14.5 The Chain Rule. dx dt

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1 SECTION 14.5 THE CHAIN RULE w lns 2 2 z w e z 37. If R is the total resistance of three resistors, connected in parallel, with resistances,,, then R 1 R 2 R If z and, changes from 1, 2 to 1.05, 2.1, compare the values of z and. 30. If z and, changes from 3, 1 to 2.96, 0.95, compare the values of z and. 31. The length and wih of a rectangle are measured as 30 cm and 24 cm, respectivel, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maimum error in the calculated area of the rectangle. 32. The dimensions of a closed rectangular bo are measured as 80 cm, 60 cm, and 50 cm, respectivel, with a possible error of 0.2 cm in each dimension. Use differentials to estimate the maimum error in calculating the surface area of the bo. 33. Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height 12 cm if the tin is 0.04 cm thick. 34. Use differentials to estimate the amount of metal in a closed clindrical can that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0.1 cm thick and the metal in the sides is 0.05 cm thick. 35. A boundar stripe 3 in. wide is painted around a rectangle whose dimensions are 100 ft b 200 ft. Use differentials to approimate the number of square feet of paint in the stripe. 36. The pressure, volume, and temperature of a mole of an ideal gas are related b the equation PV 8.31T, where P is measured in kilopascals, V in liters, and T in kelvins. Use differentials to find the approimate change in the pressure if the volume increases from 12 L to 12.3 L and the temperature decreases from 310 K to 305 K. If the resistances are measured in ohms as R 1 25, R 2 40, and R 3 50, with a possible error of 0.5% in each case, estimate the maimum error in the calculated value of R. 38. Four positive numbers, each less than 50, are rounded to the first decimal place and then multiplied together. Use differentials to estimate the maimum possible error in the computed product that might result from the rounding Show that the function is differentiable b finding values of and that satisf Definition f, Prove that if f is a function of two variables that is differentiable at a, b, then f is continuous at a, b. Hint: Show that 42. (a) The function f, lim f a, b f a, b, l 0, 0 was graphed in Figure 4. Show that f 0, 0 and f 0, 0 both eist but f is not differentiable at 0, 0. [Hint: Use the result of Eercise 41.] (b) Eplain wh and are not continuous at 0, 0. f 1 R 1 R 1 1 R 2 1 R 3 f f, 5 2 if, 0, 0 if, 0, The Chain Rule Recall that the Chain Rule for functions of a single variable gives the rule for differentiating a composite function: If f and tt, where f and t are differentiable functions, then is indirectl a differentiable function of t and 1 For functions of more than one variable, the Chain Rule has several versions, each of them giving a rule for differentiating a composite function. The first version (Theorem 2) deals with the case where z f, and each of the variables and is, in turn, a function of a variable t. This means that z is indirectl a function of t, z f tt, ht, and the Chain Rule gives a formula for differentiating z as a function of t. We assume that f is differentiable (Definition ). Recall that this is the case when f and f are continuous (Theorem ).

2 932 CHAPTER 14 PARTIAL DERIVATIVES 2 The Chain Rule (Case 1) Suppose that z f, is a differentiable function of and, where tt and ht are both differentiable functions of t. Then z is a differentiable function of t and f f Proof A change of t in t produces changes of in and in. These, in turn, produce a change of z in z, and from Definition we have where 1 l 0 and 2 l 0 as, l 0, 0. [If the functions 1 and 2 are not defined at 0, 0, we can define them to be 0 there.] Dividing both sides of this equation b t, we have If we now let t l 0, then tt t tt l 0 because t is differentiable and therefore continuous. Similarl, l 0. This, in turn, means that 1 l 0 and 2 l 0, so z lim t l 0 t f lim f t l 0 t lim lim t l 0 t 1 lim lim t l 0 t l 0 t 2 lim t l 0 t l 0 t f f z t z f f 1 2 f f f t f 0 t 1 t 2 t 0 Since we often write z in place of f, we can rewrite the Chain Rule in the form Notice the similarit to the definition of the differential: z z z z EXAMPLE 1 If z 2 3 4, where sin 2t and cos t, find when t 0. The Chain Rule gives z z cos 2t sin t It s not necessar to substitute the epressions for and in terms of t. We simpl

3 SECTION 14.5 THE CHAIN RULE 933 (0, 1) FIGURE 1 The curve =sin 2t, =cos t observe that when t 0 we have sin 0 0 and cos 0 1. Therefore, 0 32 cos 0 0 0sin 0 6 t0 The derivative in Eample 1 can be interpreted as the rate of change of z with respect to t as the point, moves along the curve C with parametric equations sin 2t, cos t. (See Figure 1.) In particular, when t 0, the point, is 0, 1 and 6 is the rate of increase as we move along the curve C through 0, 1. If, for instance, z T, represents the temperature at the point,, then the composite function z Tsin 2t, cos t represents the temperature at points on C and the derivative represents the rate at which the temperature changes along C. EXAMPLE 2 The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related b the equation PV 8.31T. Find the rate at which the pressure is changing when the temperature is 300 K and increasing at a rate of 0.1 Ks and the volume is 100 L and increasing at a rate of 0.2 Ls. If t represents the time elapsed in seconds, then at the given instant we have T 300, dt 0.1, V 100, dv 0.2. Since the Chain Rule gives P 8.31 T V dp P T dt P V dv 8.31 V dt 8.31T dv V The pressure is decreasing at a rate of about kpas. We now consider the situation where z f, but each of and is a function of two variables s and t: ts, t, hs, t. Then z is indirectl a function of s and t and we wish to find zs and zt. Recall that in computing zt we hold s fied and compute the ordinar derivative of z with respect to t. Therefore, we can appl Theorem 2 to obtain z t z t z A similar argument holds for zs and so we have proved the following version of the Chain Rule. t 3 The Chain Rule (Case 2) Suppose that z f, is a differentiable function of and, where ts, t and hs, t are differentiable functions of s and t. Then z z s s z s z t z t z t

4 934 CHAPTER 14 PARTIAL DERIVATIVES EXAMPLE 3 If z e sin, where st 2 and s 2 t, find zs and zt. Appling Case 2 of the Chain Rule, we get z z s z t z s t z t 2 e st 2 sin s 2 t 2ste st 2 coss 2 t z 2ste st 2 sin s 2 t s 2 e st 2 coss 2 t s e sin t 2 e cos 2st t e sin 2st e cos s 2 s z FIGURE 2 t z s z t s t s t Case 2 of the Chain Rule contains three tpes of variables: s and t are independent variables, and are called intermediate variables, and z is the dependent variable. Notice that Theorem 3 has one term for each intermediate variable and each of these terms resembles the one-dimensional Chain Rule in Equation 1. To remember the Chain Rule it s helpful to draw the tree diagram in Figure 2. We draw branches from the dependent variable z to the intermediate variables and to indicate that z is a function of and. Then we draw branches from and to the independent variables s and t. On each branch we write the corresponding partial derivative. To find zs we find the product of the partial derivatives along each path from z to s and then add these products: z z s s z s Similarl, we find zt b using the paths from z to t. Now we consider the general situation in which a dependent variable u is a function of n intermediate variables 1,..., n, each of which is, in turn, a function of m independent variables t 1,..., t m. Notice that there are n terms, one for each intermediate variable. The proof is similar to that of Case 1. 4 The Chain Rule (General Version) Suppose that u is a differentiable function of the n variables 1, 2,..., n and each j is a differentiable function of the m variables,,...,. Then u is a function of,,..., and t 1 t 2 t m t 1 t 2 t m u t i u 1 u 2 u 1 t i 2 t i n n t i for each i 1, 2,..., m. w z t u v u v u v u v FIGURE 3 EXAMPLE 4 Write out the Chain Rule for the case where w f,, z, t and u, v, u, v, z zu, v, and t tu, v. We appl Theorem 4 with n 4 and m 2. Figure 3 shows the tree diagram. Although we haven t written the derivatives on the branches, it s understood that if a branch leads from to u, then the partial derivative for that branch is u. With the aid

5 SECTION 14.5 THE CHAIN RULE 935 r s t r s t r s FIGURE 4 u z t of the tree diagram we can now write the required epressions: EXAMPLE 5 If u 4 2 z 3, where rse t, rs 2 e t, and z r 2 s sin t, find the value of us when r 2, s 1, t 0. With the help of the tree diagram in Figure 4, we have When r 2, s 1, and t 0, we have 2, 2, and z 0, so EXAMPLE 6 If ts, t f s 2 t 2, t 2 s 2 and f is differentiable, show that t satisfies the equation Let s 2 t 2 and t 2 s 2. Then ts, t f, and the Chain Rule gives Therefore t t s u s w u w v u t s t t u s EXAMPLE 7 If z f, has continuous second-order partial derivatives and r 2 s 2 and 2rs, find (a) zr and (b) 2 zr 2. (a) The Chain Rule gives w w s 4 3 re t 4 2z 3 2rse t 3 2 z 2 r 2 sin t f f t s t 2st f f 2st 2st f f 2st 0 z z r w u v u s t r w t t s f f z s t t 0 (b) Appling the Product Rule to the epression in part (a), we get 2 z r r2r z z 2s z 2r z 2s r z r s u z s t w u z v r w z z s z w u t z v w t t u t v f f 2s 2s f f 2t 2t z z 2r 2s

6 936 CHAPTER 14 PARTIAL DERIVATIVES FIGURE 5 z r s r s But, using the Chain Rule again (see Figure 5), we have z r z r z r 2 z 2 2r 2 z 2s z r z r z r 2 z 2r 2 z 2s 2 Putting these epressions into Equation 5 and using the equalit of the mied secondorder derivatives, we obtain 2 z z 2 2 r 2r 2r 2 z 2s 2 z 2s2r 2 2 z 4r 2 2 z 2 Implicit Differentiation 8rs 2 z 4s 2 2 z 2 The Chain Rule can be used to give a more complete description of the process of implicit differentiation that was introduced in Sections 3.6 and We suppose that an equation of the form F, 0 defines implicitl as a differentiable function of, that is, f, where F, f 0 for all in the domain of f. If F is differentiable, we can appl Case 1 of the Chain Rule to differentiate both sides of the equation F, 0 with respect to. Since both and are functions of, we obtain F F 0 But 1, so if F 0 we solve for and obtain 2 2 z 2s 2 z 6 F F F F To derive this equation we assumed that F, 0 defines implicitl as a function of. The Implicit Function Theorem, proved in advanced calculus, gives conditions under which this assumption is valid. It states that if F is defined on a disk containing a, b, where Fa, b 0, F a, b 0, and F and F are continuous on the disk, then the equation F, 0 defines as a function of near the point a, b and the derivative of this function is given b Equation 6. EXAMPLE 8 Find if

7 SECTION 14.5 THE CHAIN RULE 937 The given equation can be written as F, so Equation 6 gives The solution to Eample 8 should be compared to the one in Eample 2 in Section 3.6. F F Now we suppose that z is given implicitl as a function z f, b an equation of the form F,, z 0. This means that F,, f, 0 for all, in the domain of f. If F and f are differentiable, then we can use the Chain Rule to differentiate the equation F,, z 0 as follows: F F F z z 0 But 1 and 0 so this equation becomes F F z z 0 If Fz 0, we solve for z and obtain the first formula in Equations 7. The formula for z is obtained in a similar manner. 7 F z F z F z F z Again, a version of the Implicit Function Theorem gives conditions under which our assumption is valid. If F is defined within a sphere containing a, b, c, where Fa, b, c 0, F z a, b, c 0, and F, F, and F z are continuous inside the sphere, then the equation F,, z 0 defines z as a function of and near the point a, b, c and this function is differentiable, with partial derivatives given b (7). z z EXAMPLE 9 Find and if 3 3 z 3 6z 1. Let F,, z 3 3 z 3 6z 1. Then, from Equations 7, we have The solution to Eample 9 should be compared to the one in Eample 4 in Section z F 3 2 6z F z 3z z z 2 2 z F 3 2 6z F z 3z z z 2 2

24. ; Graph the function and observe where it is discontinuous. 2xy f x, y x 2 y 3 x 3 y xy 35. 6x 3 y 2x 4 y 4 36.

24. ; Graph the function and observe where it is discontinuous. 2xy f x, y x 2 y 3 x 3 y xy 35. 6x 3 y 2x 4 y 4 36. SECTION. LIMITS AND CONTINUITY 877. EXERCISES. Suppose that, l 3, f, 6. What can ou sa about the value of f 3,? What if f is continuous?. Eplain wh each function is continuous or discontinuous. (a) The