related rates: How fast is water level rising when...?

[tyronne]

A swimming pool is 15ft wide, 40 ft long and 3 feet deep atone end, and 10ft deep at the other. Wateris being added to it at a rate of 25ft^3/min

a) How fast is the water level rising when there is 2000ft^3 of water in the pool?

b) How fast is the water level rising when there is 3000ft^3 of water in the pool?

so dx/dt=25ft^3/min and x=2000 and 3000

Now I have no clue what to do, I got a hint stating to use similar triangles so i drew a 40 by 7 triangle but i really dont understand what this does for me

 

[stapel]

I will guess that you have defined "x" to be "the volume of water at time t", though why you're using "x" instead of "V", I don't know...?

When the volume is 2000 cubic feet, what is the height h? To find this height, don't you need to use the similar triangles? You have the fixed triangle with a height of 7 and a "base" ("top", really) of 40. The triangle you're using has height h and base [some other variable; I'd have used "x" here, but you could use, say, "b"] Both triangles have "widths" of 15, which you'll use to find the volume.

If the fixed ratio of height to base is 7/40, then what is your expression, in terms of your height variable, for this volume? Set this equal to "2000", and solve for the height. Then differentiate the volume formula, and plug in the known value for the volume, the height, and the change in volume. Solve for the change in height.

Do the same for when the volume is 3000 cubic feet.

Note: At some point, you may have to account for the pool above the three-foot-depth mark. So be sure to find the volume of the 40-by-7-by-15 part at some stage.

Eliz.

 

[galactus]

You can find the A(t), the surface area of the water when the pool has 2000 ft^3 of water in it, then divide that into dV/dt.

You need \(\displaystyle \L\\\frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}\)

\(\displaystyle \L\\2000=\frac{1}{2}bh(15)\).........[1]

Similar triangle gives us a relation between b and h:

\(\displaystyle \L\\\frac{b}{h}=\frac{7}{40}\)

Now, solve for b and sub into [1]. It'll then be entirely in terms of h.

That will give you the height of the water when there is 2000 ft^3 of water in the pool.

Then , can you finish?.