# Rate of change

1. Oct 10, 2009

### songoku

1. The problem statement, all variables and given/known data
A bathtub with shape semi-circle has radius 5 m is filled with water with rate 2 m^3/minute. How fast will the height of water in the bathtub change when it is 2 m from the bottom? Given that the volume of water when the radius = R and height = y is πy^2(R - y/3)

2. Relevant equations
differential

3. The attempt at a solution
The volume of bathtub (constant) = 2/3 πr^3

The rate of change of volume of water :
$$\frac{dV}{dt}=2\frac{m^3}{minute}$$

I have to find dy/dt. I think I have to find the relation between R and y, but I don't know how...

Thanks

2. Oct 10, 2009

### Dick

R doesn't change. R=5 m. Only V and y change. And they already gave you the relation between them. V=πy^2(R - y/3). That part of the job is done. You just have to start with the differentiation.

3. Oct 10, 2009

### songoku

Hi Dick

Oh I Think R is the radius of the water, not the bathtub. I just realized that's not possible.

Thanks a lot

4. Oct 10, 2009

### Staff: Mentor

Can you provide the exact statement of the problem? Saying that "A bathtub with shape semi-circle" doesn't clearly describe it. Is a vertical cross-section of the tub semicircular? That would probably be the most reasonable interpretation, but I suppose it could have a horizontal cross-section that is semicircular. In either case, a radius of 5 m. seems very large for a bathtub. A radius of .5 m seems more realistic.

5. Oct 10, 2009

### songoku

Hi Mark44

I got this question from my friend and that's all what he told me. I interpreted it as the vertical cross-section. Will it be different if it's horizontal cross-section? I think it will be the same.
About the radius, yeah 5 m is ridiculous for a bathtub. Maybe it's a pool

6. Oct 11, 2009

### Staff: Mentor

You are given a formula for the volume of water in the pool when the water is y feet deep, and the radius of the pool is R feet:
V = $\pi y^2(R - y/3)$

Differentiate both sides with respect to t, using the chain rule. That will give you
dV/dt = (something involving y)*dy/dt

Solve this equation for dy/dt, and substitute in all the other known quantities at the time when y = 2 feet.

7. Oct 11, 2009

### songoku

Hi Mark44

Ok Thanks a lot Mark

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook