Thread: Rate of change in a cone

1. Rate of change in a cone

I am a beginner and could use help solving this problem:

water flows out of a cone shaped reservoir at 50m^3/min. the reservoir water level 6m h when full and the radius is 45m. when the water level is at 5m, (a) how fast is the water level falling and (b) at that moment how fast is the radius changing.

So I think dv/dt = 50 and I need to find dh/dt and dr/dt, right?

I was trying to use dv/dt=1/3*pi*r^2*dh/dt + 1/3*2r*dr/dt*h.

I know the answers but I am having trouble getting there

2. Originally Posted by Charger10
I am a beginner and could use help solving this problem:

water flows out of a cone shaped reservoir at 50m^3/min. the reservoir water level 6m h when full and the radius is 45m. when the water level is at 5m, (a) how fast is the water level falling and (b) at that moment how fast is the radius changing.

So I think dv/dt = 50 and I need to find dh/dt and dr/dt, right?

I was trying to use dv/dt=1/3*pi*r^2*dh/dt + 1/3*2r*dr/dt*h.

I know the answers but I am having trouble getting there
You must use the fact that the height and the radius are proportional (because the reservoir is a cone): if $h_0=6m$ and $r_0=45m$ are the initial height and radius, then you always have $\frac{h(t)}{h_0}=\frac{r(t)}{r_0}$.

As a consequence, you can express the volume $v(t)$ in terms of $r(t)$ or in terms of $h(t)$ only. From there, you derivate $v(t)$ and substitute $h(t)=5m$ (or $r(t)=\frac{r_0}{h_0}h(t)$ because of the previous relation), in order to get $h'(t)$ and $r'(t)$.

3. Well I have r1/h1=r/h so r1/5=45/6 and r1=37.5.

I get lost subbing it in the equation above. am I using the right equation?

4. Originally Posted by Charger10
Well I have r1/h1=r/h so r1/5=45/6 and r1=37.5.

I get lost subbing it in the equation above. am I using the right equation?
What you wrote is the final value for $r$. In order to differentiate, you mustn't fix its value. You have $v(t)=\frac{1}{3}\pi r(t)^2 h(t)=\frac{1}{3}\pi r(t)^2 \frac{h_0}{r_0}r(t)=\pi\frac{h_0}{3r_0}r(t)^3$, hence $v'(t)=\pi\frac{h_0}{r_0}r(t)^2r'(t)$ and thus $r'(t)=\frac{v'(t)}{\pi\frac{h_0}{r_0}r(t)^2}$ and now, you choose $r(t)=r_1=37.5 m$. You do the same for the height: write $v(t)$ in terms of $h(t)$, differentiate, and then choose $h(t)=h_1$ to get $h'(t)$ at that time.