So I am in a calculus 2 class and am stuck on a question as I don't think I have enough information. Can someone please help me with my next step so I can solve this.
A conical tank is filled at a constant rate of $1 \text{cm}^3/\text{sec}$. How fast does the height increase if the radius is $3$ cm?
Now, I know I'm looking for $\frac{\mathrm{d}h}{\mathrm{d}t}$ and I have $\frac{\mathrm{d}V}{dt} = 1 \text{cm}^3$, how do I find dh/dt if I don't know the relationship between $h$ and $r$ in the equation $V=\frac{1}{3} \pi r^2 h$? Can anyone solve it at all? And potentially show the steps?
Let $\theta$ denote the semi-angle of the cone. Then $r = h \tan \theta$. Hence, the volume of the cone can be written as $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi h^3 \tan^2\theta$. The important thing to note here is that $\theta$ does not change with time. Upon differentiating the expression of volume with respect to $t$, we have, $$ \frac{\mathrm{d}V}{\mathrm{d}t} = \pi h^2 \tan^2 \theta \cdot \frac{\mathrm{d}h}{\mathrm{d}t} = \pi r^2 \cdot \frac{\mathrm{d}h}{\mathrm{d}t} $$ So when $\frac{\mathrm{d}V}{\mathrm{d}t} = 1 \text{cm}^3/\text{sec}$ and the radius is $3$ cm, $\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1}{9\pi}$ cm/sec.
Note that this is also what someone pointed out in their comments. When the radius is $r$, the instantaneous increase in volume is the volume of the cylinder with radius $r$ and height $\mathrm{d}h$, i.e., $ \mathrm{d}V = \pi r^2 \mathrm{d}h$, which is exactly what we obtained in the above solution.