related rates using integrals: rotating (36/2401)x^4 about y

[Sendell]

I am given the equation of a curve= (36/(2401))(x^4). Rotating this around the y-axis with the boundaries x = 0 to x = 7 creates a tank-like shape. I am asked to find how fast the depth of the water in the tank is increasing when height = 3 feet.

Now, I assume I should go about this problem as a related rates question. Usually for this sort of problem I would write a volume equation relating the variables. However, the volume equation I have is V = pi(integral[0, h]((2401h)/36)^.5)dh. Should I continue on this path and find dV/dt? Where does dh/dt come in?

Thanks.

 

[soroban]

Re: related rates using integrals

Hello, Sendell!

Your approach is correct . . .

I am given the equation of a curve: \(\displaystyle \,y \:=\:\frac{36}{2401}x^4\;\;\)Rotating this around the y-axis
with the boundaries \(\displaystyle x = 0\) to \(\displaystyle x = 7\) creates a tank-like shape.
Find how fast the depth of the water in the tank is increasing when height = 3 feet.

We need to know \(\displaystyle \frac{dV}{dt}\), the rate at which the tank is being filled.
Then we must express \(\displaystyle V\) as a function of \(\displaystyle h\), the height of the water.
Once we have \(\displaystyle V\:=\:f(h)\), we differentiate with respect to time: \(\displaystyle \,\frac{dV}{dt}\:=\:f'(h)\cdot\frac{dh}{dt}\)

The side view of the tank looks like this:

                      |
        *-----------36+-------------*
                      |
         * - - - - - h+- - - - - - *
          * ::::::::::|:::::::::::*
            *:::::::::|:::::::::*
                *:::::|:::::*
      - - - - - - - -***- - - - - - - -
                      |

Using "disks" and integrating with respect to \(\displaystyle y\), the volume is:
. . \(\displaystyle \L V \;=\;\pi\int^{\;\;\;h}_0 x^2\,dy\)

Since \(\displaystyle \L x\:=\:\frac{7}{\sqrt{6}}y^{\frac{1}{4}}\;\) we have: \(\displaystyle \L\:V\;=\;\pi\int^{\;\;\;h}_0\frac{49}{6}y^{\frac{1}{2}}\,dy\)

And we're on our way . . .

 

[Sendell]

Thank you very much for the help. Just a quick question, though... Is the derivative of the integral dV/dt = (49/6)(h^.5)? You said something about how dV/dt = f'(h)xdh/dt. Would my equation then be dV/dt = ((49/6)(h^.5))(dh/dt)?

 

[Sendell]

This question is for anyone: Taking what has been stated above into account, does dV/dt = ((49/6)(h^.5))(dh/dt)?

 

[skeeter]

yes, but don't forget the \(\displaystyle \L \pi\) in front.