Math 211, Multivariable Calculus, Fall 2011 Midterm III Solutions

Math 211, Multivariable Calculus, Fall 2011 Midterm III Solutions 1. Use the Lagrange multiplier method to find the absolute maximum and absolute minimum of the function f (x, y) = x + y subject to the constraint x2 + 2y 2 = 1. (You must use the Lagrange multiplier method to get credit for this question.) The region satisfying the constraint is closed and bounded, and the function f is continuous, so the Extreme Value Theorem implies that there is an absolute maximum and minimum. We apply the Lagrange multiplier method with g(x, y) = x2 + 2y 2 . Since f and g are differentiable everywhere, we just need to check for points where ∇f = λ∇g or ∇g = 0, where ∇f = h1, 1i , ∇g = h2x, 4yi . The only point where ∇g = 0 is (0, 0) which does not satisfy the constraint. The equation ∇f = λ∇g gives 1 = 2xλ, 1 = 4yλ. This tells us that 2y = 4xyλ = x. Substituting into the constraint we have (2y)2 + 2y 2 = 1 so 6y 2 = 1 and hence √ y = ±1/ 6. Since x = 2y, the two constrained critical points are     −2 −1 2 1 √ ,√ , √ ,√ . 6 6 6 6 We now evaluate f at each of these:   2 1 3 f √ ,√ =√ 6 6 6 which is the absolute maximum, and   −2 −1 −3 f √ ,√ =√ 6 6 6 which is the absolute minimum. 2. A function f (x, y) has ∇f = 2xey , x2 ey + y 2 − 4 . Find the critical points of f and classify them as local maxima, local minima, or saddle points. The critical points satisfy ∇f = 0 so 2xey = 0, x2 ey + y 2 − 4 = 0. Since ey cannot be zero, the first equation implies that x = 0. The second equation then tells us that y 2 = 4 so y = ±2. Therefore the two critical points are (0, 2), (0, −2). To classify these, we use the Second Derivative Test. Since fx = 2xey and fy = x2 ey + y 2 − 4, we have fxx = 2ey , fxy = fyx = 2xey , fyy = x2 ey + 2y. At (0, 2) we have fxx (0, 2) = 2e2 , fxy (0, 2) = 0, fyy (0, 2) = 4 so D(0, 2) = (2e2 )(4) − (0)2 = 8e2 > 0. Therefore, since fxx (0, 2) > 0, (0, 2) is a local minimum. Also fxx (0, −2) = 2e−2 , fxy (0, −2) = 0, fyy (0, −2) = −4 so D(0, −2) = (2e−2 )(−4) − (0)2 = −8e−2 < 0 so (0, −2) is a saddle point. 3. Let R be the quarter-disc given by x2 + y 2 ≤ 4, x ≥ 0 and y ≥ 0. Find ZZ x + y dA. R In polar coordinates, this region can be described as 0 ≤ r ≤ 2, 0≤θ≤ and so the integral is Z r=2 Z θ=π/2 Z π 2 r=2 (r cos θ + r sin θ)r dθ dr = r=0 θ=0 Zr=0 r=2 2r2 dr r=0  3 r=2 = 2r /3 r=0 16 = 3 = θ=π/2 r2 [sin θ − cos θ]θ=0 dr 4. Use a triple integral to find the volume of the region D inside the paraboloid z = x2 + y 2 and between the planes z = 1 and z = 4. If we draw a cross-section through the region we are interested in, it looks like this: It’s easiest to describe this region using cylindrical coordinates. In this case it is given by 0 ≤ θ ≤ 2π 1≤z≤4 √ 0≤r≤ z The key part here is the range of values for r given a fixed z. The √ minimum value for r is 0 and the maximum is on the surface z = xy + y 2 = r2 so r = z. (There are other possibilities: for example, one can work out the volume of the whole paraboloid from z = 0 to z = 4 and then subtract the part from z = 0 to z = 1.) The volume is given by integrating the constant function 1 over the solid region. This gives ZZZ Z θ=2π Z z=4 Z r=√z 1 dV = r dr dz dθ D θ=0 z=1 r=0 Z θ=2π Z z=4  2 r=√z = r /2 r=0 dz dθ θ=0 z=1 Z θ=2π Z z=4 z = dz dθ θ=0 z=1 2 Z θ=2π  2 z=4 = z /4 z=1 dθ θ=0 Z θ=2π 15 = dθ 4 θ=0 15π = 2 5. (a) Find the Jacobian ∂(x, y) related to the change of variables ∂(u, v) u = ex+y , v = ex−y . (b) Calculate the integral ZZ e2x dA R where R is the region bounded by the straight lines y = x, y = −x, x= ln 2 . 2 (a) We need to find x and y in terms of u and v. The two equations give x + y = ln(u) x − y = ln(v) so x= ln(u) + ln(v) , 2 y= ln(u) − ln(v) . 2 We therefore have 1 ∂x 1 ∂y 1 ∂y −1 ∂x = , = , = , = ∂u 2u ∂v 2v ∂u 2u ∂v 2v and so ∂(x, y) ∂x ∂y ∂x ∂y = − ∂(u, v) ∂u ∂v ∂v ∂u 1 −1 1 1 = − 2u 2v 2u 2v 1 −1 − = 4uv 4uv −1 = 2uv (b) The line y = x is given by x − y = 0 and so v = e0 = 1. The line y = −x is given by x + y = 0 and so u = e0 = 1. The line x = (ln 2)/2 is ln(u) + ln(v) ln 2 = 2 2 so ln(uv) = ln 2 so uv = 2 or 2 . u Therefore the region corresponding to R in the uv-plane looks like v= This is described by 1≤u≤21≤v≤ 2 . u We also have e2x = exp(ln(u) + ln(v)) = uv and so the integral becomes Z u=2 u=1 Z v=2/u v=1 −1 uv dv du = 2uv Z u=2 Z 2/u 1 dv du u=1 v=1 2 Z u=2 1 1 = − du 2 u=1 u h i u u=2 = ln(u) − 2 u=1 1 = ln(2) − 2