University:
Whitman College
Course:
MATH-225 | Calculus III
Academic year:

2020
Views:

141

Pages:

3
Author:

cebolaskj4

SOLUTIONS: Section 14.4, “Story Problems” 34. The dimensions of a closed rectangular box are measured as 80 cm, 60 cm, and 50 cm, with a possible error or 0.2 cm in each dimension. Use differentials to estimate the maximum error in calculating the surface area of the box. SOLUTION: Let the dimensions of the box be l, w and h (for length, width and height). The surface area is then: S(l, w, h) = 2lw + 2wh + 2lh = 2(lw + wh + lh) The change in area can be written as: ∆S ≈ dS = Sl dl + Sw dw + Sh dh where the partial derivatives are evaluated at l = 80, w = 60 and h = 50, and dl = dw = dh = 0.2. The partial derivatives are computed: Sl = 2(w + h) = 220 Sw = 2(l + h) = 260 Sh = 2(l + w) = 280 Substituting these in for dS, dS = 220 · 0.2 + 260 · 0.2 + 280 · 0.2 = 152 cm2 35. 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. SOLUTION: The volume of the can is V (r, h) = πr2 h Using differentials, ∆V ≈ dV = Vr dr + Vh dh with r = 4 and h = 12, dr = 0.04 and dh = 2 · 0.04 = 0.08. Compute the partial derivatives out and we get: dV = 2πrh dr + πr2 dh Substitute in the numerical values: dV = 3.84π + 1.28π ≈ 16.08 cm3 (The book rounds off a little too much) 36. Use differentials to estimate the amount of metal in a closed cylindrical 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. SOLUTION: Same volume equation as before (V = πr2 h): dV = 2πrh dr + πr2 h dh In this case, r = 2, h = 10, dr = 0.05 and dh = 2 · 0.1 = 0.2. Substituting these values in, we get: dV = 2.80π ≈ 8.8 cm3 37. A boundary stripe is 3 inches wide and is painted around a rectangle whose dimensions are 100 × 200 feet. Use differentials to approximate the number of square feet of paint in the stripe. SOLUTION: Use the area: A = xy, so that ∆A ≈ dA = Ax dx + Ay dy = y dx + x dy Notice that the change in x will be twice the width of a stripe (in feet), so dx = dy = 6 in = 1/2 ft Substituting these in, dA = 100 · 1 1 + 200 · = 150 ft2 2 2 38. Given that pressure, volume and temperature of a gas are related by: P V = 8.31T use differentials to find the approximate change in pressure if the volume increases from 12L to 12.3L and temperature decreases from 310 Kelvin to 305 Kelvin. SOLUTION: A couple of ways to go about this. Easiest way might be to first write P as a function of T, V since we want to approximate ∆P : P = 8.31 T V ⇒ dP = 8.31 8.31T dT − dV V V2 with T = 310, dT = −5, V = 12 and dV = 0.3. Substitute these in: dP = 8.31 8.31 · 310 · (−5) − · 0.3 ≈ −8.83 12 122 41. A model for the surface area of a human body is given by S = 0.109w0.425 h0.725 where w is weight (in pounds), h is in feet, and S in square feet. If the errors in measuring w and h are at most 2%, use differentials to estimate the maximum percentage error in calculating S.A. SOLUTION: The percentage error is given by: ∆S dS Sw dw + Sh dh ≈ = S S S We are also given: dw = 0.02w and dh = 0.02h. Now, Sw (0.109 · 0.425)w0.425−1 h(0.02)w dw = = 0.425 · 0.02 S S Similarly, 0.109 · 0.725w0.425 h0.725−1 0.02h Sh dh = = 0.725 · 0.02 S S Add them together to get the desired result: 0.023, or about 2.3%. For fun, did you see how much surface area you have?

Calculus III Review Solutions - Exam Solutions

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