Section 11.1: Vectors in the Plane. Suggested Problems: 1, 5, 9, 17, 23, 25-37, 40, 42, 44, 45, 47, 50

Transcription

1 Section 11.1: Vectors in the Plane Page 779 Suggested Problems: 1, 5, 9, 17, 3, 5-37, 40, 4, 44, 45, 47, 50 Determine whether the following vectors a and b are perpendicular. 5) a = 6, 0, b = 0, 7 Recall a and b are perpendicular if and only if a + b = a + b. Now a = 36 = 36 and b = 49. Since a + b = = 85, we have a + b = 85 = a + b, so a b. 7) a = i j, b = 4j + 8i Again, since a = 3 and b = 10, and a + b = 13, we have a + b = = a + b, so a and b are not perpendicular. Express i and j in terms of a and b. 9) a = i + 3j, b = 3i + 4j We want to find r, s, p, and q such that i = ra + sb and j = pa + qb. But then 1i + 0j = i = ra + sb = (r + 3s)i + (3r + 4s)j 0i + 1j = j = pa + qb = (p + 3q)i + (3p + 4q)j. 1

2 By equating coefficients, we get (1) 1 = r+3s () 0 = 3r+4s (3) 0 = p+3q (4) 1 = 3p + 4q. Subtracting (1) from (), we get 1 = r + s. Combining that with (1), we get s = 3, and so r = 4. Similarly, we get p = 3 and q =. Hence i = 4a + 3b j = 3a b. Write c in the form ra + sb where r and s are scalars. 31) a = i + j, b = i j, c = i 3j If c = ra+sb = (r+s)i+(r s)j, then equating coefficients gives us that r + s = and r s = 3. Therefore r = 1/ and s = 5/. We conclude that c = 1 a + 5 b. 35) Find a vector of length 5 with (a) the same direction as 7i 3j (b) the direction opposite that of 8i + 5j. (a) Notice that 7i 3j = 58, so the vector is what we want. (b) Since 8i + 5j = 89, the vector is the desired vector (7i 3j) 5 89 (8i + 5j) 36) For what numbers c are the vectors c, and c, 8 perpendicular?

3 Notice that c, = c +4, c, 8 = c +64, and c, + c, 8 = 4c Therefore c, and c, 8 are perpendicular if and only if 4c + 36 = (c + 4) + (c + 64) = c This happens if and only if c = 16, or c = ±4. 44) A 50-pound weight is suspended by two cables C 1 and C. C 1 makes a 45 o angle with the ceiling and C makes a 30 o angle with the ceiling. Determine the tension in each cable. Our force vector from gravity is F = 0, 50. If T 1 and T are the tensions in C 1 and C, respectively, and T 1 and T are the tension force vectors in the cables, then T 1 = T 1 cos(45 o ),T 1 sin(45 o ) = T 1, T 1 T = T cos(30 o ),T sin(30 o ) = We know the vectors sum to 0, so 3 T 1 + T = 0 T T 50 = 0. 3 T s, 1 T. From the above equation we see that T = 3 T 1. Substituting this into the lower equation, we get ( ) 3 + T 1 = 50, so T Then T = ( 3/ )T

4 Fun Problem) Show that the vectors v 1 = a 1 i + b 1 j and v = a i + b j are perpendicular if and only if a 1 a + b 1 b = 0. Since we know v 1 = a 1 + b 1, v = a + b, and v 1 + v = (a 1 + a ) + (b 1 + b ) = a 1 + a 1 a + a + b 1 + b 1 b + b, we see v 1 and v are perpendicular if and only if a 1 + b 1 + a + b = a 1 + a 1 a + a + b 1 + b 1 b + b. By canceling and dividing by, we see the above equality holds if and only if a 1 a + b 1 b = 0, which is what we wanted to show. 4

5 Section 11.: Three-Dimensional Vectors The following problems are similar to the indicated problems on pages Problems 1-6: Let a = 1,, and b = 4, 7, 4. Find (1) 3a + 6b () a b (3) a b (4) â = a/ a. (1) 3a + 6b = 3 1,, + 6 4, 7, 4 = 3, 6, 6 + 4, 4, 4 = 7, 36, 18. () Since a b = 1 4, 7, ( 4) = 3, 9, 6, we see a b = ( 3) + ( 9) + 6 = 16 = (3) a b = ( ) 7 + ( 4) = = 18. (4) Since a = 1 + ( ) + = 3, we have â = ,, = 3, 3,. 3 Problems 7-1: With a and b as above, find the angle between a and b to the nearest degree. Let α be the angle between a and b. We know that cos(α) = a b a b. We have already sen above that a b = 18 and a = 3. Now b = ( 4) = 81 = 9, so cos(α) = 18/(3 9) = /3. Therefore, α = o. 5

6 Problems 13-18: Find comp b a and compab for the vectors a and b as above. By the formula, comp b a = a b b = 18 9 = and compab = b a a = 18 3 = 6. Problems 19-4: Write the equation of the indicated spheres. Center ( 6, 7, 4) and radius 3 or We have (x ( 6)) + (x 7) + (x 4) = 3, (x + 6) + (x 7) + (x 4) = 9. Center (1,, ) and passing through the origin If a sphere with center (1,, ) passes through the origin, then the radius is the distance between the center and (0, 0, 0), which is, in our case, 3. Therefore our equation is (x 1) + (y ) + (z ) = 9. Problems 5-8: Find the center and radius of the sphere with equation x + y + z 6x + 5y + 8z = 19/4. Our method will be to complete the squares. By regrouping terms, our equation is (x 6x) + (y + 5y) + (z + 8z) = 19/4. 6

7 Completing each square, we see or (x 6x + 9) + (y + 5y + 5/4) + (z + 8z + 16) = 19/ / (x 3) + (y + 5/) + (z + 4) = 36. Therefore the center of this circle is (3, 5/, 4) and its radius is 6. Problems 9-38: Describe the graph of the given equations in geometric terms using plain, clear language. x + y = 1 The sum of squares of real numbers is always nonnegative, so this equation has no solutions. The graph of it has no points. x + y = 0 If the sum of two squares is 0, then both must be 0. The set of points satisfying this equation are exactly those points with x- and y-coordinate 0. Therefore the equation represents the z-axis. x + y = 4 For a fixed value of z 0, the set of points satisfying the equation with z- coordinate z 0 is a circle of radius centered at (0, 0,z 0 ). As z ranges over all real numbers, this circle slides along the z-axis, forming an infinitely long cylinder with radius. Problems 39-4: Determine whether the following pairs of vectors are parallel, perpendicular, or neither. a 1 = 1,, b 1 4, 1, 1 Recall a and b are perpendicular if and only if a b = 0 and are parallel if and only if a b = a b. Now a 1 b 1 = = 0, so a 1 is perpendicular to b 1. 7

8 a =, 4, 7 b 1, 1, 1 Since a b = 13 and a b = , so a and b are neither parallel nor perpendicular. a 3 = 1, 4, 16 b 3 1, 7, 8 Since a 3 b 3 = 78 and a 3 b 3 = = 78, this pair of vectors is parallel. Problems 43-44: Determine whether the following triples of points lie on a straight line. (3, 0, 6), (1, 1, 1), and ( 13, 8, 34) A triple of points P, Q and R are on a straight line if and only if PQ is parallel to QR. The above triple lie on a straight line if and only if, 1, 5 is parallel to 14, 7, 35. It is easy to see the second vector is 7 times the first, so they are indeed parallel. We conclude the three points lie on a straight line. (1,, ), (, 0, 9), and (4, 6, 7) As above, these three lie on a straight line if and only if the vectors v 1 = 1,, 11 and v, 6, are parallel. Since the first coordinate of v is twice the first coordinate of v 1, these two vectors are parallel if and only if v = v 1, which is clearly false. Hence these points do not belong to a straight line. Problems 45-48: Find the three angles of the triangle with given vertices. A = (1, 1, 1) B = (3, 4, 5) C = (4, 3, ) Let v 1 = AB, v = BC, and v 3 = CA. Let α, β, and γ be the angles at the vertices A, B, and C, respectively. We know cos(α) = v 1 v 3 v 1 v 3 = 16 =

9 cos(β) = v 1 v v 1 v = =.77 cos(γ) = v v 3 v v 3 so α = 37.4 o, β = 43.3 o, and γ = 99.3 o. = =.161, Problems 49-5: Find the direction angles of the vector represented by PQ, where P = (1, 0, 1) and Q = (, 4, 5). Let α, β, and γ be the angles between PQ and the x-, y-, and z-axes, respectively. Notice that PQ = 1, 4, 4. We know that PQ cos(α) = i PQ i = 1 33 cos(β) = cos(γ) = so α = o and β = γ = o. PQ j PQ j = 4 33 PQ k PQ k = 4, 33 Problems 53-54: Find the work W done by the force F in moving a particle in a straight line from P to Q. F = i j + k P = 0, 1, Q = 3,, 3 The book tells us that W = F PQ. Since PQ = 3, 1, 1, we see W = 6 joules. 9

10 Section 11.3: The Cross Product of Vectors Problems 1-4: Find a b, where a = 4, 6, 1 and b = 1,, 1. By Equation 5 on Page 791, i j k a b = det = i(6( 1) ( 1)) j(4( 1) 1( 1)) + k(4 6 1, so a b = 4i + 3j + k = 4, 3,. Problems 5-6: Find the cross product of the given -dimensional vectors a = a 1,a and b = b 1,b by first extending them to 3-dimensional vectors a = a 1,a, 0 and b = b 1,b, 0. a = 1, 4 b = 5, By definition, a = 1, 4, 0 and b = 5,, 0. Then i j k a b = det = i(4 0 ( )0) j( ) + k(1( ) 5 4) = k. It is important to notice this vector has i and j coefficients equal to 0. This makes sense, since a and b lie in the xy-plane, and since a b must be perpendicular to both vectors, it must be a multiple of k. Problems 7-8: Find two different unit vectors u and v both of which are perpendicular to both a = 5, 4, 1 and b =, 3, 6. 10

11 First we use the cross product to find a vector perpendicular to both a and b: i j k a b = det = 1i 8j + 7k. 3 6 Now notice that a b = 174. Therefore u = a b 174 and v = u are unit vectors, which are both perpendicular to both a and b since a b is. Problems 14-15: Find the area of the triangle with vertices P(0, 0, 1), Q(, 3, 4), and R( 6, 4, 0). Equation 10 on page 793 tells us the area A of this triangle is A = 1 PQ PR. Now PQ =, 3, 3 and PR = 6, 4, 1, so PQ PR = det Hence PQ PR = 1157, so i j k A = = 15i 16j + 6k. Problems 16-17: Find the volume V 1 of the parallelepiped with adjacent edges OP, OQ, and OR, where P = (0, 0, 1), Q = (, 3, 4), and R = ( 6, 4, 0). Then find the volume V of the pyramid determined by O, P, Q, and R. We know from Equations 17 and 18 on pages 794 and 795 that V 1 = det 3 4 =

12 = 0( ) 0( 0 ( 6)4) + 1( 4 ( 6)3) = 6 = 6. Now, for the volume of the pyramid, Example 7 on page 795 tells us that V = 1 6 V 1 = 6 6. Problem 18: Find a unit vector n perpendicular to the plane through the points P, Q, and R from above. Then find the distance from the origin to this plane by computing n OP. The vectors PQ and PR certainly lie in the plane through P, Q, and R. Recall PQ =, 3, 3 and PR = 6, 4, 1. ˆ The vector n = PQ PR is a unit vector perpendicular to PQ and PR, and so it is perpendicular to the plane containing P, Q, and R. Since we have seen that PQ PR = 15i 16j + 6k and PQ PR = 1157, we have 1 n = ( 15i 16j + 6k) Now, let C be the closest point to the origin O on the plane containing P, Q, and R. We want to find the distance from O to C, or OC. Thinking geometrically, one can reason that since OC is parallel to n, the length of OC is the length of OP in the direction of n, or comp nop. But then OC = compnop = n OP n = n OP = 6. Therefore the distance from the origin to the plane containing P, Q, and R is 6. Problems 19-: Determine whether or not the four given points A, B, C, and D are coplanar. If not, find the volume of the pyramid with these four points as its vertices, given that its volume is one-sixth that of the parallelepiped spanned by AB, AC, and AD. 1

13 A(1, 4, ) B(6, 8, ) C(4, 7, 1) D(8, 1, ) First note that AB = 5, 4, 4, AC = 3, 3, 1, and AD = 7, 8, 0. Now AB AC = det i j k = 8i + 7j + 3k, so AD ( AB AC) = = 0, so the four points are coplanar. A(1, 4, ) B(6, 8, ) C(4, 7, 1) D(14, 10, 0) Again, we start by noticing that AB = 5, 4, 4,, AC = 3, 3, 1, and AD = 13, 6,. Then AD ( AB AC) = AD 8, 7, 3 = 4, so the volume of the pyramid is 1 4 =

14 Section 11.4: Vectors, Curves, and Surfaces in Space Problems 1-4: Write parametric equations of the straight line L that passes through the point P = (1, 4, 3) and is parallel to the vector v = 6i+4j k. L is the set of vectors OP + tv = 1, 4, 3 + t 6, 4, 1 = 6t + 1, 4t 4, 3 t where t is any real number. Therefore the parametric equations for L are x = 6t + 1 y = 4t 4 z = 3 t. Problems 5-8: Write parametric equations of the straight line L that passes through the points P 1 and P. L is given parametricly in vector form by OP 1 + tp 1 P = 1, 0, 0 + t 6, 7, = 1 6t, 7t, t, so the parametric equations for L are x = 1 6t y = 7t z = t. Problems 9-14: Write both parametric and symmetric equations for the indicated straight line L. Through P = (, 3, 7) and Q = (, 6, 9) 14

15 L is the set of points OP + tpq = t +, 3t + 3, t + 7. Parametricly, L is the set of points satisfying x = t + y = 3t + 3 z = t + 7. Solving these for t gives the symmetric equations t = x = y 3 3 = z 7. Problems 15-0: Determine whether the line L 1 = x + 3 = y = z 5 is parallel to, skew to, or intersecting L. If they intersect, find the point of intersection. L = x 9 = y + 5 = z+4 L is parallel to the vector, 1, and L 1 is parallel to 3, 4,, so the two lines are not parallel. If L 1 and L intersect at a point (x 0,y 0,z 0 ), then we know x 0 = 3t 1 = t + 9 y 0 = 4t 1 6 = t 5 z 0 = t = t + 4. Adding the first and third of these, we see t 1 = 10. Substituting this in the expression for x 0, we see t = 19/. But these values contradict the other equations for y 0, so L 1 and L do not intersect. L = x 10 = y + 5 = z+5 15

16 Just as above, L is not parallel to L 1. If L 1 and L intersect in a point (x 0,y 0,z 0 ), then x 0 = t = 3t y 0 = t 1 5 = 4t 6 z 0 = t 1 5 = t + 5. Adding the first and third of these, we see t =. But then t 1 = 3. The points corresponding to these values of t are both (4,, 1), so L 1 and L intersect in the point (4,, 1). L = x 6 = y 7 = z Reading the denominators from the symmetric equations, we see that both L 1 and L are parallel to the vector 3, 4,, so L 1 and L are parallel. Problems 1-4: Write an equation of the plane P with normal vector n = 1, 1, 1 that passes through the point Q = (5, 3, 4). P is the set of points R = (x,y,z) such that QR n, or QR n = 0. That means that 0 = QR n = x 5,y + 3,z 4 1, 1, 1 = x + y + z 7. Therefore P is the set of points which is the desired equation. x + y + z = 7, Problems 5-3: Write an equation of the indicated plane P. Through Q = (1, 4, 3), R = (, 0, ), and S = (5, 3, 0) Let n = QR QS, so that n is normal to P. Then n = 3, 4, 1 4, 7, 3 = 5, 13, 37. Now, P is exactly the set of points T = (x,y,z) such that n QT = 0, or x 1,y + 4,z 3 5, 13, 37 = 0, 16

17 or 5(x 1) 13(y + 4) 37(z 3) = 0. Expanding this, we see our plane is given by the equation 5x + 13y + 37z = 74. Problems 33-34: Write an equation of the plane P that contains both the point Q = (, 4, 6) and the line L given by x = y 3 4 = z 5. L is parallel to vector 7, 4, 5 and contains the point R = ( 3, 3, ) (corresponding to t = 0). Therefore P contains the vectors QR = 5, 1, 4 and v = 7, 4, 5. Then the vector n = QR v is normal to P. Note that n = 11, 3, 13. Then P is the set of points S = (x,y,z) such that QS n = 0, or 11x + 3y 13z = 4. Problems 35-38: Determine whether the line L and the plane P intersect or ar parallel. If they intersect, find the point of intersection. P is x + 4y 3z = 1, L is { t,t, t : t R} Our strategy will be to find the normal vector to P and decide whether it is perpendicular to a vector parallel to L. Note that the points Q 1 = ( 1, 0, 0), Q = (1, 1, ), and Q 3 = (, 0, 1) all lie on P. Therefore the normal vector n to P is n = Q 1 Q Q 1 Q 3 =, 1, 3, 0, 1 = 1, 4, 3. Now, L is parallel to the vector v =, 1,. Since v n = 0, n is perpendicular to v, so v lies in P. We conclude that L is parallel to P. 17

18 P is x 4y + 13z = 10, L is { 6t + 1, t + 1, t + 1 : t R} The normal vector to P is n = 1, 4, 13. L is parallel to v = 6,,. Since n v = 1 0, v and n are not perpendicular, so v does not lie in P. Therefore L and P are not parallel. Now, let (x 0,y 0,z 0 ) be the point of intersection of L and P. Then we know that, for some t, x 0 = 6t + 1 y 0 = t + 1 z 0 = t + 1 x 0 4y z 0 = 10. Substituting the first three into the fourth, we see t = 0, which corresponds to the point (1, 1, 1). We can check that (1, 1, 1) is indeed a point of intersection for P and L. Problems 39-4: Find the angle between the planes P 1 given by x+y+3z = 13 and P given by x + 3y 4x = 9. The angle between two planes is the angle between their normal vectors. Let n 1 and n be the normal vectors to P 1 and P, respectively. Reading off coefficients, we know that n 1 = 1,, 3 and n =, 3, 4. We will find the angle α between n 1 and n using the dot product. Now 8 = n 1 n = n 1 n cos(α) = 406 cos(α), so cos(α) = , so α = cos 1 ( ) = o. Hence the intersection of P 1 and P forms a o angle. However, we always take the angle between planes to be at most 90 o. Therefore, we use the supplementary angle o as the angle between P 1 and P. 18

19 Section 11.5: Curves and Motion in Space Problems 5-10: Find the values of r (t) and r (t) for the given value of t. r(t) = e t i + sin(6t)j, t = π/ Since the derivative of v(t) = x(t)i + y(t)j is v (t) = x (t)i + y (t)j, we have that r (t) = (e t ) i + (sin(6t)) j = te t i + 6 cos(6t)j. Therefore, r (t) = (r (t)) = (te t ) i + (6 cos(6t)) j = (e t + 4t e t )i 36 sin(6t)j. Substituting in t = π/, we have r (π/) = πe π /4 i 6j and r (π/) = ( + π)e pi /4 i. Problems 11-16: Find the velocity and acceleration vectors and the speed at time t = π/ for the given position vector. r(t) = t, sin(t),e t Let v(t) and a(t) be the velocity and acceleration vectors, so that v(t) = r (t) and a(t) = r (t). Then and v(t) = 1, cos(t),e t a(t) = 0, sin(t),e t. Substituting in t = π/, we have v(π/) = 1, 0,e π/ and a(π/) = 0, 1,e π/. Also, the speed at t = π/ is v(π/) = e π = e π

20 Problems 17-0: Calculate the following integral. π/ 0 cos(t)i + (t 3 t )jdt Since we can integrate componentwise, ( π/ ) ( π/ ) π/ cos(t)i + (t 3 t )jdt = cos(t) dt i + t 3 t dt j 0 ( = sin(t) π/ 0 ) ( i + 0 t 4 /4 t 3 /3 π/ 0 0 ) ( π 4 j = i + 64 π3 4 ) j. Problems 1-4: Apply Theorem of Page 808 to compute the derivative D t [u(t) v(t)]. u(t) = 6ti + t j v(t) = e i + 3 t j Theorem tells us that D t [u(t) v(t)] = u (t) v(t) + u(t) v (t). Therefore, we have D t [u(t) v(t)] = (6i + tj) (e i + 3 t j) + (6ti + t j) (ln(3)3 t j) = (6e i + t 3 t j) + (ln(3)t 3 t j) = 6e i + (1 + ln 3)t 3 t j. Problems 5-34: For a particle moving in xyz-space, we know the acceleration vector a(t) = 14, 7t 7 e t, 16e 4t, the initial position and the initial velocity r 0 = r(0) = 0, 1, 0, v 0 = v(0) = 1, 1, 0 0

21 Find its position vector r(t) at time t = 4. The velocity vector v(t) = a(t) dt, and so v(t) = 14t + C 1, 9t 8 e t + C, 4e 4t + C 3. Using the initial velocity v(0), we can see that C 1 = 1 and C = C 3 = 0, so v(t) = 14t + 1, 9t 8 e t 4e 4t. Now, using that the position vector r(t) is v(t) dt, we have r(t) = 7t + t + D 1,t 9 e t + D,e 4t + D 3 for some constants D 1,D,D 3. Using what we know about r(0), it is easy to see that D 1 = D = D 3 = 0. Hence r(t) = 7t + t,t 9 e t,e 4t. Problem 36: Use the equations of Theorem to calculate D t [u(t) v(t)] and D t [u(t) v(t)] if u(t) = t,t,t 3 and v(t) = t, cos(t) sin(t), sin(t). Theorem tells us that D t [u(t) v(t)] = u (t) v(t) + u(t) v (t) = 1, t, 3t t, cos t sin t, sin t + t,t,t 3 ln() t, cos t sin t, cos t = t, t cos t sin t, 3t sin t + ln()t t,t (cos t sin t),t 3 cos t = (1 + ln()t) t, t cos t sin t + t (cos t sin t), 3t sin t + t 3 cost. Also, D t [u(t) v(t)] = u (t) v(t) + u(t) u (t) = ( 1, t, 3t t, cos t sin t, sin t ) + ( t,t,t 3 ln t, cos t sin t, cos t ) = t sin t 3t t, 3t t sin t, cos t sin t t t + t cost t 3 (cos t sin t), ln()t t t cos t,t(cos t sin t) ln()t t. 1

22 Problem 43: Let a projectile fired at an inclination of 30 o have range (horizontal distance traveled) 100 meters. What is the initial velocity v 0 of the projectile? We will consider this problem in only two dimensions, x (horizontal) and y (vertical). We know the acceleration vector of the projectile is 0, 9.8. Therefore the velocity vector v(t) = C 1, 9.8t+C for constants C 1 and C by integration. But since v(0) = v 0 cos(30 o ),v 0 sin(30 o ), C 1 = v 0 cos(30 o ) and C = v 0 sin(30 o ). Integrating the velocity vector to get the position vector r(t), we see r(t) = v 0 cos(30 o )t + D 1,v 0 sin(30 o )t 4.9t + D. If we call the initial position (0, 0), then we can set D 1 = D = 0. We know that, if t is the time when the projectile lands, that r(t ) = 100, 0. Solving this in the y-coordinate, we have v 0 sin(30 o )t 4.9t = 0, so t = v 0 sin(30 o ) 4.9. But then, looking at the x-coordinate, 100 = v 0 cos(30 o )t = v 0 cos(30 o ) sin(30 o ), 4.9 so we have that v 0 = meters per second. Problem 44: What is the maximum height achieved by the projectile of the last example? As we have already seen, the height of the projectile at time t is v 0 sin(30 o )t 4.9t = t 4.9t. We know how to maximize this function; it is maximized when t = = At this time, the height is meters.

23 Section 11.6: Curvature and Acceleration Page 830 Problems 1-6: Find the arc length of the given curve. x = 3 t, y = t 3, z = t from t = 0 to t = 5 The arc length is the integral 5 5 x + y + z dt = 9t + 9t dt 0 = 5 0 3t + 1 dt = (t 3 + t) = 130. x = t sin t, y = t cos t, z = 3 t3/ from t = 0 to t = 1 The arc length is = (t cos t + sint) + ( t sin t + cos t) + ( t) dt 1 ( ) t 1 t t dt = t + 1 dt = + t = Problems 7-1: Find the curvature of the given plane curve at the indicated point. y = x sin x at (π, 0) Recall that κ = cos t, we have κ = y (1+y ) 3/. Since y = x cos x + sinx and y = t sin t + cos t t sin t (1 + t cos t + sin t + t cos t sin t) 3/. 3

24 At the point (π, 0), κ = (1+π ) 3/. y = x 4 at (3, 81) Using the formula above and that y = 4x 3 and y = 1x, we have which is κ = at the indicated point. 1x (1 + 16x 6 ) 3/, Problems 13-16: Find the point or points on the curve y = e x at which the curvature is maximized. Again, since y = y = e x, the formula gives that κ = e x (1 + e x ) 3/. To find when κ is maximized, we differentiate: κ = ex (1 + e x ) 3/ 3e x 1 + e x (1 + e x) 3. Now, κ = 0 when the numerator is 0. Dividing the numerator through by e x 1 + e x, we see the numerator is 0 when 1 = e x, or when x = ln. This gives y = (. Therefore the curvature is maximized at ln, ). Problems 17-1: Find the unit tangent and normal vectors to y = x 4 at the point (1, 1). We can parametrize this function as x = t, y = t 4 so that our function is the set of points t,t 4. Differentiating with respect to t, we see the tangent vector to this function is 1, 4t 3. Taking the unit vector in this direction, we see T = 1, 4t3 16t Now, by inspection, the vector n = 4t 3, 1 is normal to the curve. Since y = 1x > 0, we know the function is concave up, so N = ˆn = 4t3, 1 16t

25 Evaluating each of these at the point (1, 1), we have T(1) = , 4 and N = , 1. Problems -6: The position vector of a particle moving in the plane is given by r(t) = 3 sin(πt)i + 3 cos(πt)j. Find the tangential and normal components of the acceleration vector. Equation on Page 84 tells us that a = dv dt T + κv N, where the first term on the right is the tangential component and the second term is the normal component of acceleration. Now, T = 1 3π 3π cos(πt), 3 sin(πt). Also, we see v(t) = 3π, so dv = 0. Hence the tangential component of acceleration is 0. dt For the normal component, we notice that since there is no tangential component of acceleration, the normal component is simply a(t) = π r(t). Problems 9-31: Find the equation of the osculating circle for the plane curve y = 1 x at the point (0, 1). To determine the osculating circle, we need to know the normal vector N and the curvature κ. Now κ(x) = y (1 + y ) = 3/ (1 + 4x) 3/, so κ(0) =. Therefore the radius of curvature is 1/. Now the vector 1, 0 is tangent to the curve at this given point by differentiating, and since the curve is concave down, a normal vector at the point is 0, 1. Hence the center of the osculating circle is 0, , 1 = 0, 1/. Combining all this, we get that the equation of the osculating circle is x + (y 1/) = 1/4. 5

26 Problems 3-41: Find the curvature κ and the tangential and normal components of acceleration of the space curve with position vector r(t) = t,t, 3 t3. First note the velocity and acceleration vectors are v(t) = 1, t, t a(t) = 0,, 4t. Now the tangential component of acceleration is compva = a v v = 0 + 4t + 8t t + 4t 4 = 4t. The normal component of acceleration is a(t) projva = a(t) (compva)v = 0,, 4t 4t 1, t, t = 4t, 8t, 4t 8t 3. Problems 4-45: Find the unit vectors T and N for the curve sin(t), cos(t),t at the point (0, 1, π). Taking derivatives, the tangent vector to the curve is given by t(t) = cos(t), sin(t), 1 which has length. Hence the unit tangent vector is T(t) = 1 cos(t), sin(t), 1, which is 0, 1, 1 at our point (which corresponds to t = π). 6

27 Taking a second derivative, the acceleration vector is given by The vector a(t) = sin(t), cos(t), 0. orth T a = a a T T = sin(t), cos(t), 0 0T = a T is normal to the curve. Hence the unit normal vector is which equals 0, 1, 0 at our point. N = â = a = sin(t), cos(t), 0, 7

28 Section 11.7: Cylinders and Quadric Surfaces Page 839 Problems 1-30: Describe the graphs of the given equations. x + z = 13 This is a cylinder extending infinitely far in either direction with radius 13 centered around the y-axis. x + 4y = z For any fixed nonnegative value of z, say z 0, the equation x + 4y = z 0 describes an ellipse in the plane z = z 0. Letting z grow, this ellipse gets larger. This is an elliptical paraboloid. 3x 4y + 8z = π This is a plane. y = cos(x) In two dimensions, this is something like a wavy line. Since there is no restriction on z, this equation represents a wavy plane in three dimensions. Problems 31-40: The equation of a curve in one of the coordinate planes is given. Write an equation for the surface generated by revolving this curve around the indicated axis. x = z x-axis A point (x,y,z) lies on this surface of revolution if and only if its distance to the x-axis, namely y + z, is equal to the z-coordinate of the 8

29 plane curve. Therefore the equation of the surface is gotten by replacing z in the equation of the plane curve with y + z. This gives us x = y + z, or x = y + z, as the equation of the surface. This is a circular paraboloid. y = sin(x) x-axis A point (x,y,z) lies on the surface of revolution if and only if its distance to the x-axis, namely y + z, equals the y-coordinate of the plane curve. Therefore the equation of the surface is gotten by replacing y with y + z in the equation of the curve. This gives us y + z = sin(x) as the equation of the surface. Problems 41-48: Describe the traces of the given surfaces in planes of the indicated type. x + 3z = 8 in planes parallel to any coordinate plane In planes parallel to the xy-plane, z is fixed, say z 0. Then the equation is x = 8 3z 0, which describes two parallel lines. Similarly, in planes parallel to the yz-plane, the trace with be two parallel lines. In planes parallel to the xz-plane, y is fixed but irrelevant; the equation x + 3z = 8 describes an ellipse in that plane. x = yz in planes parallel to the yz-plane and to the xy-plane In planes parallel to the yz-plane, x is some fixed value x 0, and the equation yz = x 0 describes a hyperbola in that plane. In planes parallel to the xy-plane, z is some fixed value z 0, so the equation x = z 0 y describes a line through the origin with slope z 0. x = sin (y) (y 1) + z 1 in the xz-plane In the xz-plane, y = 0, so the equation reduces to x = z or x = z, which is a V shape. 9

30 Section 11.8: Cylinders and Spherical Surfaces Page 845 Problems 1-6: Find the rectangular coordinates of the point with the given cylindrical coordinates. (1,π/, 3) We know x = r cos(θ) = cos(π/) = 0 and y = r sin(θ) = 1. Therefore this point is (0, 1, 3) in rectangular form. (,π/4, 1) Again, x = r cos(θ) = = and y = r sin(θ) =. Hence this point is (,, 1) in rectangular form. Problems 7-1: Find the rectangular coordinates of the points with the given spherical coordinates. (3,π/, 0) We know x = ρ sin(φ) cos(θ) = 3 sin(π/) cos(0) = 3 y = ρ sin(φ) sin(θ) = 3 sin(π/) sin(0) = 0 z = ρ cos(θ) = 3 cos(π/) = 0, so this point is (3, 0, 0) in rectangular form. (6, π/3, π/4) Again, 3 x = ρ sin(φ) cos(θ) = 6 sin(π/3) cos(π/4) = 6 =

31 3 y = ρ sin(φ) sin(θ) = 6 sin(π/3) sin(π/4) = 6 = 3 6 z = ρ cos(θ) = 6 cos(π/3) = 6/ = 3, so this point is ( 3 6, 3 6, 3) in rectangular form. Problems 13-: Find both the cylindrical coordinates and the spherical coordinates of the point P with the given rectangular coordinates. (3, 4, 5) First we will find the cylindrical coordinates. We know r = = 5. Since 3 = x = r cos(θ), we have cos(θ) = 3/5. Since θ lies in the first quadrant, θ.97. Hence this point is (5,.97, 5) in cylindrical form. Now we find the spherical coordinates. We know ρ = = 5. Since z = ρ cos(φ), we have cos(φ) = 1, so φ = π/4. Then 3 = x = ρ sin(φ) cos(θ) = 5 cos(θ). Since θ lies in the first quadrant, θ.97. Therefore the spherical coordinates are (5, π,.97). 4 (,, 1) We again start with the cylindrical coordinates. We have r = ( ) + =. Since = x = r cos(θ) and = y = r sin(θ), we see that θ = 3π 4. Therefore the cylindrical coordinates are (, 3π 4, 1). Now, for the spherical coordinates, note that ρ = ( ) = 3. Then 1 = z = 3 cos(φ), so φ Then = x = 3 sin(φ) cos(θ) = cos(θ) = y = 3 sin(φ) sin(θ) = sin(θ), so θ = 3π. Therefore this point is (3, 1.31, 3π ) in spherical form. 4 4 Problems 3-38: Describe the graph of the given equation. r = 6 In any plane z = z 0, this is a circle of radius 6 centered at the origin. Hence the whole curve is a cylinder of radius 6 centered about the z-axis. 31

32 ρ = 6 Since ρ is the distance of a point from the origin, this is a sphere of radius 6 centered at the origin. ρ = 3 cos(φ) Multiplying by ρ, we see ρ = 3ρ cos(φ) = 3z. Since we know that ρ = x + y + z, we equate to get or x + y + z = 3z x + y + ( z 3 ) = 9 4. Hence the equation is a sphere centered at (0, 0, 3/) with radius 3/. r 5r Solving this quadratic tells us r 3. Since, for any fixed r 0, the equation r = r 0 is a cylinder of radius r 0 centered at the z-axis, our equation is a thick pipe, or, if you d rather, a hollowed-out Combo. Problems 39-44: Convert the given equation both to cylindrical and to spherical coordinates. x + y + z = 16 Using that x = r cos(θ) and y = r sin(θ), the equation becomes 16 = r cos (θ) + r sin (θ) + z = r + z, so the equation is r + z = 16 in cylindrical form. For spherical form, we plug in the formulas for x, y, and z: 16 = ρ sin (φ) cos (θ) + ρ sin (φ) sin (θ) + ρ cos (φ) = ρ sin (φ) + ρ cos (φ) = ρ, 3

33 so the equation is ρ = 16. z = x y For cylindrical form, we have so the equation becomes z = r cos (θ) r sin (θ), z r = cos (θ) sin (theta). In spherical form, we again plug in the equations for x, y, and z to get ρ cos(φ) = ρ sin (φ) cos (θ) ρ sin (φ) sin (φ). Dividing through by ρ sin (φ), the equation of our curve becomes cos(φ) ρ sin (φ) = cos (θ) sin (θ). Problems 45-5: Describe the surface or solid described by the given equations and/or inequalities. 0 φ π/4 ρ 5 This is an equation in spherical form. Since ρ measures the distance of a point from the origin, all points in this solid have distance at most 5 from the origin. Furthermore, since φ π/4, we have z = ρ cos(φ) 5 cos(π/4) = 5. Hence this solid is all points (x,y,z) whose distance from the origin is at most 5 with z 5. This is a ball with the top and bottom caps cut off. r = 6 4 z 5 This equation is in cylindrical form. Now r = 6 describes a circle in the xy-plane of radius 6 centered at the origin. The condition on z tells us this equation is a cylinder (with no bases) of height 1 with radius 6 centered at the z-axis. Its shape is like a pipe. 33