Chapter 2 - Derivatives - 2.9 Linear Approximations and Differentials - 2.9 Exercises - Page 193: 31

(a) If $x$ is the edge length, then $V$ = $x^{3}$ $dV$ = $3x^{2}dx$ when $x$ = $30$ and $dx$ = $0.1$ $dV$ = $3(30)^{2}(0.1)$ = $270$ so the maximum possible error in computing the volume of the cube is about $270$ $cm^{3}$ The relative error is calculated by dividing the change in $V$, $ΔV$ by $V$. we approximate $ΔV$ with $dV$ Relative error = $\frac{ΔV}{V}$ $\approx$ $\frac{dV}{V}$ = $\frac{3x^{2}dx}{x^{3}}$ = $3\frac{dx}{x}$ = $3(\frac{0.1}{30})$ = $0.01$ Percentage error = relative error $\times$ $100%$ = $0.01\times100$% = $1$% (b) $S$ = $6x^{2}$ $dS$ = $12xdx$ when $x$ = $30$ and $dx$ = $0.1$ $dS$ = $12(30)(0.1)$ = $36$ so the maximum possible error in computing the surface area of the cube is about $36$ $cm^{2}$ Relative error = $\frac{ΔS}{S}$ $\approx$ $\frac{dS}{S}$ = $\frac{12xdx}{6x^{2}}$ = $2\frac{dx}{x}$ = $2(\frac{0.1}{30})$ = $0.006$ Percentage error = relative error $\times$ $100%$ = $0.006\times100$% = $0.6$%

(a) If $x$ is the edge length, then $V$ = $x^{3}$ $dV$ = $3x^{2}dx$ when $x$ = $30$ and $dx$ = $0.1$ $dV$ = $3(30)^{2}(0.1)$ = $270$ so the maximum possible error in computing the volume of the cube is about $270$ $cm^{3}$ The relative error is calculated by dividing the change in $V$, $ΔV$ by $V$. we approximate $ΔV$ with $dV$ Relative error = $\frac{ΔV}{V}$ $\approx$ $\frac{dV}{V}$ = $\frac{3x^{2}dx}{x^{3}}$ = $3\frac{dx}{x}$ = $3(\frac{0.1}{30})$ = $0.01$ Percentage error = relative error $\times$ $100%$ = $0.01\times100$% = $1$% (b) $S$ = $6x^{2}$ $dS$ = $12xdx$ when $x$ = $30$ and $dx$ = $0.1$ $dS$ = $12(30)(0.1)$ = $36$ so the maximum possible error in computing the surface area of the cube is about $36$ $cm^{2}$ Relative error = $\frac{ΔS}{S}$ $\approx$ $\frac{dS}{S}$ = $\frac{12xdx}{6x^{2}}$ = $2\frac{dx}{x}$ = $2(\frac{0.1}{30})$ = $0.006$ Percentage error = relative error $\times$ $100%$ = $0.006\times100$% = $0.6$%