# Derivative of the cone volume

"A cone-shaped tank is mounted vertically with its pointed end down. It has a height of 10 m and a radius of circular base 5 m. Water flows at a rate of 8m^3 / min. How fast does the water level rise when the water is 5 m high?

I calculated tanu=5/10=0,5, where u is the angle between the height h=10 and the side of the cone.

Thus r'(t) =0,5 and r(t) =0,5*5=2,5 m, where r(t) is the radius of cone's circular base when the water is 5 m high.

Thus I use the formula of the cone volume V(t) = (pi/3)*(r(t)^2)*h(t)

Then I find the derivative of V(t) and using the fact that V'(t) =8, r(t) =2.5, r'(t) =0.5, I solve the equation and find h'(t).

Is my way of thinking correct? Thank you very much in advance.

• I am not sure how you got $r('t) = 0.5$. That does not seem right. It is better to convert volume in terms of height and then differentiate. Commented Oct 21, 2020 at 21:47

Ratio of radius to the height of the cone $$= \frac{R}{H} = \frac{1}{2}$$ and this remains same at all height.

Now at a given height $$h$$, $$\, r = \frac{h}{2}$$

So, $$V = \frac{\pi}{3}r^2h = \frac{\pi}{12}h^3$$

$$\displaystyle \frac{dV}{dt} = \frac{\pi}{12} \times 3h^2 \frac{dh}{dt}$$

At height $$h = 5$$ and given the rate of volume change -

$$\displaystyle 8 = \frac{\pi}{4} (5)^2 \frac{dh}{dt}$$

$$\displaystyle \frac{dh}{dt} = \frac{32}{25\pi}$$ meter/min.