Question

question: an inverted cone has a height of 11 inches and a radius of 18 inches. the volume of the inverted cone is decreasing at a rate of 541 cubic inches per second, with the height being held constant. what is the rate of change of the radius, in inches per second, when the radius is 5 inches? remember that the volume of a cone is v = 1/3πr²h. round your answer to the nearest hundredth. (do not include any units in your answer.) provide your answer below:

Explanation:

Step1: Differentiate volume formula

$V=\frac{1}{3}\pi r^{2}h$, since $h$ is constant, $\frac{dV}{dt}=\frac{2}{3}\pi h r\frac{dr}{dt}$

Step2: Substitute known values

Given $\frac{dV}{dt}=- 541$, $h = 11$, $r = 5$. Then $-541=\frac{2}{3}\pi\times11\times5\times\frac{dr}{dt}$

Step3: Solve for $\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{-541\times3}{2\times\pi\times11\times5}\approx - 4.69$

Answer:

$-4.69$