Swimming Pools and Related Rates along with Implicit Differentiation

Salazar
Messages
8
Reaction score
0

Homework Statement



A swimming pool is 40 feet long, 20 feet wide, 4 feet deep at the shallow end, and 9 feet deep at the deep end. Water is being pumped into the pool at 10 cubic feet per minute.
a. When the water is 3 feet deep at the shallow end, at what rate is the water level rising? b. When the water is 3 feet deep at the deep end, at what rate is the water level rising?

Homework Equations


None Given.

The Attempt at a Solution



I had [dV/dt] and the volume for part a is V = L*W*H and I had L and W, which were 40' and 20' respectively. So V = 800H and I used implicit differentiation to get [dV/dt] = 800[dH/dt], so [dH/dt] = 1/80 feet per minute.

For part be it seems straightforward, but I do not know how to get the length, since it has change. Picture of the pool is uploaded.
 

Attachments

  • Picture (Device Independent Bitmap) 1.jpg
    Picture (Device Independent Bitmap) 1.jpg
    5.3 KB · Views: 552
Last edited:
Physics news on Phys.org
Counting h from the top of the pool you have then
V = LWH for 0 < H < 4 ft and
V = something else for 4 ft < H < 9ft.
 
I understand that, but what is such something else for L, length?
 
The volume of the lower half is a triangles area times the width of the pool; you can give the triangle for example by one of the angles and the height, you do not need to use length explicitly at all.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top