homework14_solutions

Hour 14 - Partial Derivatives Homework 1. (Problem Set 14, #1) Verify that Clairaut’s Theorem holds for f ( x, y ) = e x sin y by finding both partial derivatives f xy and f yx . Solution. f x = (sin y ) e x sin y , so f xy = (cos y ) e x sin y + (sin y ) e x sin y ( x cos y ). On the other hand, f y = ( x cos y ) e x sin y , so f yx = (cos y ) e x sin y + ( x cos y ) e x sin y sin y . Clairaut’s Theorem is indeed satisfied, as f xy = f yx . 2. (Problem Set 14, #2) Open this GeoGebra applet . It shows the graph of a function f ( x, y ), together with a point that you can move around using the sliders. Initially, the point is at (3 , − 2 , f (3 , − 2)). (a) Click the appropriate checkbox to see f x (3 , − 2) . Is f x (3 , − 2) positive, negative, or 0? Solution. As you move through (3 , − 2 , f (3 , − 2)) in the positive x -direction, f is decreasing. So, f x (3 , − 2) is negative . (b) Click the appropriate checkbox to see f y (3 , − 2) . Is f y (3 , − 2) positive, negative, or 0? Solution. The slice of z = f ( x, y ) in x = 3 has a local maximum at y = − 2, so f y (3 , − 2) = 0 . (c) Keep the x -coordinate of the point at 3 , but use the slider to change the y -coordinate to 0 . Is f y (3 , 0) positive, negative, or 0? Solution. f y (3 , 0) is negative because, as you go through the point (3 , 0 , f (3 , 0)) in the positive y -direction, f is decreasing. (d) Keep the y -coordinate of the point at 0 , but use the slider to change the x -coordinate to 2 . 5 . Now, increase the x -coordinate from 2 . 5 to 3 . 5 (while leaving the y -coordinate constant). As you increase x from 2 . 5 to 3 . 5 , is f y increasing, decreasing, or staying constant? Solution. f y is increasing : as we increase x , f y changes from more negative to less negative. (e) Using your answer to #3(d), which of the following is a valid conclusion? A. f yx (3 , 0) < 0 B. f yx (3 , 0) > 0 C. f yy (3 , 0) < 0 D. f yy (3 , 0) > 0 Solution. We saw that, as we increased x (going through the point where x = 3, y = 0), f y increased. This means the rate of change of f y with respect to x was positive, so f yx (3 , 0) > 0. Thus, B. is correct. 1

(f) Is ∂ 2 f ∂y 2 (3 , − 2) positive or negative? Solution. As we move in the positive y -direction through (3 , − 2 , f (3 , − 2)), f y is decreasing (it’s positive before (3 , − 2 , f (3 , − 2)) and negative after). So, f yy is negative . Another way to think about f yy is that, if we keep x constant (that is, look at f (3 , y )) and think about f as a function of just y , the second derivative f yy tells us about concavity. The slice of z = f ( x, y ) in x = 3 is concave down at y = − 2, which is another way of seeing that f yy (3 , − 2) is negative. 3. (Problem Set 14, #3) Interpreting partial derivatives. (a) The price P in dollars to purchase a used car is a function of its original cost C in dollars and its age A in years. i. What are the units of ∂P ∂C and ∂P ∂A ? Solution. The units of ∂P ∂C are dollars/dollar . (This can be thought of as unit-less, but I find it easier to interpret ∂P ∂C as a rate by keeping the units of dollars/dollar.) The units of ∂P ∂A are dollars/year . ii. Do you expect ∂P ∂C to be positive or negative? Why? Solution. Fox a fixed age car (like all cars that are 5 years old), we expect that the higher the original cost, the higher the cost to purchase it used. That is, we expect P to be an increasing function of C , so ∂P ∂C should be positive . iii. Do you expect ∂P ∂A to be positive or negative? Why? Solution. For a fixed original price (like all cars that originally cost $ 20,000), we expect that, the older the used car, the cheaper it should be. That is, we expect P to be a decreasing function of A , so ∂P ∂A should be negative . (b) As we’ve discussed several times and you’ve undoubtedly experienced, if you’re standing outside, both the temperature and the wind speed affect how cold you feel. The wind-chill index gives the perceived temperature as a function of actual temperature and wind speed. The wind-chill index is modeled by the function W = 13 . 12 + 0 . 6215 T − 11 . 37 v 0 . 16 + 0 . 3965 Tv 0 . 16 where T is the temperature in ◦ C and v is the wind speed in km/hour. (1) When T = − 15 ◦ C and v = 30 km/hour, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 0 . 1 ◦ C? What if the wind speed increases by 2 km/hour? Solution. We should be able to use the partial derivative ∂W ∂T to tell us how W changes with a small change in temperature T . Let’s calculate it: ∂W ∂T = 0 . 6215 + 0 . 3965 v 0 . 16 (1) See https://www.nytimes.com/2017/09/14/science/maurice-bluestein-who-modernized-the-wind-chill-index-dies- at-76.html for an explanation of how two scientists came up with this formula. (The formula in that article looks different because it uses temperature in ◦ F and wind speed in mph.) 2

(b) f x ( A ) Solution. If you move in the positive x -direction through A , you’re moving from where f is between 1 and 2 to where f is between 2 and 3, so f is increasing. Thus, f x ( A ) is positive . (c) f x ( B ) Solution. If you move in the positive x -direction through B , f switches from increasing to decreasing at B , so the instantaneous rate of change at B is 0 . (d) f yy ( A ) Solution. Since f yy = ( f y ) y , asking whether f yy ( A ) is positive or negative is asking whether f y is increasing or decreasing in the positive y -direction. If we move in the positive y -direction through A , f y is negative and becoming more negative (because the vertical spacing between the contours is decreasing as we move up). So, f y is decreasing, which means f yy ( A ) is negative . (e) f xx ( B ) Solution. Asking whether f xx ( B ) is positive is asking whether f x is increasing as we move in the positive x -direction through B . Here, f x is positive to the left of B and negative to the right, so it’s decreasing, which makes f xx negative at B . (f) f xy ( C ) Solution. If we think of f xy as ( f x ) y , this is asking whether f x is increasing or decreasing as we move in the positive y -direction through C . As we move in the positive y -direction through C , f x is positive and become more positive (because the horizontal spacing between contours is decreasing). That means f x is increasing, so f xy is positive . 5. (Problem Set 14, #5) The graph of f ( x, y ) = 6 e − ( x − 1) 2 − ( y − 2) 2 intersects the plane x = 2 in a curve C . (a) Find the tangent line to the curve C at the point ( 2 , 4 , 6 e 5 ) . Solution. Our curve C is the slice of z = f ( x, y ) in x = 2. If we draw this slice in a 2-dimensional picture (involving just y and z ), the slope of the line tangent to the slice at y = 4 is f y (2 , 4), which we can calculate: f y ( x, y ) = 6 e − ( x − 1) 2 − ( y − 2) 2 [ − 2( y − 2)] f y (2 , 4) = 6 e − 1 − 4 ( − 4) = − 24 e − 5 “Slope” is an ambiguous notion in R 3 , but since we’re looking here at the cross-section in x = 2, we can think of our curve as just a relationship between y and z . When we picture it that way, then we think of the tangent line as being in R 2 , so it makes perfect sense to talk about its slope: 4

4 y z slice of z = f ( x, y ) in x = 2 In particular, if we move along the tangent line so that our y -coordinate increases by 1, then our z -coordinate decreases by 24 e − 5 (while our x -coordinate stays constant). The vector that describes this displacement is ⟨ 0 , 1 , − 24 e − 5 ⟩ . So, ⟨ 0 , 1 , − 24 e − 5 ⟩ is parallel to the tangent line, and one parameterization of the tangent line is 2 , 4 , 6 e 5 + t 0 , 1 , − 24 e − 5 . (b) Explain your method to #5(a) in complete sentences. Solution. See #5(a). 5