Question

Water runs into a conical tank at the rate of 9 ft 3/min. the tank is standing, inverted, and has a height of 10 feet and a base diameter of 10 feet. At what rate is the radius of the water in the tank increasing when the radius is two feet

Answer 1

Answer:

0.36ft/min

Step-by-step explanation:

We are given that

$\frac{dv}{dt}=9ft^3/min$

Diameter of  tank,d=10ft

Radius,$r=\frac{d}{2}=\frac{10}{2}=5ft$

Height of tank,h=10 ft

We have to find the rate at which radius of the water in the tank increasing when r=2 ft

$\frac{h}{r}=\frac{10}{5}=2$

$h=2r$

Volume of conical tank=$V=\frac{1}{3}\pi r^2 h$

Substitute the values

$V=\frac{1}{3}\pi r^2(2r)=\frac{2}{3}\pi r^3$

Differentiate w.r.t t

$\frac{dV}{dt}=2\pi r^2\frac{dr}{dt}$

Substitute the values

$9=2\pi(2)^2\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{9}{2\pi(2)^2}=\frac{9}{8\pi}=0.36ft/min$