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.
The change in the volume of the water is the difference in the fill and leak rates (taking the leak rate to be a positive number).
$$\frac{dV}{dt} = \frac{dF}{dt} - \frac{dL}{dt}$$
Thus, the leak rate is the difference in the fill rate and the actual rate of increase in the volume of the water.
$$\frac{dL}{dt} = \frac{dF}{dt} - \frac{dV}{dt}$$
The volume of the water when its height is $h$ and its radius is $r$ is
$$V = \frac{1}{3}\pi r^2h$$
Since the inverted conical tank has a diameter of $10~\text{ft}$ at the top, its radius at the top is $5~\text{ft}$.
By similar triangles, the ratio of the radius of the water to the height of the water in the cone is
$$\frac{r}{h} = \frac{5~\text{ft}}{20~\text{ft}} = \frac{1}{4} \implies r = \frac{h}{4}$$
Thus, we can express the volume of the water as a function of $h$ by substituting $h/4$ for $r$, which yields
$$V(h) = \frac{1}{3}\pi\left(\frac{h}{4}\right)^2h = \frac{1}{48}\pi h^3$$
Differentiating with respect to time yields
$$\frac{dV}{dt} = \frac{1}{16}\pi h^2 \frac{dh}{dt}$$
We are given that
$$\frac{dh}{dt} = 1~\frac{\text{in}}{\text{min}} = \frac{1}{12}~\frac{\text{ft}}{\text{min}}$$
when $h = 16~\text{ft}$. Thus, the rate at which the volume is increasing is
$$\frac{dV}{dt} = \frac{1}{16}\pi (16~\text{ft})^2\left(\frac{1}{12}~\frac{\text{ft}}{\text{min}}\right) = \frac{4}{3}\pi~\frac{\text{ft}^3}{\text{min}}$$
Since the fill rate is $8~\text{ft}^3/\text{min}$, the leak rate is
$$\frac{dL}{dt} = \frac{dF}{dt} - \frac{dV}{dt} = 8~\frac{\text{ft}^3}{\text{min}} - \frac{4}{3}\pi~\frac{\text{ft}^3}{\text{min}} = \left(8 - \frac{4}{3}\pi\right)~\frac{\text{ft}^3}{\text{min}}$$
Best Answer
$V = \pi r^2 \frac{h}{3}$
diameter $= 3h \Rightarrow r = \frac{3}{2}h$
$V(h) = \pi \cdot \frac{9}{4}h^2 \cdot \frac{h}{3}= \frac{3}{4}\pi h^3$
Height is a function of time since the height increases as sand piles on.
I.e $V(h) = (V \circ h)(t)$
Let $t_0$ be such that $h(t_0) = 15$. You just need to solve:
$$10=\frac{d}{dt}\Bigr|_{t = t_0} ( V \circ h)(t) = V'(h(t_0)) \cdot h'(t_0) $$