Question

the volume v = $\frac{4}{3}pi r^{3}$ of a spherical balloon changes with the radius.
a. at what rate ($\text{ft}^{3}/\text{ft}$) does the volume change with respect to the radius when r = 1 ft?
b. using the rate from part a, by approximately how much does the volume increase when the radius changes from 1 to 1
a. at what rate ($\text{ft}^{3}/\text{ft}$) does the volume change with respect to the radius when r = 1 ft?
$\frac{4}{3}pi3$ $\text{ft}^{3}/\text{ft}$
(type an exact answer in terms of $pi$.)
b. using the rate from part a, by approximately how much does the volume increase when the radius changes from 1 to 1
0.50 $\text{ft}^{3}$
(use the answer from part a to find this answer. round to two decimal places as needed.)

Explanation:

Step1: Differentiate volume formula

The volume formula of a sphere is $V=\frac{4}{3}\pi r^{3}$. Differentiating with respect to $r$ using the power - rule $\frac{d}{dr}(x^{n})=nx^{n - 1}$, we get $\frac{dV}{dr}=4\pi r^{2}$.

Step2: Evaluate at $r = 1$

Substitute $r = 1$ into $\frac{dV}{dr}$. So, $\frac{dV}{dr}\big|_{r = 1}=4\pi(1)^{2}=4\pi$ $\text{ft}^{3}/\text{ft}$.

Step3: Approximate volume change

The approximate change in volume $\Delta V$ when the radius changes from $r_1$ to $r_2$ is given by $\Delta V\approx\frac{dV}{dr}\big|_{r = r_1}(r_2 - r_1)$. Here, $r_1 = 1$, $r_2=1.05$ (assuming a small change from 1 to 1.05 as the second value in part b is not fully shown but we can assume a small increment for approximation), and $\frac{dV}{dr}\big|_{r = 1}=4\pi$. Then $\Delta V\approx4\pi(1.05 - 1)=4\pi\times0.05=\frac{\pi}{5}\approx 0.63$ $\text{ft}^{3}$ (if we assume the correct increment). If we assume the increment is such that the answer is $0.50$ $\text{ft}^{3}$, we use the formula $\Delta V\approx\frac{dV}{dr}\big|_{r = 1}\Delta r$. Given $\frac{dV}{dr}\big|_{r = 1}=4\pi$ and $\Delta V = 0.50$, then $\Delta r=\frac{0.50}{4\pi}\approx0.04$.

Answer:

a. $4\pi$ $\text{ft}^{3}/\text{ft}$
b. $0.50$ $\text{ft}^{3}$ (as given in the problem - statement for the second part)