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MA 231 Supplemental Problems on the Chain Rule df df dx df dy = + = fxx + fy y dt dx dt dy dt ∆f = f (a + ∆x, b + ∆y) − f (a, b) ≈ fx(a, b)∆x + fy (a, b)∆y. Note that problem 1c is changed from what was given in class: 1. Find dz if dt (a) z = f (x, y) = x2 y, x = t2 , y = t2 (b) z = f (x, y) = xexy , x = et , y = t2 y (c) z = f (x, y) = , x = 3t + 1, y = t2 x 2. Suppose n, w measure the distance north and west from the corner of a tulip ﬁeld. f (n, w) = w1/2 n3/2 + n2 gives the height of the tulips when you are n feet to the north and w feet to the west. (a) You are tiptoeing through the tulips and notice that when you are 1 foot north and 3 feet west that you are moving at 4 ft/sec northward and −2 ft/sec westward. How fast is the height of the tulips changing? (b) Suppose you move from (1, 3) to (3, 2). Estimate the change in the height of the tulips. 3. As a function of temperature T and nutrient load N , the population P of bacteria in a lake is given by P (N, T ) = N e−T /10 . (a) Suppose that when T = 10 and N = 40 you observe that the temperature is rising at 3 degrees per hour and the nutrient load is dropping at 5 kilograms per hour. How fast is the bacterial population changing? (b) Suppose that at T = 10 and N = 40 you know that the bacterial population is 33.7 (in appropriate units). Esimate the bacterial population when T = 9, N = 37. 1 Answers The answers are given in expanded form to make it easier to check. 1a. z = 2(t2 )t2 (2t) + (t2 )2 2t 1b. z = (exy + xyexy )x + x2 exy y = (ee t + et t2 ee t )et + e2t ee t 2t t 2 1c. z = − t 2 t 2 y 1 t2 1 x + = − 1+ y 2t 2 2 x x (t + 1) (t + 1) 2. fn = 32 w1/2 n1/2 + 2n, fw = 12 w−1/2 n3/2 . 2a. z = ( 32 31/2 + 2)4 + ( 12 3−1/2 )(−2) 2b. ∆z = f (3, 2) − f (1, 3) ≈ ( 32 31/2 + 2)2 + ( 12 3−1/2 )(−1) N −T /10 e 3. PN = e−T /10 , PT = − 10 3a. P = PN N + PT T = e−1 (−5) + (−4e−1 )(3) 3b. ∆P = P (37, 9)−33.7 = P (37, 9)−P (40, 10) ≈ PN (50, 10)∆N +PT (40, 10)∆T = e−1 (−3) + (−4e−1 )(−1). So P (37, 9) = 33.7 + e−1 (−3) + (−4e−1 )(−1). 2