A water tank has the shape of an inverted right circular cone with base radius 3 meters and height 6 meters. Water is being pumped into the tank at the rate of 12 meters3/sec. Find the rate, in meters/sec, at which the water level is rising when the water is 2 meters deep. Give 2 decimal places for your answer. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Respuesta :

Answer:

the rate of change is 3.8m/s

Step-by-step explanation:

The volume of the right circular cone is

[tex]V= \frac{\pi*r^2h}{3}.[/tex]

The rate of change of volume is

[tex]\frac{dV}{dt} =\frac{d}{dt}( \frac{\pi*r^2h}{3})=\frac{\pi}{3} \frac{d}{dt}(r^2h)=\frac{\pi}{3}(2rh\frac{dr}{dt}+r^2\frac{dh}{dt})[/tex]

In order to proceed further we have to define r in terms of h so tht we can compute the derivative above.

The ratio between h and r is

[tex]\frac{h}{r}=\frac{6}{3} =2[/tex]

Therefore [tex]r=\frac{h}{2}[/tex].

We plug that into the derivative above and get:

[tex]2rh\frac{dr}{dt}+r^2\frac{dh}{dt}=\frac{h^2}{2}\frac{dh}{dt}+\frac{h^2}{4}\frac{dh}{dt}=\frac{3h^2}{4}\frac{dh}{dt}.[/tex]

Thus

[tex]\frac{dV}{dt}=\frac{\pi}{3}\frac{3h^2}{4}\frac{dh}{dt}[/tex]

Now for the numerical part.

The rate of change of volume [tex]\frac{dV}{dt}[/tex] is [tex]12\frac{m^3}{s}[/tex], so when the water is 2 meters deep [tex]h=2[/tex], therefore:

[tex]12=\frac{\pi*3*2^2}{4*3} \frac{dh}{dt} \\\\\therefore \boxed{ \frac{dh}{dt}=3.8m/s.}[/tex]

Answer:

3.82 meters/sec

Step-by-step explanation:

The guy above is correct he just forgot to put the third decimal place