Multivariable Calculus, Fall 2011 Midterm II Practice Exam 2 Solutions

Math 211, Multivariable Calculus, Fall 2011 Midterm II Practice Exam 2 Solutions 1. In a certain ski slope, the elevation (in meters) of the land at the points with coordinates (x, y) (also measured in meters) is given by the function y2 x2 − 100 50 √ Which direction could a skier at the point (25 3, −12.5) go in order to descend at an angle of 45 degree? Give your answer in the form of the angle between the direction the skier should head and due North. (Assume that the vector h1, 0i points due North and h0, 1i points due East.) h(x, y) = 1000 − To descend at an angle of 45 degrees we want the slope in the direction the skier goes to be −1. That is, the directional derivative of h in that direction should be −1. The gradient vector of h is ∇h = h−x/50, −y/25i. At the given point, we have √ √ ∇h(25 3, −12.5) = h− 3/2, 1/2i. We therefore want to find a unit vector u such that √ h− 3/2, 1/2i · u = −1. If the angle between u and the gradient vector is θ, we would have √ |h− 3/2, 1/2i||u| cos θ = −1. √ Now |u| = 1 since u must be a unit vector, and |h− 3/2, 1/2i| = 1 by calculation, so we must have cos θ = −1 so θ = π. This tells us that u should point exactly opposite the gradient vector, that is, in the direction of √ h 3/2, −1/2i. We can now draw a diagram to find the angle between this vector and due North (i.e. the vector h1, 0i. We see this is 120 degrees. So the skier should head at an angle of 120 degrees from due North. 2. Find the linear approximation to the function f (x, y) = cos(3xy) − sin(3xy) at the point (1, π) and use it to estimate the value of f (1, π − 0.01). We have ∂f = −3y sin(3xy) − 3y cos(3xy) ∂x 1 and ∂f = −3x sin(3xy) − 3x cos(3xy). ∂y Therefore ∂f (1, π) = −3π sin(3π) − 3π cos(3π) = 3π ∂x and ∂f (1, π) = −3 sin(3π) − 3 cos(3π) = 3. ∂y Since we also have f (1, π) = −1, the linear approximation is l(x, y) = −1 + 3π(x − 1) + 3(y − π). Therefore, we get the estimate f (1, π − 0.01) ≈ −1 + 3(−0.01) = −1.03. 3. In each of the following cases, find a function f of two variables that satisfies the given condition. (You should justify your answers as precisely as possible.) (a) f is defined at all points and is continuous at all points except (0, 0) (b) f is continuous at all points and is differentiable at all points except (0, 0) (a) There are many possibilities. For example, the function ( 1 if (x, y) 6= (0, 0); f (x, y) = 2 if (x, y) = (0, 0). This is continuous at points not equal to (0, 0) because in a region around any such point, it is the constant, continuous function 1. But the limit of f as (x, y) → (0, 0) is 1 which is not equal to f (0, 0), so f is not continuous at (0, 0). (Notice that the explanation must have reference to the definition of being continuous, namely that the limit as (x, y) approaches (0, 0) either is or is not equal to f (0, 0).) (b) Again there are many possibilities but it is harder to justify. A good one to keep in mind is f (x, y) = (x2 + y 2 )1/3 . This is defined everywhere by a single formula involving power functions and addition, so is continuous everywhere. At points other than (0, 0), the partial derivatives are given by ∂f 2x = 2 ∂x 3(x + y 2 )2/3 and 2y ∂f = . 2 ∂y 3(x + y 2 )2/3 These are both continuous everywhere they are defined, which is everywhere except (0, 0). So, the partial derivatives of f are continuous at and around any point that is not (0, 0), so f is differentiable there. 2 On the other hand, if we try to find the partial derivatives at (0, 0), we get ∂f f (h, 0) − f (0, 0) h2/3 1 (0, 0) = lim = lim = lim 1/3 . h→0 h→0 h h→0 h ∂x h This limit does not exist because h1/3 → 0 as h → 0. Therefore, the partial derivative at (0, 0) does not exist and so f is not differentiable at (0, 0). 4. A particle moves with position vector at time t given by r(t) = hsin(2t), 2t, cos(2t + π)i. Find the tangential and normal components of the acceleration of the particle. The tangential component of acceleration is given by aT = v 0 where v = |v| is the speed of the path. We have v(t) = r0 (t) = h2 cos(2t), 2, −2 sin(2t + π)i and so q v(t) = 4 cos2 (2t) + 4 + 4 sin2 (2t + π). Now sin(2t + π) = − sin(2t), so sin2 (2t + π) = sin2 (2t). Therefore, the speed is q √ √ v(t) = 4 cos2 (2t) + 4 + 4 sin2 (2t) = 8 = 2 2. This is constant, so the tangential component of the acceleration is zero. The normal component of acceleration is given by aN = κv 2 where κ is the curvature. Here it is easiest to use the formula |r0 (t) × r00 (t)| κ(t) = . |r0 (t)|3 We have r00 (t) = h−4 sin(2t), 0, −4 cos(2t + π)i and so r0 (t) × r00 (t) = h−8 cos(2t + π), −8 sin2 (2t) − 8 cos2 (2t), 8 sin(2t)i and so 0 00 |r (t) × r (t) = q 64 cos2 (2t) + 64 + 64 sin2 (2t) = The curvature is therefore √ 8 2 1 κ(t) = √ = . 2 (2 2)3 The normal component of acceleration is therefore aN = κv 2 = 4. 3 √ √ 128 = 8 2. 5. The following diagram shows a selection of level curves for a function f (x, y): (a) Estimate the value of each of the partial derivatives of f at the point (3, 4) (marked with a cross). (b) Give a vector u such that the directional derivative at (3, 4) of f in the direction u is approximately equal to 0. (c) Draw on the above diagram an arrow to represent the gradient vector of f at the point (3, 4). (Only the direction of your arrow is important.) (d) Draw a graph to illustrate the y = 4 cross-section of the function f . (Be as accurate as possible based on the information given.) What is the connection between this graph and one of your answers to part (a)? (a) ∂f ∂f (3, 4) ≈ 0.5, (3, 4) ≈ −1. (In (c) below we realize that these are not that accu∂x ∂y rate, as the ratio is not quite right, but this is good enough based on the diagram.) (b) The vector u should be tangent to the level curve at (3, 4), for example u ≈ h1, 2i. (c) The gradient vector at (3, 4) should be perpendicular to the level curve at (3, 4), and in the direction of increasing f . (This actually suggests that our guesses in part (a) were a bit off as the absolute value of ∂f should be greater than that of ∂f .) ∂x ∂y 4 (d) The connection is that the slope of the tangent line to the cross-section at x = 3 should be equal to the partial derivative ∂f (3, 4) which is about 0.5. ∂x 5