The volume is given by:
$V=\frac{1}{3}\pi r^2h$
You are given $\frac{dV}{dt}$, so you will have to take the derivative of the volume function with respect to time. Keep in mind that the radius and the height are NOT constants, they are variables. However, they are proportional:
$\frac{h_1}{r_1}=\frac{h_2}{r_2}$
This is because, if you draw a triangle straight down in the cone, you'll get similar triangles, as I, a true artist, have demonstrated below:
This is true for RIGHT cones.
We are given the cone's actual height and radius, and so we know that:
$\frac{h}{r}=\frac{20}{10}=2$
Hence:
$r=\frac{h}{2}$
I found what r is because I want to use it in the equation because we are seeking for the change in the height over time. Let's plug it into the original equation, then:
$V=\frac{1}{3}\pi (\frac{h}{2})^2h=\frac{1}{12}\pi h^3$
Take the derivative of this with respect to time:
$\frac{dV}{dt}=\frac{1}{4}\pi h^2 * \frac{dh}{dt}$
This by the way always happens, we just never see it. Take for instance:
$y= 5x^2+x$
When you differentiate, the answer is really:
$\frac{dy}{dx}=10x*\frac{dx}{dx}+\frac{dx}{dx}$
But $\frac{dx}{dx} = 1$ so we never show it.
Back to what I was saying, the next step is to determine $\frac{dh}{dt}$, which is what we are trying to solve. So we need the instantaneous height of the cone when it's 1/8 filled. 1/8 filled means the cone's volume is 1/8 of its original volume. Its original volume is (and you can use $V=\frac{1}{3}\pi r^2h$, but I find the function of volume against height only easier since there's only 1 variable):
$V=\frac{1}{12}\pi h^3$
$V_{full}=\frac{1}{12}\pi (20)^3 =$ some value I stored on my calculator.
Multiply that by 1/8 and you get the volume of the tank at that time. Use the equation to find the instantaneous height:
$V_{instantaneous}=\frac{1}{12}\pi h^3$
The height is 10. You can now plug it into the derivative equation we had:
$\frac{dV}{dt}=\frac{1}{4}\pi h^2 * \frac{dh}{dt}$
We are given that $\frac{dV}{dt}=2$, and we now know the instantaneous height. The answer is then:
$\frac{dh}{dt}=\frac{2}{25 \pi} \approx 0.025 \frac{ft}{min}$
Let me know if I made any mistakes before you down vote me into oblivion.
Best Answer
You have a couple of errors: as you suspected, the height isn’t quite right, but the cross-sectional area is also wrong.
The radius of the tank at height $y$ is $\frac25y$, so the cross-sectional area is $\frac{4\pi}{25}y^2$. To raise the cross-sectional slice of thickness $dy$ at height $y$ to the top of the tank, you must raise it $15-y$ feet, doing $$dW=64\cdot\frac{4\pi}{25}y^2(15-y)dy=\frac{256\pi}{25}(15y^2-y^3)dy$$ foot-pounds of work. Thus, it takes
$$\frac{256\pi}{25}\int_0^{10}(15y^2-y^3)dy$$
foot-pounds of work to raise all of the water to the top of the tank.
There are $64\cdot\frac13\cdot16\pi\cdot10=\frac{10240\pi}3$ pounds of water in the tank; they all fall a distance of $115-18=97$ feet after being raised to the top of the tank, doing $97\cdot\frac{10240\pi}3=\frac{993,280\pi}3$ foot-pounds of work in the process. Thus, the net work required is
$$\frac{256\pi}{25}\int_0^{10}(15y^2-y^3)dy-\frac{993,280\pi}3\text{ foot-pounds}\;.$$
You can of course set it up as a single integral: the slice at height $y$ is first raised $15-y$ feet and then dropped $97$ feet, so it is ‘raised’ a total of $15-y-97=-y-82$; your error was getting the wrong algebraic sign on $y$. If you do this, the answer is
$$\frac{256\pi}{25}\int_0^{10}(-y^3-82y^2)dy$$
foot-pounds.