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 $r = 3$ ft?
b. using the rate from part a, by approximately how much does the volume increase when the radius changes from 3 to 3.3 ft?
a. at what rate ($ft^{3}/ft$) does the volume change with respect to the radius when $r = 3$ ft?
$\square ft^{3}/ft$
(type an exact answer in terms of $\pi$.)

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 $\frac{d}{dr}(r^{n})=nr^{n - 1}$, we get $\frac{dV}{dr}=4\pi r^{2}$.

Step2: Evaluate the derivative at $r = 3$

Substitute $r = 3$ into $\frac{dV}{dr}$. So, $\frac{dV}{dr}\big|_{r = 3}=4\pi(3)^{2}=36\pi$.

Step3: Approximate the volume change

The differential $dV=\frac{dV}{dr}dr$. We know from part a that $\frac{dV}{dr}=36\pi$ when $r = 3$, and $dr=3.3 - 3=0.3$. Then $dV=(36\pi)\times0.3 = 10.8\pi$.

Answer:

a. $36\pi$
b. $10.8\pi$