Math 13 Fall 2008 Exam 2

Math 13 Fall 2008: Exam 2 Name: Instructions: There are 5 questions on this exam of which you must do 4. Each problem is scored out of 8 points for a total of 32 points. You may not use any outside materials(eg. notes or calculators). You have 50 minutes to complete this exam. Remember to fully justify your answers. Score: Circle below the 4 problems you wish to be graded. Otherwise, I will grade the first 4 completed problems 1 2 3 4 5 1 Problem 1. Determine the following limits or show that they do not exist. (a) x4 y 4 (x,y)→(0,0) (x2 + y 4 )3 lim (b) 2x4 y + y4 lim (x,y)→(0,0) x4 Proof. (a) Considering the path y = 0 we have x4 y 4 0 = lim = 0. (x,y)→(0,0) (x2 + y 4 )3 (x,0)→(0,0) x2 lim Considering the path x = y 2 y 12 1 y 12 = lim = 4 3 y→0 8y 12 8 (y 2 ,y)→(0,0) (2y ) lim Since the limit along two different paths to (0, 0) do not agree, then the limit does not exist. (b) We will show that this limit is 0. Method 1: using the definition of limits: Using¯ the ²−δ ¯ definition of limits we choose ² > 0 and need to find a δ > 0 such that if ¯ 2x4 y ¯ then ¯ x4 +y4 ¯ < ². So we consider ¯ ¯ ¯ 2x4 y ¯ ¯ ¯ ¯ x4 + y 4 ¯ < ². p x2 + y 2 < δ Then since x4 and y 4 are non-negative we have x4 and so x4 ≤1 + y4 ¯ ¯ p p ¯ 2x4 y ¯ 2 2 2 ¯ ¯ ¯ x4 + y 4 ¯ ≤ 2 |y| = 2 y ≤ 2 x + y So we may choose δ = 2² . Alternatively, we could use the Squeeze Theorem to have −2 |y| ≤ 2x4 y ≤ 2 |y| . + y4 x4 Since −2 |y| converges to 0 and 2 |y| converges to 0 as (x, y) → (0, 0) we have that converge to 0. 2 2x4 y x4 +y 4 must also Method 2: using the Squeeze Theorem: We have ¯ ¯ ¯ 2x4 y ¯ ¯ ¯ ¯ x4 + y 4 ¯ ≤ |2y| so we have −2 |y| ≤ 2x4 y ≤ 2 |y| . x4 + y 4 We have lim (x,y)→(0,0) −2 |y| = lim −2 |y| = 0 lim (x,y)→(0,0) y→0 2 |y| = lim 2 |y| = 0 y→0 so by the squeeze theorem we have 2x4 y = 0. + y4 lim (x,y)→(0,0) x4 3 Problem 2. Let w = f (x, y, z) = x3 + y 3 − z. (a) Find the rate of change of f at the point (1, 1, 2) along the line of decreasing x. (b) Find ∂w ∂t x−1 3 = y−1 2 = 2−z 2 in the direction for x(s, t) = sin(st) y(s, t) = ln(st) z(s, t) = set . Proof. (a) We have that the line is in the direction h3, 2, −2i and so we have u = ­ ® have ∇f = 3x2 , 3y 2 , −1 and so we have √1 17 h−3, −2, 2i. We also √ 1 −9 − 6 − 2 √ Du f (1, 1, 2) = h3, 3, −1i · √ h−3, −2, 2i = = − 17. 17 17 (b) We have ∂w ∂f ∂x ∂f ∂y ∂f ∂z = + + ∂t ∂x ∂t ∂y ∂t ∂z ∂t s 2 = (3x )(s cos(st)) + 3y 2 − set st 3(ln(st))2 2 = 3s sin (st) cos(st) + − set . t 4 Problem 3. Classify the critical points of f (x, y) = x4 + y 3 − 32x − 27y − 1. Proof. We have fx = 4x3 − 32 = 0 fy = 3y 2 − 27 = 0 so we have x3 = 8 y2 = 9 Which gives the points (2, 3), (2, −3). Taking the second derivatives we have fxx = 12x2 fyy = 6y fxy = 0. So we have D(2, 3) = 24 · 18 > 0 D(2, −3) = 24 · (−18) < 0 Since D > 0 and fxx > 0 we have (2, 3) is a relative min. Since D < 0 we have (2, −3) is a saddle point. 5 Problem 4. Find the maximum and minimum for 3x2 + 4xy for points on the disk x2 + y 2 ≤ 14. Proof. First we find the critical points. So we have 6x + 4y = 0 and 4x = 0 so we have x = 0 and hence y = 0 which is the critical point (0, 0). This point is within the disk. Using Lagrange Multipliers we have 6x + 4y = λ2x 4x = λ2y x2 + y 2 = 14 Method 1: solving first √ for λ: If y = 0 we have x = ± 14 by the constraint equation, but x = 0 by the second equation, so y = 0 is not possible. So we have, λ = 2x y and y(6x + 4y) = 2x(2x) x2 + y 2 = 14. Which is 6xy + 4y 2 = 4x2 x2 + y 2 = 14. and the first equation factors as 2(y + 2x)(2y − x) = 0, in other words, x = 2y or y = −2x. If x = 2y we have 5y 2 = 14 q p p p p and hence y = ± 14 14/15), (−2 14/5, − 14/5). 5 , which gives the two points (2 14/5, If y = −2x then we have 5x2 = 14 q p p p p and hence x = ± 14 5 , which gives the two points (− 14/5, 2 14/5), ( 14/5, −2 14/5). Method 2: solving first for x: We have x = λy/2 and so 3λy + 4y = λ2 y λ2 y 2 /4 + y 2 = 14. We can factor the first equation y(λ − 4)(λ + 1) = 0 This gives the three solutions y = 0, λ = 4, λ = −1. If y = 0, then the second equation cannot be satisfied, so that does not yeidl any solutiosn. If λ = 4 we get 5y 2 = 14 6 and so y = ± then we have p p p p p 14/5, which gives the two points (2 14/5, 14/15), (−2 14/5, − 14/5). If λ = −1 5y 2 /4 = 14 p p p p p and hence y = 2 14/5, which gives the two points (− 14/5, 2 14/5), ( 14/5, −2 14/5). So we have f (0, 0) = 0 r ! 14 14 , −2 = −14 f 5 5 à r r ! 14 14 f − ,2 = −14 5 5 à r r ! 14 14 f 2 , = 56 5 5 à r r ! 14 14 f −2 = 56 ,− 5 5 Ãr So we have à r r ! r ! 14 14 14 14 Min at , −2 and − ,2 is − 14 5 5 5 5 à r à r r ! r ! 14 14 14 14 Max at 2 , and −2 ,− is 56 5 5 5 5 Ãr 7 Problem 5. Let a < 0. Consider the surface xyz = a. (a) Find the tangent plane to the surface at (x0 , y0 , z0 ). (b) The tangent plane forms a tetrahedron (pyramid with a triangular base) in the first octant. Show that the volume of this tetrahedron is 92 a. (Hint: the volume of a tetrahedron is given by 13 (area of the base triangle)(height).) Proof. (a) We have ∇f = hyx, xz, xyi and so ∇f (x0 , yo , z0 ) = hy0 x0 , x0 z0 , x0 y0 i and the tangent plane is given by y0 z0 (x − x0 ) + x0 z0 (y − y0 ) + x0 y0 (z − z0 ) = 0. which is y0 z0 x + x0 z0 y + x0 y0 z = 3a (b) We know that (0, 0, 0) is a vertex of the tetrahedron. To find the other three, we set two fo x, y, z equal to zero and solve for the third. So for the point on the x-axis we get y0 z0 x = 3a = 3x0 y0 z0 which is x = 3x0 . Similarly for the y-axis and z-axis. So the points are the coordinates axes are given by (0, 0, 0), (0, 0, 3z0 ), (0, 3y0 , 0), (3x0 , 0, 0). So we have the volume as µ ¶ 1 1 27x0 y0 z0 9 (3x0 )(3y0 ) (3z0 ) = = a. 3 2 6 2 8