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Calculus 3 Lia Vas The Chain Rule Recall the chain rule for functions of single variable: If y = f (x) and x = g(t), then y 0 (t) = y 0 (x) · x0 (t) ( or dy dx dy = ). dt dx dt The chain rule for function z = f (x, y) with x = g(t) and y = h(t) is z 0 (t) = zx · x0 (t) + zy · y 0 (t) or dz ∂z dx ∂z dy = + . dt ∂x dt ∂y dt The chain rule for function z = f (x, y) with x = g(s, t) and y = h(s, t) is zs = zx · xs + zy · ys , or zt = zx · xt + zy · yt ∂z ∂z ∂x ∂z ∂y = + ∂s ∂x ∂s ∂y ∂s ∂z ∂z ∂x ∂z ∂y = + ∂t ∂x ∂t ∂y ∂t Implicit functions. In some cases, an implicit function F (x, y, z) = 0 cannot be solved for z. To find the derivatives zx and zy in these cases, use implicit differentiation. Differentiating the equation with respect to x we obtain Fx + Fz zx = 0. Solving for zx gives us zx = − FFxz . Differentiating the equation with respect to y we obtain Fy + Fz zy = 0. Solving for zy gives us zy = − FFyz . Thus, the partial derivatives zx and zy are given by: zx = − Fx Fz and zy = − Fy Fz Practice problems. 1. Find the indicated derivatives. (a) z = 3x2 + 2xy − 5y 2 , x = 2 + t2 , y = 1 − t3 ; z 0 (t) when t = 0. (b) z = x ln(x + 2y), x = cos t, y = sin t; z 0 (t) when t = 0. (c) z = ex y + xy 2 , x = st, y = s2 + t2 ; zs and zt at (1, 1). (d) z = x2 + xy, x = et cos s, y = et sin s; zs and zt at (π, 0). (e) xy 2 + yz 2 + zx2 = 3; zx and zy at (1, 1, 1). (f) x − yz = cos(x + y + z); zx and zy at (0, 1, −1). 2. The pressure of 1 mole of an ideal gas is increasing at a rate of 0.05 kPa/s and the temperature is rising at a rate of 0.15 K/s. The pressure P , volume V and temperature T are related by the equation P V = 8.31T. Find the rate of change of the volume when the pressure is 20 kPa and temperature 320 K. 3. The number N of bacteria in a culture depends on temperature T and pressure P which depend on time t in minutes. Assume that 3 minutes after the experiment started, the pressure is increasing at a rate of 0.1 kPa/min and the temperature at a rate of 0.5 K/min. The number of bacteria changes at the rates of 3 bacteria per kPa and 5 bacteria per Kelvin. Find the rate at which the number of bacteria is increasing 3 minutes after the experiment started. 4. The number of flowers N in a closed environment depends on the amount of sunlight S that the flowers receive and the temperature T of the environment. Assume that the number of flowers changes at the rates NS = 2 and NT = 4. If the temperature depends on time as 1 8 T (t) = 85 − 1+t 2 and the amount of sunlight decreases on time as S(t) = t , find the rate of change of the flower population at time t = 2 days. 5. The temperature at a point (x, y) is T (x, y),√measured in degrees Celsius. A bug crawls so that its position after t seconds is given by x = 1 + t, y = 2 + 31 t where x and y are measured in centimeters. The temperature function satisfies Tx (2, 3) = 4 and Ty (2, 3) = 3. Determine how fast the temperature is rising on the bug’s path after 3 seconds. 6. If the trajectory of an object is given by a parametric curve, its speed at a point can be found as the length of the tangent vector at that point. Assume that an object moves in √ space away from its initial position so that after t hours it is at x = 2t and y = 2t2 −1 and z = 3t + 1 miles from its initial position. Find the speed of that object 5 hours after it started moving. 7. An object moves in xy-plane so that after t hours it is at x = x(t) and y = y(t) miles from its initial position. Suppose that after 2 hours, the object has the velocity so that x0 (2) = 3 and y 0 (2) = 5. Find the speed of the object after 2 hours. Solutions. ∂z dx ∂z dy = ∂x + ∂y = (6x + 2y)(2t) + (2x − 10y)(−3t2 ). When t = 0, x = 2 + 02 = 2 and 1. (a) dz dt dt dt y = 1 − t3 = 1 = 0 = 1. Thus, z 0 (0) = (12 + 2)(0) + (4 − 10)0 = 0. ∂z dx ∂z dy x 2x = ∂x + ∂y = (ln(x + 2y) + x+2y )(− sin t) + x+2y (cos t). When t = 0, x = cos 0 = 1 (b) dz dt dt dt 1 2 0 and y = sin 0 = 0. Thus, z (0) = (ln(1 + 0) + 1+0 )(0) + 1+0 (1) = 0 + 2 = 2. (c) zs = zx xs +zy ys = (ex y+y 2 )t+(ex +2xy)2s, zt = zx xt +zy yt = (ex y+y 2 )s+(ex +2xy)2t. When s = 1, t = 1, x = 1(1) = 1 and y = 12 + 12 = 2. Thus zs (1, 1) = (e1 2 + 22 )1 + (e1 + 2(1)2)2 = 4e + 12 and zt (1, 1) = (e1 2 + 22 )1 + (e1 + 2(1)2)2 = 4e + 12. (d) zs = zx xs + zy ys = (2x + y)(−et sin s) + x(et cos s), zt = zx xt + zy yt = (2x + y)(et cos s) + x(et sin s). When s = π and t = 0, x = e0 cos π = −1 and y = e0 sin π = 0. Thus, zs (π, 0) = (−2)(0) + (−1)(−1) = 1 and zt (π, 0) = (−2)(−1) + (−1)(0) = 2. (e) Use the formulas for derivatives of implicit function. Let F = xy 2 + yz 2 + zx2 − 3. Then 2 +2xz Fy Fx = y 2 + 2xz, Fy = 2xy + z 2 , Fz = 2yz + x2 and so zx = − FFxz = − y2yz+x = 2 and zy = − F z 2 2xy+z 1+2 2+1 − 2yz+x 2 . At (1, 1, 1), zx = − 2+1 = −1 and zy = − 2+1 = −1. (f) Use the formulas for derivatives of implicit function. Let F = x − yz − cos(x + y + z). Then Fx = 1 + sin(x + y + z), Fy = −z + sin(x + y + z), Fz = −y + sin(x + y + z) and so 1+sin(x+y+z) −z+sin(x+y+z) 1+0 zx = − FFxz = − −y+sin(x+y+z) and zy = − FFyz = − −y+sin(x+y+z) . At (0, 1, −1), zx = − −1+0 = 1 and 1+0 zy = − −1+0 = 1. 2. We are given the rates dP = 0.05 kPa/s, dt dV need to find the rate dt at (P0 , T0 ). dT dt = 0.15 K/s at P0 = 20 kPa and T0 = 320 K. We Since the derivatives are with respect to P and T , treat V as the dependent variable and . By the chain solve for V in terms of P and T to obtain the dependence formula V = 8.31T P dV ∂V dP ∂V dT 8.31T ∂V −8.31T rule, dt = ∂P dt + ∂T dt . From V = P we find that ∂P = P 2 and that ∂V = 8.31 . ∂T P −8.31(320) ∂V At P0 = 20 and T0 = 320, we calculate that ∂P = = −6.648 liters/kPa and that 400 ∂V 8.31 dV = 20 = 0.4155 liters/K. Thus, dt = (−6.648)(0.05) + (0.4155)(0.15) = −.27 liter per ∂T second. 3. We are given the rates P 0 (3) = 0.1, T 0 (3) = 0.5, NP = 3, and NT = 5. By the chain rule formula, N 0 (t) = NT T 0 + NP P 0 . Thus N 0 (3) = 5 · 0.5 + 3 · 0.1 = 2.8 bacteria/minute. 4. We are given the rates NS = 2 and NT = 4. The remaining two rates S 0 and T 0 needed for the chain rule formula N 0 = NS S 0 + NT T 0 can be found from the formulas for S and T. S= 1 t ⇒ S0 = T = 85 − 8 1+t2 −1 . t2 At t = 2, S 0 = ⇒ T0 = 16t . (1+t2 )2 −1 . 4 At t = 2, T 0 = 32 . 25 32 Thus, N 0 (2) = NS S 0 + NT T 0 = 2 −1 + 4 25 = 4.62 flowers/day. 4 5. We are given the rates Tx (2, 3) = 4 and Ty (2, 3) = 3 and the problem is asking for the rate at t = 3. dT dt dy dx Using chain rule, dT = ∂T + ∂T . x0 = 2√11+t and y 0 = 13 . At t = 3, x0 = 2√1 4 = 14 and y 0 = 13 . dt ∂x dt ∂y dt √ When t = 3, x = 1 + 3 = 2 and y = 2 + 1 = 3 so the point (2, 3) at which the derivatives Tx = 4 14 + 3 31 = 1 + 1 = 2 degrees Celsius and Ty are given corresponds exactly to t = 3. Thus, dT dt per second. 6. x0 = 2, y 0 = 4t and z 0 = √3 . 2 3t+1 When t = 5, x0 = 2, y 0 = 20 and z 0 = 38 . Thus, the tangent vector is h2, 20, 38 i. The speed is the length |h2, 20, 83 i| = q 22 + 202 + ( 83 )2 = 20.1 miles/hour. √ 7. At t = 2, hx0 , y 0 i = h3, 5i. The speed = length of h3, 5i = 32 + 52 = 5.83 miles/ hour.

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