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?
36\pi 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 3 to 3.3 ft?
ft^{3}
(use the answer from part a to find this answer. round to two decimal places as needed.)

Explanation:

Step1: Recall the derivative formula for volume

The volume of a sphere is $V = \frac{4}{3}\pi r^{3}$. The derivative of $V$ with respect to $r$ is $V'=\frac{dV}{dr}=4\pi r^{2}$.

Step2: Find the rate of change at $r = 3$

Substitute $r = 3$ into $V'$. So $V'(3)=4\pi(3)^{2}=36\pi$ $\text{ft}^3/\text{ft}$. This is the answer for part a.

Step3: Use the linear - approximation formula

The linear - approximation formula is $\Delta V\approx V'(r_0)\Delta r$. Here, $r_0 = 3$, $\Delta r=3.3 - 3=0.3$, and $V'(r_0)=36\pi$. Then $\Delta V\approx36\pi\times0.3$.

Step4: Calculate the value of $\Delta V$

$\Delta V\approx36\pi\times0.3 = 10.8\pi\approx 33.93$ $\text{ft}^3$.

Answer:

33.93