# Question 497d1

Aug 5, 2015

$\frac{111}{250} \pi {r}^{3}$

#### Explanation:

$V = \frac{4}{3} \pi {r}^{3}$
$\implies \frac{\mathrm{dV}}{\mathrm{dr}} = \frac{4}{\cancel{3}} \pi \cancel{3} {r}^{2} = 4 \pi {r}^{2}$

We know that the change in $r$ is 10%, so $\delta r = \frac{r}{10}$.

Therefore, as a linear approximation, we may use:
$\delta V = 4 \pi {r}^{2} \delta r = 2 \cancel{4} \pi {r}^{2} \frac{r}{5 \cancel{10}} = \frac{2}{5} \pi {r}^{3}$.

Of course, you may also calculate the exact increase by substitution of $r + \delta r$ into the formula directly.

$t i l \mathrm{de} V = \frac{4}{3} \pi {\left(r + \delta r\right)}^{3} = \frac{4}{3} \pi {r}^{3} + \frac{4}{3} \pi \left(3 {r}^{2} \delta r + 3 r {\left(\delta r\right)}^{2} + {\left(\delta r\right)}^{3}\right)$
$\implies t i l \mathrm{de} V = {V}_{0} + 2 \cancel{4} \pi {r}^{2} \frac{r}{5 \cancel{10}} + \cancel{4} \pi r {r}^{2} / \left(25 \cancel{100}\right) + \frac{\cancel{4}}{3} \pi {r}^{3} / \left(250 \cancel{1000}\right)$
$\implies \delta V = \left(\frac{2}{5} + \frac{1}{25} + \frac{1}{250}\right) \pi {r}^{3} = \frac{111}{250} \pi {r}^{3}$
Of course, this is very close to the approximation we get using the differential.

Aug 5, 2015

The amount of the increase is: $0.331$ times the original volume

The proportion or percent increase is: 0.331 = 33.1%

#### Explanation:

${V}_{1} = \frac{4}{3} \pi {r}^{3}$

When the radius increases 10%, the new radius is $1.1 r$ so the new volume is:

${V}_{2} = \frac{4}{3} \pi {\left(1.1 r\right)}^{3} = \frac{4}{3} \pi \left(1.331\right) {r}^{3}$

$= 1.331 \left(\frac{4}{3} \pi {r}^{3}\right) = 1.331 {V}_{1}$

The amount of the increase is:

${V}_{2} - {V}_{1} = 0.331 {V}_{1}$

The proportion or percent increase is:

(V_2-V_1)/V_1 = 0.331 = 33.1%#