Question

the volume $v = \frac{4}{3}\pi r^{3}$ of a spherical balloon changes with the radius.
a. at what rate ($ft^{3}/ft$) 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 1.2 ft?
a. at what rate ($ft^{3}/ft$) does the volume change with respect to the radius when $r = 1$ ft?
$\frac{4}{3}\pi3$ $ft^{3}/ft$
(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 1.2 ft?
$2.51$ $ft^{3}$
(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: Use linear approximation

The linear approximation formula is $\Delta V\approx\frac{dV}{dr}\Delta r$. Here, $\frac{dV}{dr}=4\pi$ (from part a) and $\Delta r=1.2 - 1=0.2$. Then $\Delta V\approx4\pi\times0.2 = 0.8\pi\approx2.51$ $\text{ft}^{3}$.

Answer:

a. $4\pi$ $\text{ft}^{3}/\text{ft}$
b. $2.51$ $\text{ft}^{3}$