A cone shaped container has a diameter of $0.6m$ and height of $0.5m$. Water is poured into the container with a constant rate of $0.2m^{3}s^{-1}$.
Calculate the rate of change in height of the water level when its height reaches $0.4m$.
Ok, so $\frac {dV}{dt} = 0.2$
V = $\frac 1 3 \pi r^2h$
I want to find $\frac {dh}{dt}$
so can it be found by doing this?
$\frac {dV}{dt} = \frac{dV}{dh} . \frac{dh}{dt}$
if so, then how can $\frac {dV}{dh}$ be calculated?
$\frac {dV}{dh} = \frac1 3 \pi r^2$
This is as far as I can go. I don't know the value of $r$ when the height is $0.4m$.
Update:
Since $\tan \theta = \frac {0.3}{0.5}$
$\theta = \tan^{-1} 0.6 = 30.96$
When, $ h = 0.4$
$r = \tan (30.96) \times 0.4 = 0.24$
$\frac {dV}{dh} = \frac 1 3 \pi (0.24)^2 = 0.0603$
So, $\frac {dh}{dt} = \frac{0.2}{0.0603} = 3.316$
But the real answer is $1.105$. Could it be that $\frac{dV}{dh} = \pi r^2$ and not $\frac 1 3 \pi r^2$ ? Or have I made some seriously wrong calculations?
The red circled triangle has sides $R$ (the radius of the sphere), $r$ (the radius of the water surface) and $R-h$ (where $h$ is the water height). Pythagoras tells you that $R^2 = r^2 + (R-h)^2.$
Best Answer
Using similar triangles, we have that $\frac{r}{h}=\frac{.3}{.5}$, so $r=\frac{3}{5}h$.
Then $\displaystyle V=\frac{1}{3}\pi r^2h=\frac{\pi}{3}\big(\frac{3}{5}h\big)^2h=\frac{3\pi}{25}h^3,\;\;$ so $\displaystyle\frac{dV}{dt}=\frac{9\pi}{25}h^2\frac{dh}{dt}$.
When $h=.4$, this gives $\displaystyle.2=\frac{9\pi}{25}\big(.4\big)^2\frac{dh}{dt},\;\;$ so $\displaystyle\frac{dh}{dt}=\frac{125}{36\pi}\approx1.105 \;m/s$