Question

question 27 - 1 point a sphere is partially filled with air. the radius of the sphere is 8 cm. if the volume of the sphere is decreasing at a rate of 534 cubic cm per minute, what is the rate, in cm per minute, at which the radius of the sphere is changing when the radius is 2 cm? remember that the volume of a sphere is v = 4/3πr³. round your answer to the nearest hundredth. (do not include any units in your answer.) provide your answer below.

Explanation:

Step1: Differentiate volume formula

$V=\frac{4}{3}\pi r^{3}$, so $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$

Step2: Substitute values

Given $\frac{dV}{dt}=- 534$, $r = 8$, solve for $\frac{dr}{dt}$: $-534=4\pi(8)^{2}\frac{dr}{dt}$

Step3: Calculate $\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{-534}{4\pi\times64}\approx - 0.66$

Answer:

$-0.66$