13 CALCULUS OF VECTORVALUED FUNCTIONS


 Terence Douglas
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1 CALCULUS OF VECTORVALUED FUNCTIONS. VectorValued 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 zplane? The projection of the path onto the zplane is the curve traced b ti + e t k t,,e t. This is the curve z e in the zplane.. 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 zplane is cos t,, sin t. Since + z cos t + sin t, the projection is a circle in the zplane. 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 zplane 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 zplane 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. VectorValued 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 vectorvalued 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 VECTORVALUED 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 vectorvalued functions a) f) with the space curves i) vi) in Figure. Match the space curves in Figure 8 with the following vectorvalued 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. VectorValued 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 zplane. We conclude that the function traces the circle of radius, centered at the point,, 4), and contained in the zplane. 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 VECTORVALUED FUNCTIONS LT CHAPTER 4) In Eercises 9 and, let Use a computer algebra sstem to plot the projections onto the  and zplanes 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 zplane 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. VectorValued 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 zcoordinate 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 vectorvalued 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 inthezplane 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 VECTORVALUED 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 zplane is traced b cos t,, sin t which is the unit circle in the zplane. The projection of the curve on the zplane 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. VectorValued 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 VECTORVALUED FUNCTIONS LT CHAPTER 4). The line through the origin whose projection on the plane is a line of slope and whose projection on the zplane 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 zplane 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 zplane The horizontal circle of radius with center,, 4) The circle is parallel to the zplane and centered at,, 5), hence the coordinates of the points on the circle are. The projection of the circle on the zplane 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 + intheplane, 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 inthezplane, 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. VectorValued 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 VECTORVALUED 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 VectorValued 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 VectorValued Functions LT Section 4.) Preliminar Questions. State the three forms of the Product Rule for vectorvalued functions. The Product Rule for scalar multiple ft)of a vectorvalued 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 VECTORVALUED 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 vectorvalued function is defined as the limit of the difference quotient, just as in the scalarvalued case. The statement is true. The derivative of a vectorvalued 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 scalarvalued case.. There are two Chain Rules for vectorvalued functions: one for the composite of two vectorvalued functions and one for the composite of a vectorvalued and a scalarvalued function. This statement is false. A vectorvalued 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 vectorvalued 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 scalarvalued function ft), the composition rf t)) has a meaning, and there is a Chain Rule for differentiabilit of this vectorvalued 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 vectorvalued function is the slope of the tangent line, just as in the scalar case. The statement is false. The derivative of a vectorvalued function is again a vectorvalued 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 vectorvalued 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 vectorvalued 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 VectorValued 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 VECTORVALUED 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 VectorValued 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 VECTORVALUED 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 VectorValued 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 Vectorvalued 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 vectorvalued 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 VECTORVALUED 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 VectorValued 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 VECTORVALUED 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 vectorvalued function. Find the of the equation r in threespace. 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 VectorValued 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 VECTORVALUED 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 vectorvalued 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 scalarvalued function ft)and differentiable vectorvalued 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 vectorvalued 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 vectorvalued 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 VECTORVALUED 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 VECTORVALUED 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 zcomponent, 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 VECTORVALUED 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,
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