Math 13 Fall 2008 Final Exam

Math 13 Fall 2008: Final Exam Instructions: There are 10 questions on this exam of which you must do 8. Each problem is scored out of 8 points for a total of 64 points. You may not use any outside materials (eg. notes or calculators). You have 3 hours to complete this exam. Remember to fully justify your answers. Mark clearly in your blue book which problems are to be graded. ­ ® Problem 1. A particle moves through space with position r(t) = t, t2 , 34 t3/2 . ¡ ¢ (1) Find the symmetric and equations for the tangent line at 1, 1, 43 . ¡ parametric ¢ (2) Find the curvature at 1, 1, 43 . Proof. ­ ® (1) We have T is in the direction of r0 (t) = 1, 2t, 2t1/2 . At t = 1 this is h1, 2, 2i so we have the line ¿ À 4 r(t) = 1 + t, 1 + 2t, + 2t 3 z− y−1 = 2 2 (2) We know the curvature is given by x−1= κ= So we compute 4 3 |r0 × r00 | |r0 |3 D E r0 (t) = 1, 2t, 2t1/2 D E r00 (t) = 0, 2, t−1/2 At t = 1 we have r0 (1) = h1, 2, 2i r00 (1) = h0, 2, 1i ¯ 0 ¯ ¯r (t)¯ = 3. Then we compute the cross product to get h1, 2, 2i × h0, 2, 1i = h−2, −1, 2i So we have κ= 3 1 = . 27 9 Problem 2. For the two planes S1 : 4x + y − z = 4 and S2 : 2x + 2y + z = 3. (1) Find the line of intersection of S1 and S2 . (2) Find the angle between S1 and S2 . Proof. 1 ¤ 2 (1) The direction of the line of intersection is given by the cross product h4, 2, −1i × h2, 2, 1i = h3, −6, 6i . We can also write this direction as h1, −2, 2i. A point of intersection is (0, 73 , 53 ). So we have the line ¿ À 7 5 r(t) = t, − 2t, + 2t . 3 3 (2) The angle is given by 8+2−1 9 1 cos θ = √ √ = √ = √ 18 9 9 2 2 so we have θ = π4 . ¤ Problem 3. Determine the following two limits (if they exist). 5x2 y . x2 +y 2 5x2 y lim(x,y)→(0,0) x4 +y2 . (1) lim(x,y)→(0,0) (2) Proof. (1) We have ¯ ¯ ¯ 5x2 y ¯ ¯ ¯ ¯ x2 + y 2 ¯ ≤ |5y| and hence 2 y −5 |y| ≤ x5x 5 |y| 2 +y 2 ≤ ↓y→0 ↓ ↓y→0 5x2 y 0 ≤ lim(x,y)→(0,0) x4 +y2 ≤ 0 so by the squeeze theorem the limit exists and takes the values 0. (2) We take the path x = 0 to get lim (0,y)→(0,0) 0 =0 0 + y2 and the path y = x2 to get 5x4 5 = 4 2 2x 2 (x,x )→(0,0) lim Since these two paths give different limiting values, the limit does not exist. ¤ Problem 4. The plane 8x − 8y − 2z = C is tangent to the surface z = x2 + 2y 2 at a certain point (x0 , y0 , z0 ). Find (x0 , y0 , z0 ) and the constant C. 3 Proof. The tangent plane to the surface is given by h2x, 4y, −1i so we must have 2x0 (x − x0 ) + 4y0 (y − y0 ) − (z − z0 ) = 0 The normal to the plane is given by h8, −8, −2i which is parallel to h4, −4, −1i so we must have x0 = 2 and y0 = −1. Finding z0 from the equation for the surface we z0 = 6. We find C using the fact that (x0 , y0 , z0 ) must also be on the plane, so we get C = 8(2) − 8(−1) − 2(6) = 16 + 8 − 12 = 12. So we have C = 12 and (x0 , y0 , z0 ) = (2, 1, 4). ¤ Problem 5. Classify the critical points for f (x, y) = x4 + y 4 − 4xy + 1. Proof. We find the critical points by solving fx = 4x3 − 4y = 0 fy = 4y 3 − 4x = 0. Which is the same as x3 = y y3 = x Which has the three solutions (1, 1), (−1, −1), and (0, 0). Using the second derivative test we compute fxx = 12x2 fxy = −4 fyy = 12y 2 So we have discriminants D(1, 1) = 144 − 16 > 0 D(−1, −1) = 144 − 16 > 0 D(0, 0) = 0 − 16 < 0. So we have (0, 0) is a saddle point. For the other two points was have fxx > 0 so we have a local minimums at (1, 1) and (−1, −1). ¤ Problem 6. Find the point(s) on the surface 2x2 +3y 2 +6z 2 = 36 the sum of whose coordinates is maximal. Proof. This is a Lagrange Multiplier problem with function f (x, y, z) = x + y + z and constraint x2 + y 2 + z 2 − 36 = 0. So we have the system 1 = λ4x 1 = λ6y 1 = λ12z 36 = 2x2 + 3y 2 + 6z 2 . 4 Since λ cannot be 0 and still satisfy the equations we must have 4x = 6y = 12z. Hence we have 1 and z = y. 2 3 x= y 2 Now using the constraint equation we have 9 3 36 = y 2 + 3y 2 + y 2 2 2 Which is 72 = 18y 2 and hence y = ±2. So we have the two point (3, 2, 1) and (−3, −2, −1). So the maximum occurs for (3, 2, 1). ¤ Problem 7. Find the moment of inertia about the x-axis for the triangle bounded by 3x+4y = 24, x = 0, and y = 0 and having constant density 1. Proof. Z Ix = 0 Z Ix = 0 8 Z 6−(3/4)x 0 6 Z 8−(4/3)y Z 8 y 2 dydx = − Z 0 6 y 2 dxdy = 0 0 9 (−8 + x)3 dx = 64 Z 8 72 − 27x + 0 27x2 9x3 − dx = 144 8 64 8 y4 4 8y 2 − y 3 dy = ( y 3 − )60 = 72(8) − 36(12) = 72(2) = 144 3 3 3 ¤ Problem 8. Find the volume bounded by y = x2 + 2z 2 and y = 4 − x2 . Proof. The intersection is along x2 + z 2 = 2 so we have in (alternative) polar coordinates Z 2π √ 2 Z V = 0 0 1 √ 4r − 2r3 drdθ = 2π(2r2 − r4 )0 2 = 4π. 2 ¤ Problem 9. Evaluate Z 1 −1 Z √ 1−x2 √ − 1−x2 Z 2−x2 −y 2 x2 +y 2 (x2 + y 2 )3/2 dzdydx. 5 Proof. Changing to Cylindrical we have Z 1 −1 Z √ 1−x2 √ − 1−x2 Z Z 2−x2 −y 2 2 2 3/2 (x + y ) 2π Z 1 Z 2−r2 dzdydx = x2 +y 2 0 Z 2π Z 1 = 0 Z r4 (2 − r2 − r2 )drdθ 0 2π Z = 0 Z r3 rdzdrdθ r2 0 1 2r4 − 2r6 drdθ 0 2π = 0 2 2 ( − )dθ 5 7 8π = . 35 ¤ Problem 10. A closed path C is traversed counter-clockwise beginning at (−1, −1) and moving to (2, 2) along the curve y = x2 − 2 then returning to the point (−1, −1) along the curve y = x. Find Z x2 −xydx + dy 2 C in two ways. (1) Parameterizing the curves and computing the line integral directly. (2) Using Green’s Theorem. Proof. ­ ® (1) We parameterize C1 : t, t2 − 2 , −1 ≤ t ≤ 2 and C2 : h2 − t, 2 − ti , 0 ≤ t ≤ 3. So we have Z Z 2 Z 3 t2 (2 − t)2 x2 2 −t(t − 2) + (2t)dt + −(2 − t)2 (−1) + (−1)dt −xydx + dy = 2 2 2 −1 0 C Z 2 Z 3 3 3 = −t + 2t + t dt + 4 − 4t + t2 − (2 − 2t + t2 /2)dt −1 Z 0 3 t2 2 − 2t + dt 2 0 · ¸3 3 t = 3 + 2t − t2 + 6 0 9 = 3 + (6 − 9 + ) 2 9 = . 2 £ ¤2 = t2 −1 + 6 (2) Using Green’s Theorem we have Z Z 2Z x x2 −xydx + dy = x − (−x)dydx 2 −1 x2 −2 C Z 2Z x = 2xdydx −1 2 Z = −1 2 x2 −2 2x(x − x2 + 2)dx Z = −2x3 + 2x2 + 4xdx −1 · 4 ¸2 2x3 x 2 + 2x = − + 2 3 −1 16 1 2 = (−8 + + 8) − (− − + 2) 3 2 3 16 2 3 = + − 3 3 2 3 9 =6− = . 2 2 ¤