Derivatives: Swimming Pool

1. Apr 8, 2009

andrew.c

1. The problem statement, all variables and given/known data

A rectandular swimming pool is 10m wide and 20m long. The bottom of the pools is a sloping plane with the depth of the pool varying along the length of the pool from 1m at the shallow end to 5m at the deep end. Water is being pumped into the pool at the rate of $$30 m^3 / min$$. Shat that the total volume of water in the pook when the depth of thewater at the deep end is hm (where $$0 \leq h \leq 4$$) satisfies

$$V = 25h^2$$

Hence find the rate at which the water level is rising when the water is 3m deep at the deep end.

2. Relevant equations
n/a
3. The attempt at a solution

Equation to describe volume at h...

When 4m full, the 'length' filled by the water is 20m, therefore..

\begin{align*} V &= \frac{1}{2}LHW\\ &=\frac{1}{2}(5h)(h)(10)\\ &=(2.5h^2)(10)\\ &=25h^2\\ \end{align*}

Then, i tried to differentiate this expression to find h' (difference in height), but this came to 0.

Ideas?

2. Apr 8, 2009

Tom Mattson

Staff Emeritus
Say what? If $V(h)=25h^2$ then $\frac{dV}{dh}=50h$. So $\frac{dV}{dt}=?$

3. Apr 8, 2009

andrew.c

Is the final answer 1/150 m/s?

4. Apr 8, 2009

Happyzor

I'm a newb but I'll try to help -_-.

Try differentiating $$V=25h^2$$ with respect to $$t$$. Since you know $$dV/dt$$ and $$h$$, you can figure out the answer.

Last edited: Apr 8, 2009
5. Apr 9, 2009

andrew.c

$$V=25h^2$$

\begin{align*} differentiated &=\frac{dV}{dt} (25h^2)\\ &=\frac{dV}{dh}(25h^2) \frac{dh}{dt}\\ &=(50h) \frac{dh}{dt}\\ \frac{dh}{dt} &= 50h\\ sub. in value for h to find dh/dt\\ \frac{dt}{dh} &= 50(3)\\ \frac{dt}{dh} &= 150\\ \frac{dh}{dt} &= 1/150\\ \end{align*}

6. Apr 9, 2009

andrew.c

Does that look even a little bit right to anyone???

7. Apr 9, 2009

Billy Bob

Not right.

For one thing, it looks like you are writing $$\frac{dV}{dt}(25h^2)$$ when you mean $$\frac{d}{dt}(25h^2)$$.

Here is better notation.

\begin{align*} \frac{dV}{dt}&=\frac{dV}{dh}\frac{dh}{dt}\\ &=\frac{d}{dh}(25h^2) \frac{dh}{dt} \end{align*}

What's next?

8. Apr 9, 2009

andrew.c

\begin{align*} &= 50h \frac{dh}{dt} \\ \end{align*}

Then move dh/dt over to the other side, where it is 'dividing', i.e $$\frac{1}{\frac{dh}{dt}} = 50h$$

so

$$\frac{dt}{dh} = 50h$$

but we want dh/dt, so needs to be rearranged,

$$\frac{dh}{dt} = \frac{1}{50h}$$

then I subbed in 3 for H to get 1/150

9. Apr 9, 2009

Billy Bob

Yes, now write it as part of an equation. Then what?

10. Apr 9, 2009

andrew.c

I editted above post ^^ Accidently hit reply before I was finished.

11. Apr 9, 2009

Billy Bob

No, in your first step, the other side is not "1." What is the other side?

12. Apr 9, 2009

andrew.c

dV/dt? and does that = 30?

13. Apr 9, 2009

Billy Bob

Bingo

14. Apr 9, 2009

andrew.c

Thanks for all the help!

Ta muchly!
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