LESSON 21: DIFFERENTIALS OF MULTIVARIABLE FUNCTIONS MATH FALL 2018

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1 LESSON 21: DIFFERENTIALS OF MULTIVARIABLE FUNCTIONS MATH FALL 2018 ELLEN WELD 1. Quick Review of Differentials Ex 1. Consider the function f(x) x. We know that f(9) 9 3, but what is f(9.1) 9.1? Obviously, if you have a calculator this is easy. But instead of just using a calculator, we ll use differentials. Let x 9 and x + 9.1, that is,.1. is the actual change in the input x. Our goal is to approximate how this change in the input affects the output function, that is, f(9.1) f(x + ). For this, we use calculus. Write y f(x + ) f(x) f(9.1) f(9) y is the actual change the function f(x), which is our goal. In an ideal world, we could compute this directly for any given. But, in general, this is difficult to compute even with a calculator so we settle for an approximation of y instead. (1) Observe that the equation y f(x + ) f(x) looks a lot like a derivative. In fact, the only difference between equation (1) and an actual derivative is that we need to take the limit as 0. Because limits deal with things getting really close together, if our is small we can make an approximation of y using this derivative. We can write this like More helpfully, we have y f(x + ) f(x) (2) y f (x). f (x) dy dx. This just means that we can approximate the change in the function by taking the change in the input and multiplying it by the derivative of the function. Let s apply this to the example above. Since f(x) x, we have Hence, by equation (2), f (x) 1 2 x y f (9) (.1).1 2(3) 1 60.

2 2 ELLEN WELD So, if , we can add 9 to both sides to get Using a calculator, we find So our approximation is pretty good. 9.1 }{{} Note 1. We call dx and dy differentials. By the nature of derivatives (because we would assume that 0), the smaller is, the better the approximation of y. Think of as the actual change and d as the infinitesimal change. This is why we use dx in an integral and not because is too big. 2. Differentials of Multivariable Functions We can apply much of this thinking to functions of more than 1 variable as well. This time, however, we consider how changes in x and y affect z f(x, y). Our notation will be essentially the same and our goal will be to approximate The total differential is given by z f(x +, y + y) f(x, y). dx + x y dy f x(x, y)dx + f y (x, y)dy. We can use this formula to approximate z (remember, z is the actual change in z). As with before, we think of dx and y dy especially for and y small. Hence, (3) z + x y y. We call this equation the incremental approximation formula for functions of two variables. Ex 2. Suppose we have z f(x, y) x 2 + y 2. Then if x 3, y 4, f(3, 4) (3) 2 + (4) What if we wanted to find f(3.1, 3.8)? Take and y Next, note that f x (x, y) ( ) x2 + y x 2 f y (x, y) ( ) x2 + y y 2 2x 2 x 2 + y 2 x x2 + y 2 2y 2 x 2 + y 2 y x2 + y 2.

3 Therefore, by equation (3) above, AN UNOFFICIAL GUIDE TO MATH FALL z x x2 + y + y 2 x2 + y y 2 3 (3)2 + (4) (.1) (3)2 + (4) (.2) (.1) (.2) So we can write z f(x +, y + y) f(x, y) f(3.1, 3.8) f(3, 4) and adding f(3, 4) to both sides, we get f(3.1, 3.8) f(3, 4) + z 5 + (.1) 4.9. Plugging it into a calculator, (3.1) 2 + (3.8) So our approximation wasn t too far off. Note 2. Try to choose your x and y as simply as possible. This isn t exact to begin with (it s inherently an approximation) so make choices that make your life easier. In the previous Ex, we took x and y to be the whole numbers because that s easier than taking x 3.1 and y 3.8. Once you choose your x and y, your z is always given by z f(x +, y + y) f(x, y). 3. Solution to In-Class Examples Example 1. Use increments to estimate the change in z at (1, 1) if x 3x + y and 9y given.01 and y.02. y Solution: We use our incremental approximation formula. z (1, 1) + (1, 1) y x y [3(1) + ( 1)] + [9( 1)] }{{}}{{}}{{}}{{} (.01) x (1, 1) 2(.01) 9(.02).02 9(.02) (.02) y (1, 1) y

4 4 ELLEN WELD 8(.02).16. Example 2. Suppose that when a babysitter feeds a child x donuts and y pieces of cake, the child needs to run x 2 y + 7 laps in the backyard to be able to go to bed before the parents get home. If one evening the babysitter gives the child 3 donuts and 2 pieces of cake and the next time babysitting, 3.5 donuts and 1.5 pieces of cake, estimate the difference in the number of laps the child will need to run. Solution: Take x 3, y 2. Then and y Next, we need to find the derivatives with respect to x and y. Write Thus, z x 2xy 2 x 2 y + 7 xy x2 y + 7 y x 2 2 x 2 y + 7. xy x2 y x 2 2 x 2 y + 7 y (3)(2) (3)2 (2) + 7 (.5) + (3) 2 2 (3) 2 (2) + 7 (.5) 6 9 (.5) (.5) Example 3. A company produces boxes with square bases. Suppose they initially create a box that is 10 cm tall and 4 cm wide but they want to increase the box s height by.5 cm. Estimate how they must change the width so that the box stays the same volume. Solution: Because we are told these boxes have a square base, the formula for volume is V hw 2 where h is the height and w is the width. We are told h 10, w 4, h.5 and V 0 (because we want the volume of the box to stay the same). Now, we know that So, applying our formula we have V h w2 and V w 2wh. V V V h + h w w V (w 2 ) h + (2wh) w

5 AN UNOFFICIAL GUIDE TO MATH FALL [(4) 2 ](.5) + [2(10)(4)] w w. So we need to solve for w given w. We conclude that w This tells us that the width decreases by 1 10 cm. Example 4. Suppose the function S W 2 F + F 2 W describes the number of fern spores (in millions) released into the air where F is the number of ferns in an area and W is the speed of the wind in miles per hour. Suppose F 56 and W 10 with maximum errors of 2 ferns and 3 miles per hour. Find the approximate relative percentage error in calculating S. Round your answer to the nearest percent. Solution: Here, we think of the relative errors as our. Let F ±2 and W ±3. We are essentially trying to figure out how changing the inputs (in the sense of correcting the error) changes the number of spores released. We know that By our formula, S F W 2 + 2F W and S W 2W F + F 2. S (W 2 + 2F W ) F + (2W F + F 2 ) W [ (56)(10)](±2) + [2(10)(56) ](±3) ±( (560))(2) ± (2(560) )(3) ±2440 ± 12, 768. Now, we need to consider the 4 different possibilities that we get from the ± signs. Write , , , , , , , , 208. To find the maximum error, we re looking for is the largest number in absolute value. So we say S 15, 208. Finally, to determine the relative error, we take Thus, our answer is 41%. S S 15, 208 (10) 2 (56) + (56) 2 (10) 15, , This tells us that our formula is not very good as a model because small changes in the input (i.e., the errors) lead to large changes in the output.