What is the Derivative of a Swimming Pool with a Sloping Bottom?

[andrew.c]

Homework Statement

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.

Homework Equations

n/a

The Attempt at a Solution

Equation to describe volume at h...

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

<br /> \begin{align*}<br /> V &amp;= \frac{1}{2}LHW\\<br /> &amp;=\frac{1}{2}(5h)(h)(10)\\<br /> &amp;=(2.5h^2)(10)\\<br /> &amp;=25h^2\\<br /> \end{align*}<br />

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

Ideas?

 

[quantumdude]

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

 

[andrew.c]

Is the final answer 1/150 m/s?

 

[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.

 

[andrew.c]

V=25h^2

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

 

[andrew.c]

Does that look even a little bit right to anyone?

 

[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.

<br /> \begin{align*}<br /> \frac{dV}{dt}&amp;=\frac{dV}{dh}\frac{dh}{dt}\\<br /> &amp;=\frac{d}{dh}(25h^2) \frac{dh}{dt}<br /> \end{align*}<br />

What's next?

 

[andrew.c]

<br /> \begin{align*}<br /> &amp;= 50h \frac{dh}{dt} \\<br /> \end{align*}<br />

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

 

[Billy Bob]

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

 

[andrew.c]

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

 

[Billy Bob]

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

 

[andrew.c]

dV/dt? and does that = 30?

 

[Billy Bob]

Bingo

 

[andrew.c]

Thanks for all the help!

Ta muchly!
x