Even numbered problems and the “story problems” Section 14.5 - Solution

Even numbered problems and the “story problems”: Section 14.5 12. Use the Chain Rule to find ∂z/∂s and ∂z/∂t: z = tan(u/v) u = 2s + 3t v = 3s − 2t SOLUTION: You should draw a tree diagram to help keep your variables straight. We’ll need all the partial derivatives, so we’ll go ahead and compute them all: z = tan(u/v) u = 2s + 3t v = 3s − 2t us = 2 vs = 3 zu = sec2 (u/v) · (1/v) 2 2 zv = sec (u/v) · (−u/v ) ut = 3 vt = −2 Now, write out what we need, substitute and simplify a bit: ∂z 2 sec2 (u/v) −3u sec2 (u/v) 2v − 3u = zu · us + zv · vs = + = sec2 (u/v) 2 ∂s v v v2   And for the other one: ∂z 3 sec2 (u/v) 2u sec2 (u/v) 3v + 2u = zu · ut + zv · vt = + = sec2 (u/v) 2 ∂t v v v2   Side Remark: You might notice that these partial derivatives are written in u, v and not in s, t- As it happened, s, t ended up not appearing at all! It is OK to leave in this form unless you’re asked to back substitute to get expressions in s and t (by using u = 2s + 3t and v = 3s − 2t) 14. Let W = F (u, v) where u is a function of s, t and v is a function of s, t. We also have the table of values: v(1, 0) = 3 u(1, 0) = 2 us (1, 0) = −2 vs (1, 0) = 5 Fu (2, 3) = −1 ut (1, 0) = 6 vt (1, 0) = 4 Fv (2, 3) = 10 Find Ws (1, 0) and Wt (1, 0). SOLUTION: Nice problem! Go directly from our notation into the numbers. First, Ws = Fu us + Fv vs ⇒ Ws (1, 0) = (−1)(−2) + (10)(5) = 52 And Wt = Fu ut + Fv vt ⇒ Wt (1, 0) = (−1)(6) + (10)(4) = 34 35. The temperature at a point (x, y) is T (x, y) measured in degrees Celsius. A bug crawls so taht its position after t seconds is x= √ 1+t 1 y =2+ t 3 where x, y are measured in centimeters. The temperature function satisfies Tx (2, 3) = 4 and Ty (2, 3) = 3. How fast is the temperature rising on the bug’s path after 3 seconds? SOLUTION: √ We should interpret the question to mean: Find dT /dt when t = 3, which means x = 4 = 2 and y = 2 + 1 = 3: dT = Tx (2, 3) x0 (3) + Ty (2, 3) y 0 (3) dt Side note: In this case, x and y were functions of one variable only, so the notation x0 is meaningful here, and it means to differentiate in time. We only need x0 (t) = 21 (1 + t)−1/2 , so x0 (3) = 1 4 and y 0 (t) = 13 . Substitute: dT 1 1 =4· +3· =2 dt 4 3 In terms of the bug, the temperature is rising at an (instantaneous) rate of 2◦ C per second at t = 3. 38. The radius of a right circular cone is increasing at a rate of 1.8 inches per second, while its height is decreasing at a rate of 2.5 inches per second. At what rate is the volume of the cone changing when the radius is 120 inches and the height is 140 inches. SOLUTION: The volume of the cone (see the front notes of the book) is: 1 V (r, h) = πr2 h 3 Where we think of r and h as functions of time (so then volume is also a function of time). Therefore, dV dr dh = Vr · + Vh dt dt dt We compute each of these quantities and then substitute: Vr = (2/3)πrh Vr (120, 140) = 11, 200π Vh = (1/3)πr2 Vh (120, 140) = 4, 800π dV = 11200π · 1.8 + 4800π(−2.5) = 8160π in3 /sec dt 39. A box is given, and at a certain instant the dimensions are l = 1, and w = h = 2. Quantities l, w are decreasing at a rate of 2 meters per second (everything is in meters), while h is decreasing at a rate of 3 meters per second. At that instant find the rates at which the following are changing: (a) Volume SOLUTION: The relationship is V = lwh, where l, w, h are functions of time: dV dl dw dh = Vl + Vw + Vh = whl0 + lhw0 + lwh0 = 6 dt dt dt dt (b) Surface Area: SOLUTION: A = 2(lw + lh + wh), and dA = Al l0 + Aw w0 + Ah h0 = 10 dt (c) The length of a diagonal (HINT: It is easier if we think of each of these as functions of time): L(t)2 = l(t)2 + w(t)2 + h(t)2 We want dL/dt: 2L dl dw dh dL = 2l + 2w + 2h =0 dt dt dt dt 48. If z = f (x, y), x = s + t and y = s − t, show that zx2 − zy2 = zs zt Compute the partials of z with respect to s and t, then simplify the right-hand side of the equation: ∂z ∂z ∂x ∂z ∂y = + = zx + zy ∂s ∂x ∂s ∂y ∂s ∂z ∂z ∂x ∂z ∂y = + = zx − zy ∂t ∂x ∂t ∂y ∂t Therefore, zs zt = (zx + zy )(zx − zy ) = zx2 − zy2 49. Show that any function of the form z = f (x − at) + g(x + at) is a solution of the “wave equation”, 2 ztt2 = a2 zxx Hint: Let u = x + at, v = x − at. SOLUTION: Following the hint, we take z to be a function of u and v, z = f (u) + g(v), u = x + at v = x − at Therefore, treat f, g as functions of u, v, and differentiate using the chain rule. We recall that ztt = ftt + gtt zxx = fxx + gxx so we’ll compute these separately and put them together at the end. We’ll also go ahead and compute the partials of u and v: ut = a ux = 1 vt = −a vx = 1 First, the derivatives with respect to t: ft = ∂f ∂u ∂f ∂v + = fu (a) + (0)(a) = afu ∂u ∂t ∂v ∂t Where fv = 0 since f is a function of u only. Differentiating again, ∂fu ∂u ∂fu ∂v ftt = (afu )t = a + ∂u ∂t ∂v ∂t ! = a(fuu (a) + 0) = a2 fuu Do the same thing with g(v) (now gu = 0): gt = ∂g ∂u ∂g ∂v + = 0 + gv (−a) = −agv ∂u ∂t ∂v ∂t Differentiating again, ∂gv ∂u ∂gv ∂v + gtt = (−agv )t = −a ∂u ∂t ∂v ∂t ! = −a(0 − agvv ) = a2 gvv So far we have shown that: ztt = ftt + gtt = a2 (fuu + gvv ) To finish this exercise, show that a2 zxx is actually the right hand side of the equation above. Start with fx , and gx then compute the second derivatives (its gets faster with practice!). Here is some shortcut notation that makes it a bit easier: fx = fu ux + fv vx = fu + 0 = fu fxx = (fu )x = (fu )u ux + (fu )v vx = fuu + 0 = fuu And gx = gu ux + gv vx = 0 + gv = gv gxx = (gv )x = (gv )u ux + (gv )v vx = 0 + gvv = gvv Therefore, zxx = fuu + gvv , which is what we wanted.