Chapter 15 - Differentiation in Several Variables - 15.4 Differentiability and Tangent Planes - Exercises - Page 790: 35

(a) Using the linear approximation, Eq. (5) we show that $\frac{{\Delta V}}{V} \approx \frac{{2\Delta r}}{r} + \frac{{\Delta h}}{h}$ (b) The percentage increase in $V$ is $6\%$ if $r$ and $h$ are each increased by $2\%$. (c) A $1\%$ error in $r$ leads to a greater error in $V$ than a $1\%$ error in $h$.

(a) The volume of a cylinder of radius $r$ and height $h$ is given by $V = \pi {r^2}h$. So, the partial derivatives with respect to $r$ and $h$ are $\frac{{\partial V}}{{\partial r}} = 2\pi rh$ ${\ \ }$ and ${\ \ }$ $\frac{{\partial V}}{{\partial h}} = \pi {r^2}$, respectively. Using the linear approximation, Eq. (5) the change in $V$ is estimated to be $\Delta V \approx \frac{{\partial V}}{{\partial r}}\Delta r + \frac{{\partial V}}{{\partial h}}\Delta h$ Substituting $\frac{{\partial V}}{{\partial r}}$ and $\frac{{\partial V}}{{\partial h}}$ in $\Delta V$ gives $\Delta V \approx 2\pi rh\Delta r + \pi {r^2}\Delta h$ Since $V = \pi {r^2}h$, so $2\pi rh = \frac{{2V}}{r}$ and $\pi {r^2} = \frac{V}{h}$. Therefore, $\Delta V \approx \frac{{2V}}{r}\Delta r + \frac{V}{h}\Delta h$ Hence, $\frac{{\Delta V}}{V} \approx \frac{{2\Delta r}}{r} + \frac{{\Delta h}}{h}$ (b) From part (a) we obtain $\frac{{\Delta V}}{V} \approx \frac{{2\Delta r}}{r} + \frac{{\Delta h}}{h}$ If $r$ and $h$ are each increased by $2\%$, we have $\frac{{\Delta r}}{r} = \frac{{\Delta h}}{h} = 0.02$ So, $\frac{{\Delta V}}{V} \simeq 2\cdot0.02 + 0.02 = 0.06$ Thus, the percentage increase in $V$ is $6\%$ if $r$ and $h$ are each increased by $2\%$. (c) From part (a) we obtain $\frac{{\Delta V}}{V} \approx \frac{{2\Delta r}}{r} + \frac{{\Delta h}}{h}$ 1. Suppose there is no error in $h$ but a $1\%$ error in $r$. Then $\frac{{\Delta r}}{r} = 0.01$ ${\ \ }$ and ${\ \ }$ $\Delta h = 0$ $\frac{{\Delta V}}{V} \approx 2\cdot0.01 = 0.02$ Thus, the estimated error in $V$ is $2\%$. 2. Suppose there is no error in $r$ but a $1\%$ error in $h$. Then $\frac{{\Delta h}}{h} = 0.01$ ${\ \ }$ and ${\ \ }$ $\Delta r = 0$ $\frac{{\Delta V}}{V} \approx 0.01$ Thus, the estimated error in $V$ is $1\%$. From these results we conclude that a $1\%$ error in $r$ leads to a greater error in $V$ than a $1\%$ error in $h$.