Multivariable Calculus, Midterm III Practice Exam 2 Solutions

Math 211, Multivariable Calculus, Fall 2011 Midterm III Practice Exam 2 Solutions 1. (a) What does it mean to say that (a, b) is a saddle point of the function f (x, y)? (b) Find the critical points of the function f (x, y) = x3 − xy + y 2 . (c) For each critical point, decide if it is a local maximum, local minimum or saddle point. (a) It means that (a, b) is a strict local maximum for f when restricted to one crosssection through (a, b) and a strict local minimum for f when restricted to another cross-section through (a, b). (b) The critical points occur when ∂f ∂x = 0 and 3x2 − y = 0, ∂f ∂y = 0, that is −x + 2y = 0. Therefore x = 2y and so 3(2y)2 − y = 0 and so y(12y − 1) = 0. This implies that y = 0 or y = 1 and ( 16 , 12 ). 1 . 12 Therefore, the two critical points are (0, 0) (c) We use the Second Derivative Test. We have 2 D = fxx fyy − fxy = (6x)(2) − (−1)2 = 12x − 1. 1 1 ) = 2, fxx ( 16 , 12 )> Therefore D(0, 0) = −1, so (0, 0) is a saddle point, and D( 61 , 12 1 0, so ( 16 , 12 ) is a local minimum. 2. Find the absolute maximum of the function f (x, y) = x − y 2 on the region x2 + y 2 ≤ 1. (Make sure you explain how you know that your answer is the absolute maximum.) We know there is an absolute maximum by the Extreme Value Theorem because the region x2 + y 2 ≤ 1 is closed and bounded. The possible points are: • points where ∇f = 0 (but there are no such points because ∇f = h1, −2yi) • points where f is not differentiable (none) 1 • points on the boundary x2 + y 2 = 1 To find the maximum value of f on the boundary we can use the Lagrange Multiplier Method with g(x) = x2 + y 2 . The possibilities are • points on the boundary where ∇f = λ∇g for some λ • points on the boundary where ∇g = 0 (there are none of these because ∇g = h2x, 2yi so is only zero at (0, 0) which is not on the boundary) • points where g is not differentiable (none) The possible points then are where −2y = 2yλ. 1 = 2xλ, The first equation tells us that x 6= 0 so, λ = −2y = 1 2x and so y x or y(1 + 2x) = 0. Therefore, either y = 0, in which case x = ±1 (from x2 + y 2 = 1) or x = − 12 , in which √ case y = ± 23 . So there are four candidates: √ √ (1, 0), (−1, 0), (−1/2, 3/2), (−1/2, − 3/2). It’s easy to see that f (x, y) is negative when x is negative, so the absolute maximum must occur at (1, 0) with f (1, 0) = 1. 3. Let R be the triangular region with vertices (0, 0), (0, 1), (2, 2). Calculate ZZ xy dA. R We can express this region as 1 x ≤ y ≤ x + 1. 2 0 ≤ x ≤ 2, The integral is therefore Z x=2 Z y= 1 x+1 Z 2 x=2 xy dy dx = x=0 y=x x=0 x=2 Z = Zx=0 x=2  y= 1 x+1 x y 2 /2 y=x2 dx x ((x/2 + 1)2 − x2 ) dx 2 (x/2 + x2 /2 − 3x3 /8) dx x=0  2 x=2 = x /4 + x3 /6 − 3x4 /32 x=0 = = 1 + 8/6 − 3/2 = 5/6 2 4. Let D be the solid cylinder whose ends are given by the planes z = −1 and z = 2, and whose curved surface is given by x2 + y 2 = 1. Calculate ZZZ z dV. D The region D can be expressed in cylindrical coordinates as −1 ≤ z ≤ 2, The integral is therefore Z z=2 Z θ=2π Z 0 ≤ θ ≤ 2π, r=1 Z 0 ≤ r ≤ 1. z=2 θ=0 θ=2π z zr dr dθ dz = z=−1 Z θ=0 z=−1 z=2 r=0 1 dθ dz 2 Z = πz dz = z=−1 π[z 2 /2]z=2 z=−1 = 3π 2 5. The hyperbolic coordinates of a point in the xy-plane are the variables s, t given (when x > 0) by y s = xy, t = . x (a) Find the Jacobian ∂(x,y) ∂(s,t) for the change of variables from x, y to s, t. (b) Let R be region in the xy-plane bounded by the lines y = x and y = 2x, and the curves y = x1 and y = x2 . Sketch a diagram of the region R. (c) Use hyperbolic coordinates to calculate the integral ZZ xy dA. R (a) (This question should have specified to assume that x, y > 0 to avoid problems when y = 0.) We first need to find expressions for x and y in terms of s and t. The second equation gives y = xt so then the first gives s = x2 t and so r x= s . t (We take the positive square-root because we are told in the question that these coordinates only work for the region where x > 0.) 3 We then get y = xt = √ st. We then have ∂(x, y) ∂x ∂y ∂x ∂y = ( )( ) − ( )( ) ∂(s, t) ∂s ∂t ∂t ∂s √ √ √ 1 − s t s √ − √ = √ 2 st 2 t 2t3/2 2 s 1 1 = + 4t 4t 1 = 2t (b) The picture looks like: (c) The four boundary curves become, respectively, t = 1, t = 2, s = 1, s = 2, so the integral is Z s=2 Z s=2 Z t=2 s s dt ds = [ln t]t=2 t=1 ds s=1 t=1 2t s=1 2 Z s=2 s = ln 2 ds s=1 2  s=2 = ln 2 s2 /4 s=1 3 ln 2 = 4 4