sin 6 v, rt), gt + tv, Step. Solve for v. The projectile hits the point 4, on the ground if there eists a time t such that rt) 4, ; that is,, gt + tv, 4, Equating components, we obtain tv 4, gt + tv The first equation ields t 8 v. Now substitute in the second equation and solve, using g 9.8m/s : ) 8 ) v v v Ma 6, ) 8 4 v 4.9 v ) v , v m/s We obtain v m/s. 7. During a short time interval [.5,.5], the path of an unmanned sp plane is described b A speciall trained mouse runs counterclockwise in a circle of radius.6 m on the floor of an elevator with speed. m/s while the elevator ascends from ground level along the z-ais) at a speed of m/s. Find the mouse s acceleration vector as a function of time. rt) Assume that, 7the circle t,4 ist t centered at the origin of the -plane and the mouse is at,, ) at t. A laser is fired in the tangential direction) toward the z-plane at time t. Which point in the z-plane does the laser beam hit? Notice first that b differentiating we get the tangent vector: r t),, t, t r ),, and the tangent line to the path would be: ls) r) + sr ), 6, 9 + s,, + s, 6 s, 9 s If the laser is fired in the tangential direction toward the z-plane means that the -coordinate will be zero - this is when s /. Therefore, Hence, the laser beam will hit the point, /, 8). l/), /, 8 A force F t + 4, 8 4t in newtons) acts on a -kg mass. Find the position of the mass at t sifitis located at 4, 6) at t and has initial velocit, in m/s.

83 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) 9. Find the unit tangent vector to rt) sin t,t,cos t at t π. The unit tangent vector at t π is We differentiate rt) componentwise to obtain: Therefore, Tπ) r π) r π) r t) cos t,, sin t ) r π) cos π,, sin π,, We compute the length of r π): Substituting in ) gives: r π) ) + + Tπ),,. Calculate κ) for rt) ln t,t. Find the unit tangent vector to rt) t, tan t,t at t. Recall, κt) r t) r t) r t) Computing derivatives we get: r t) t,, r ),, r ) r t) t,, r ), Computing the cross product we get: r ) r i j k ),, and r ) r ). Therefore, κ) r ) r ) r ) ) / In Eercises Calculate and κ 4, π ) write 4 for rt) the acceleration tan t,sec vector t,cos t. a at the point indicated as a sum of tangential and normal components.. rθ) cos θ,sin θ, θ π 4 First note here that: vθ) r θ) sin θ, cos θ aθ) r θ) cos θ, 4 sin θ At t π/4 we have: Thus, v r π/4), a r π/4), 4 a v, 4, v + Ma 6,

84 Chapter Review Eercises Recall that we have: Net, we compute a N and N: T v v, / a T a v v / /, a N N a a T T, 4,, 4 This vector has length: a N a N N 4 and thus, N a NN, 4, a N 4 Finall, we obtain the decomposition, a, 4 T + 4N where T, and N,. 5. At a certain rt) time t t, the, t t path,t of a moving particle is tangent to the -ais in the positive direction. The particle s speed at time t is 4 m/s, and its acceleration, t vector is a 5, 4,. Determine the curvature of the path at t. We are given that the particle is moving tangent to the -ais with speed 4 m/s, so then: and a r 5, 4,. Recall the formula for curvature: r, 4, κ r r r First calculate the cross product: r r i j k 4 48,, 5 4 Then the length of r and r r : r 4, r r so then for curvature we get: κ r r r Parametrize the osculating circle to at Parametrize the osculating circle to 4. at. First differentiate twice: and at the point 4 we get: f ), f ) 4 / f 4) 4, f 4) Ma 6,

85 4 CHAPTER CALCULUS OF VECTOR-VALUED FUNCTIONS LT CHAPTER 4) Step. Find the radius Then recall the formula for curvature: and evaluating at 4 we have: κ4) [ + 6 κ) ] / f ) [ + f )) ] / 7 6 ) / 6 / 7 / 7 / Therefore the radius of the osculating circle is R 7/. Step. Find N at 4 First we will parametrize the curve f) as: r),, r4) 4, and differentiate: r ), / Note here that the vector /, is orthogonal to r ) for all values of and points in the direction of the bending of the curve. Computing the unit normal to the curve, using the vector orthogonal to r ) we get: N) /, 4 + Step. Find the center Q Now to find the center Q of the osculating circle:, N4) OQ r4) + κ N4) 4, + 7/ 4, 4, + 4 4, 7 4, +, 4 5, The center of the osculating circle is Q 5, ). Step 4. Parametrize the osculating circle Then parametrizing the osculating circle we get: , 4 7 4, 5 ct), + 7/ cos t,sin t 9. Suppose the orbit of a planet is an ellipse of eccentricit e c/a and period T Figure ). Use Kepler s Second Law to show that the time required to travel from A to B is equal to If a planet has zero mass m ), then Newton s laws of motion reduce to r t) and the orbit is a straight line rt) r + tv, where r r) and v r ) Figure ). Show that the area swept out b the radial vector at time t is At) r v t and thus Kepler s Second Law continues to hold the rate is constant). 4 + e ) T π Ma 6,

86 Chapter Review Eercises 5 B b Sun A a A O c, ) B FIGURE B the Law of Equal Areas, the position vector pointing from the sun to the planet sweeps out equal areas in equal times. We denote b S the area swept b the position vector when the planet moves from A to B, and t is the desired time. Since the position vector sweeps out the whole area of the ellipse πab) in time T, the Law of Equal Areas implies that: S πab t T t TS πab We now find the area S as the sum of the area of a quarter of the ellipse and the area of the triangle ODB. That is, ) Substituting in ) we get: t S πab 4 T bπa + c) 4πab + OD OB Tπa+ c) 4πa πab 4 + cb b πa + c) 4 T 4 + ) c T π a 4 + e ) π A a S b O Dc, ) Sun B The period of Mercur is approimatel 88 das, and its orbit has eccentricit.5. How much longer does it take Mercur to travel from A to B than from B to A Figure )? Ma 6,