Exercise 8.8 — Book-Back Answer Key
Exercise 8.8 — Book-Back Answer Key
Each answer below was solved independently; all fifteen agree with the official key. The single best option is given with a one-line reason.
- Option 2 — 0.4%. Area error is twice the radius error: \( 2 \times \tfrac{0.02}{10} \times 100 = 0.4\% \).
- Option 2 — \( \tfrac{1}{5} \). An \(n\)th root scales the percentage error by \( \tfrac{1}{n} \); here \( n = 5 \).
- Option 2 — \( 2xu \). \( \partial_{x}e^{x^{2}+y^{2}} = 2x\,e^{x^{2}+y^{2}} = 2xu \).
- Option 4 — 1. The two partials are \( \tfrac{e^{x}}{e^{x}+e^{y}} \) and \( \tfrac{e^{y}}{e^{x}+e^{y}} \); they sum to 1.
- Option 3 — \( y x^{y-1} \). With \(y\) fixed, \( x^{y} \) differentiates by the power rule.
- Option 2 — \( (1+xy)e^{xy} \). \( \partial_{y} = x e^{xy} \), then \( \partial_{x} \) of that gives \( (1+xy)e^{xy} \).
- Option 4 — 4.8 cu.cm. \( dV = 3x^{2}\,dx = 3(16)(0.1) = 4.8 \).
- Option 2 — \( 12x_{0}\,dx \). \( dS = 12x\,dx \) evaluated at \( x_{0} \).
- Option 4 — \( 0.03x^{3}\ \text{m}^{3} \). \( dV = 3x^{2}\,dx \) with \( dx = 0.01x \).
- Option 1 — \( 6e^{2t} + 5\sin t - 4\cos t\sin t \). Chain rule: \( 6x\,e^{t} + (-5+4y)(-\sin t) \) at \( x=e^{t},\,y=\cos t \).
- Option 2 — \( \tfrac{1}{(x+1)^{2}}\,dx \). Quotient rule gives \( f'(x) = \tfrac{1}{(x+1)^{2}} \).
- Option 3 — \( -7 \). \( \partial_{x}u = 2x + 3y = 8 - 15 = -7 \) at \( (4,-5) \).
- Option 2 — \( -x + \tfrac{\pi}{2} \). \( g(\tfrac{\pi}{2})=0,\ g'(\tfrac{\pi}{2})=-1 \), so \( L(x) = -(x - \tfrac{\pi}{2}) \).
- Option 4 — 0. All quadratic and cross terms cancel in \( w_{x} + w_{y} + w_{z} \).
- Option 1 — \( z - x \). \( f_{x} = y + z,\ f_{z} = y + x \), so \( f_{x} - f_{z} = z - x \).