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MATH 2004 Homework Solution Han-Bom Moon Homework 7 Model Solution Section 14.4. 14.4.5 Find an equation of the tangent plane to z = x sin(x + y) at (−1, 1, 0). f (x, y) = x sin(x + y) fx = sin(x + y) + x cos(x + y), fy = x cos(x + y) f (−1, 1) = (−1) sin(−1 + 1) = 0, fx (−1, 1) = sin(−1+1)+(−1) cos(−1+1) = −1, fy (−1, 1) = (−1) cos(−1+1) = −1 Tangent plane: z = f (−1, 1) + fx (−1, 1)(x − (−1)) + fy (−1, 1)(y − 1) z = −(x + 1) − (y − 1) or x+y+z =0 14.4.6 Find an equation of the tangent plane to z = ln(x − 2y) at (3, 1, 0). f (x, y) = ln(x − 2y) fx = 1 , x − 2y fy = 1 2 · (−2) = − x − 2y x − 2y f (3, 1) = ln(3 − 2 · 1) = ln 1 = 0 fx (3, 1) = 1 = 1, 3−2·1 fy (3, 1) = − 2 = −2 3−2·1 Tangent plane: z = f (3, 1) + fx (3, 1)(x − 3) + fy (3, 1)(y − 1) z = (x − 3) − 2(y − 1) or x − 2y − z = 1 1 MATH 2004 Homework Solution Han-Bom Moon 14.4.11 Explain why f (x, y) = 1 + x ln(xy − 5) is differentiable at (2, 3). Then find the linearization L(x, y) of the function at that point. fx = ln(xy − 5) + x · fy = 1 xy · y = ln(xy − 5) + xy − 5 xy − 5 x x2 ·x= xy − 5 xy − 5 At (2, 3), note that the denominator xy − 5 of fractional parts is nonzero. So both fx and fy are continuous because they are sums and compositions of continuous functions. Therefore f (x, y) is differentiable at (2, 3). f (2, 3) = 1 + 3 ln(2 · 3 − 5) = 1 + 3 ln 1 = 1 fx (2, 3) = ln(2 · 3 − 5) + 2·3 6 = ln 1 + = 6 2·3−5 1 22 =4 2·3−5 L(x, y) = f (2, 3) + fx (2, 3)(x − 2) + fy (2, 3)(y − 3) = 1 + 6(x − 2) + 4(y − 3) p 14.4.21 Find the linear approximation of the function f (x, y, z) = x2 + y 2 + z 2 at p (3, 2, 6) and use it to approximate the number (3.02)2 + (1.97)2 + (5.99)2 . fy (2, 3) = 1 f (x, y, z) = (x2 + y 2 + z 2 ) 2 1 x 1 fx = (x2 + y 2 + z 2 )− 2 · 2x = p 2 x2 + y 2 + z 2 Similarly, y z fy = p , fz = p . x2 + y 2 + z 2 x2 + y 2 + z 2 p √ f (3, 2, 6) = 32 + 22 + 62 = 49 = 7 3 3 = 2 2 7 +2 +6 2 2 fy (3, 2, 6) = √ = 2 2 2 7 3 +2 +6 6 6 fz (3, 2, 6) = √ = 7 32 + 2 2 + 6 2 fx (3, 2, 6) = √ 32 L(x, y, z) = f (3, 2, 6) + fx (3, 2, 6)(x − 3) + fy (3, 2, 6)(y − 2) + fz (3, 2, 6)(z − 6) 2 6 3 = 7 + (x − 3) + (y − 2) + (z − 6) 7 7 7 p Approximation of (3.02)2 + (1.97)2 + (5.99)2 = f (3.02, 1.97, 5.99) is 3 2 6 L(3.02, 1.97, 5.99) = 7 + (3.02 − 3) + (1.97 − 2) + (5.99 − 6) 7 7 7 0.06 −0.06 −0.06 0.06 = 7+ + + =7− ; 6.99142857 7 7 7 7 2 MATH 2004 Homework Solution Han-Bom Moon 14.4.29 Find the differential of R = αβ 2 cos γ. ∂R = β 2 cos γ ∂α ∂R = 2αβ cos γ ∂β ∂R = −αβ 2 sin γ ∂γ ∂R ∂R ∂R dα + dβ + dγ ∂α ∂β ∂γ = β 2 cos γ dα + 2αβ cos γ dβ − αβ 2 sin γ dγ dR = 14.4.32 If z = x2 − xy + 3y 2 and (x, y) changes from (3, −1) to (2.96, −0.95), compare the values of ∆z and dz. ∆z = (2.96)2 − (2.96)(−0.95) + 3(−0.95)2 − 32 − 3(−1) + 3(−1)2 = −0.7189 ∂z = 2x − y, ∂x ∂z = −x + 6y ∂y dz = (2x − y) dx + (−x + 6y) dy x = 3, y = −1, dx = 2.96 − 3 = −0.04, dy = −0.95 − (−1) = 0.05 dz = (2 · 3 − (−1))(−0.04) + (−3 + 6 · (−1))(0.05) = −0.73 14.4.33 The length and width of a rectangle are measured ad 30 cm and 24 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle. Length = l, width = w ⇒ area = A = wl dA = w dl + l dw l = 30, w = 24, |dl| ≤ 0.1, |dw| ≤ 0.1 |dA| = |w dl + l dw| ≤ w|dl| + l|dw| ≤ 30 · 0.1 + 24 · 0.1 = 5.4 Approximation of the maximum error = 5.4 cm2 14.4.41 A model for the surface area of a human body is given by S = 0.1091w0.425 h0.725 , where w is the weight (in pounds), h is the height (in inches), and S is measured in square feet. If the errors in measurement of w and h are at most 2%, use differentials to estimate the maximum percentage error in the calculated surface area. ∂S ∂S dw + dh ∂w ∂h = (0.1091)(0.425)w−0.575 h0.725 dw + (0.1091)(0.725)w0.425 h−0.275 dh dS = = 0.0463675w−0.575 h0.725 dw + 0.0790975w0.425 h−0.275 dh 3 MATH 2004 Homework Solution Han-Bom Moon |dw| ≤ 0.02w, |dh| ≤ 0.02h |dS| = |0.0463675w−0.575 h0.725 dw + 0.0790975w0.425 h−0.275 dh| ≤ 0.0463675w−0.575 h0.725 0.02w + 0.0790975w0.425 h−0.275 0.02h = 0.0025093w0.425 h0.725 0.0025093w0.425 h0.725 |dS| = = 0.023 S 0.1091w0.425 h0.725 Therefore the approximation of the maximum percentage error in the calculated surface area is 2.3%. 4