Question

the volume v = \frac{4}{3}\pi r^{3} of a spherical balloon changes with the radius.
a. at what rate (in^{3}/in) does the volume change with respect to the radius when r = 11 in?
b. using the rate from part a, by approximately how much does the volume increase when the radius changes from 11 to 11.1 in?
a. at what rate (in^{3}/in) does the volume change with respect to the radius when r = 11 in?
\square in^{3}/in
(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 $\frac{d}{dr}(x^{n})=nx^{n - 1}$, we get $\frac{dV}{dr}=4\pi r^{2}$.

Step2: Evaluate at $r = 11$

Substitute $r = 11$ into $\frac{dV}{dr}$. So, $\frac{dV}{dr}\big|_{r = 11}=4\pi(11)^{2}=484\pi$ $\text{in}^3/\text{in}$.

Step3: Use differential approximation for part b

The differential $dV=\frac{dV}{dr}dr$. We know $\frac{dV}{dr}=484\pi$ from part a, and $dr=11.1 - 11=0.1$ in. Then $dV = 484\pi\times0.1 = 48.4\pi$ $\text{in}^3$.

Answer:

a. $484\pi$
b. $48.4\pi$