Question

A cone has a radius of 5 ft and a height of 15 ft. It is empty and is being filled with water at a constant rate of 15 15 ft 3 ft3 / sec sec . Find the rate of change of the radius of the surface of the water when the radius of the surface of the water is 2 2 ft. (You must also include the units)

Answer 1

Answer:

$\frac{dr}{dt}=0.11062ft/sec$

Step-by-step explanation:

From the question we are told that:

Radius $r=5ft$

Height $H=15ft$

Rate $R=15ft/3sec =5ft/s$

Surface Radius $R_{surf}=2.2ft$

Generally the equation for Volume is mathematically given by

 $V=\frac{1}{3}\pi*r^2h$

Since radius to height ratio gives

 $\frac{r}{h}=\frac{5}{15}$

 $\frac{r}{h}=\frac{1}{3}$

 $h=3r$

Therefore

 $V=\frac{1}{3}\pi*r^2(3r)$

 $V=\pi r^3$

Generally the equation for Change of Volume is mathematically given by

 $\frac{dv}{dt}=\pi \frac{d}{dt}(r^3)$

 $\frac{dv}{dt}=\pi 3*r^2 \frac{dr}{dt}$

 $\frac{dv}{dt}=\pi 3*(2.2)^2 \frac{dr}{dt}$

 $\frac{dr}{dt}=\frac{5}{45.62}$

 $\frac{dr}{dt}=0.11062ft/sec$