# Check this problem out!

• May 25th 2007, 04:03 AM
((ECHO))
Check this problem out!
Im reviewing for my calculus final exam and I was just going through some problems that I didnt understand and never got back to them. Im trying to solve this one out and I cant get it right.

A swimming pool is 22 ft wide, 60 ft long, 2 ft deepat the shallow end, and 10 ft deep at its deepest point. If the pool is being filled with water at a rate of 0.8ft^3/min, how fast is the water level rising when the depth at the deepest point is 7 ft??

I keep getting .00115ft/min and i know thats not the right answer.
Can some one give me a hand?
• May 25th 2007, 09:41 AM
Jhevon
Quote:

Originally Posted by ((ECHO))
Im reviewing for my calculus final exam and I was just going through some problems that I didnt understand and never got back to them. Im trying to solve this one out and I cant get it right.

A swimming pool is 22 ft wide, 60 ft long, 2 ft deepat the shallow end, and 10 ft deep at its deepest point. If the pool is being filled with water at a rate of 0.8ft^3/min, how fast is the water level rising when the depth at the deepest point is 7 ft??

I keep getting .00115ft/min and i know thats not the right answer.
Can some one give me a hand?

i got something close to that, 0.00069264, is that the correct answer?
• May 26th 2007, 03:09 AM
behemoth100
Quote:

Originally Posted by Jhevon
i got something close to that, 0.00069264, is that the correct answer?

I also got something similar: .00071294893. Not sure if the difference between my answer and Jhevon's is due to rounding errors during working though.
• May 26th 2007, 03:38 AM
behemoth100
My apologies I repeated my calculation and got 0.00069264069 as well. Rounding error the first time:

Let the height = h. Let S be the length of the slope of the pool as the depth changes from 2 feet to 10 feet. Then Cos(theta) = 8/60
theta = Cos(^-1) (8/60)
= 82.33774434 * (let * be degrees) here on referred to as A
Thus Cos (A) = h/S

Let the length of the triangular prism (where the whole triangular prism has dimensions: Length 60 feet, height 8 feet, width 22 feet, slope 60.5309838 feet)
(SQUARE ROOT OF: ((h^2)/cos^2 (A) - h^2)) gives LENGTH L. (Pythagorus' theorem)

Thus volume of triangular prism given by:
V = (1/2)(h)((SQUARE ROOT OF: ((h^2)/cos^2 (A) - h^2)))22
= 11(h^2)((SQUARE ROOT OF: (1- cos^2 (A))/cos^2 (A)))
dV/dt = .8
dh/dt = (dh/dV)(dV/dt)
= (dh/dV) .8
dV/dh = 22h((SQUARE ROOT OF: (1- cos^2 (A))/cos^2 (A)))
dh/dV = 1/22h((SQUARE ROOT OF: (1- cos^2 (A))/cos^2 (A)))
dh/dt = (.8)1/22h((SQUARE ROOT OF: (1- cos^2 (A))/cos^2 (A)))
where A = 82.33774434 * (let * be degrees)
Let h = 7
dh/dt = 0.00069264069
• May 26th 2007, 05:51 AM
Soroban
Hello, ((ECHO))!

I get the same answer is behemoth and Jhevon . . .

Quote:

A swimming pool is 22 ft wide and 60 ft long.
It is 2 ft deep at the shallow end, and 10 ft deep at the other end.
If the pool is being filled with water at a rate of 0.8 ft³/min,
how fast is the water level rising when the depth at the deepest point is 7 ft?

Code:

                        60   -  * - - - - - - - - - - - - - - - - - - - - *   :  |                                        |   2  |                                        | 2   :  |                60                      |   - A + - - - - - - - - - - - - - - - - - - - - * B   :  |                                  *   :  |            x                *   : C +-----------------------* D   8  |:::::::::::::::::*   :  |h :::::::::*   :  |:::::*   - E *

Since $\displaystyle \Delta BAE \sim \Delta DCE$, we have: .$\displaystyle \frac{x}{h} = \frac{60}{8}\quad\Rightarrow\quad x = \frac{15}{2}h$ .[1]
. . Hence: .$\displaystyle \frac{dx}{dt} = \frac{15}{2}\!\cdot\!\frac{dh}{dt}$ . [2]

The volume of water is: .$\displaystyle V \:=\:\frac{1}{2}xh(22) \:=\:11xh$

. . Then: .$\displaystyle \frac{dV}{dt}\;=\;11x\!\cdot\frac{dh}{dt} + 11h\!\cdot\!\frac{dx}{dt}$ . [3]

When $\displaystyle h = 7$, [1] gives us: .$\displaystyle x \:=\:\frac{15}{2}(7) \:=\:\frac{105}{2}$

Substitute into [3]: .$\displaystyle 0.8 \;=\;11\left(\frac{105}{2}\right)\left(\frac{dh}{d t}\right) + 11(7)\left(\frac{15}{2}\!\cdot\!\frac{dh}{dt}\righ t)$

Hence: .$\displaystyle 1155\left(\frac{dh}{dt}\right) \:=\:0.8\quad\Rightarrow\quad \frac{dh}{dt}\:=\:\frac{0.8}{1155}$

Therefore: .$\displaystyle \frac{dh}{dt} \:=\:0.000692641$ ft/min.

• May 26th 2007, 07:13 AM
Jhevon
Quote:

Originally Posted by Soroban
Hello, ((ECHO))!

I get the same answer is behemoth and Jhevon . . .

Code:

                        60   -  * - - - - - - - - - - - - - - - - - - - - *   :  |                                        |   2  |                                        | 2   :  |                60                      |   - A + - - - - - - - - - - - - - - - - - - - - * B   :  |                                  *   :  |            x                *   : C +-----------------------* D   8  |:::::::::::::::::*   :  |h :::::::::*   :  |:::::*   - E *

Since $\displaystyle \Delta BAE \sim \Delta DCE$, we have: .$\displaystyle \frac{x}{h} = \frac{60}{8}\quad\Rightarrow\quad x = \frac{15}{2}h$ .[1]
. . Hence: .$\displaystyle \frac{dx}{dt} = \frac{15}{2}\!\cdot\!\frac{dh}{dt}$ . [2]

The volume of water is: .$\displaystyle V \:=\:\frac{1}{2}xh(22) \:=\:11xh$

. . Then: .$\displaystyle \frac{dV}{dt}\;=\;11x\!\cdot\frac{dh}{dt} + 11h\!\cdot\!\frac{dx}{dt}$ . [3]

When $\displaystyle h = 7$, [1] gives us: .$\displaystyle x \:=\:\frac{15}{2}(7) \:=\:\frac{105}{2}$

Substitute into [3]: .$\displaystyle 0.8 \;=\;11\left(\frac{105}{2}\right)\left(\frac{dh}{d t}\right) + 11(7)\left(\frac{15}{2}\!\cdot\!\frac{dh}{dt}\righ t)$

Hence: .$\displaystyle 1155\left(\frac{dh}{dt}\right) \:=\:0.8\quad\Rightarrow\quad \frac{dh}{dt}\:=\:\frac{0.8}{1155}$

Therefore: .$\displaystyle \frac{dh}{dt} \:=\:0.000692641$ ft/min.

This is the approach i used, behemoth100 did something different using trig functions. it seemed to work out though:D