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kirza4 [7]
3 years ago
11

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
Mathematics
1 answer:
Phoenix [80]3 years ago
8 0

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

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