# Help with Differential equation

1. Feb 2, 2013

### Hey there

A hemispherical bowl of radius a has its axis vertical and is full of water. At time t=0 water starts running out of a small hole in the bottom of the bowl so that the depth of water in the bowl at t is x. The rate at which the volume of water is decreasing is proportional to x. Given that the volume of water in the bowl when the depth is x is $$\pi(ax^2-\frac{1}{3}x^3)$$ show that there is a positive constant k such that

$$\pi(2ax-x^2)\dfrac{dx}{dt}=-kx$$
My method
$$-\dfrac{dV}{dt}\propto x$$
$$\dfrac{dV}{dt}=-kx$$
$$\dfrac{dV}{dt}=\dfrac{dV}{dx} \dfrac{dx}{dt}$$
$$V=\pi(ax^2-\frac{1}{3}x^3)$$
$$\dfrac{dV}{dx}=\pi(2ax-x^2)$$
$$\pi(2ax-x^2)\dfrac{dx}{dt}=-kx$$

(ii) Given that the bowl is empty after a time T, show that

$$k=\dfrac{3 \pi a^2}{2T}$$

My method (I'm not sure how to answer this part).

$$\pi(2ax-x^2)\dfrac{dx}{dt}=-kx$$
$$\displaystyle \int (2\pi a-\pi x )dx=\displaystyle \int-k dt$$
$$2\pi ax-\frac{\pi x^2}{2}=-kt + C$$

x=0 when t=T

I'm not sure how to go about showing $$k=\dfrac{3 \pi a^2}{2T}$$with the information in the question.

Can you help me?

Thanks.

Last edited: Feb 2, 2013
2. Feb 2, 2013

### Dick

There's a mistake in your integration in part (ii). The integral of 2*pi*a*dx is not 2*pi*a. Fix that first. Then use that at t=0, x=a (the hemisphere was initially full) to find the value of C. Once you've found C put x=0 and solve for t.

Last edited by a moderator: Feb 2, 2013
3. Feb 2, 2013

### Hey there

I got $$C=2a^2(\pi-\frac{1}{4})$$

So

$$2 \pi ax-\frac{\pi x^2}{2}=-kt+2a^2(\pi -\frac{1}{4})$$

when x=0 t=T

$$kT=2a^2(\pi-\frac{1}{4})$$

$$k=\dfrac{2a^2(\pi-\frac{1}{4})}{T}$$

$$k=\dfrac{2a^2 \pi-\frac{1}{2}a^2}{T}$$

$$k=\dfrac{4a^2 \pi-a^2}{2T}$$

$$k=\dfrac{a^2(4 \pi -1)}{2T}$$

This doesn't equal $$k=\dfrac{3\pi a^2}{2T}$$

Where have I gone wrong?

Last edited: Feb 2, 2013
4. Feb 2, 2013

### Dick

If I put x=a into $2 \pi ax-\frac{\pi x^2}{2}$, which is what you did I hope, I don't get $2a^2(\pi-\frac{1}{4})$.

5. Feb 2, 2013

### Hey there

I simplified it wrong then.

I'll have another look.

6. Feb 2, 2013

### Hey there

Oh my god. This is incredibly simple. I over complicated it I guess. So I have

$$C=\dfrac{3\pi a^2}{2}$$

$$2\pi ax- \dfrac{\pi x^2}{2}=-kx+\dfrac{3 \pi a^2}{2}$$

x=0 when t=T

$$kT=\dfrac{3 \pi a^2}{2}$$

$$k=\dfrac{3 \pi a^2}{2T}$$

Got it.

7. Feb 2, 2013

### Dick

Yeah, you had the right idea all along. Some simple errors were tripping you up.