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Isosceles Triangle related rates problem

by hks118
Tags: isosceles, rates, triangle
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hks118
#1
Mar6-10, 04:33 PM
P: 19
1. The problem statement, all variables and given/known data
A trough is 9 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft3/min, how fast is the water level rising when the water is 8 inches deep?

Variables:
b=5 ft
h=1 ft
l=9 ft
2. Relevant equations
v=(1/2)bhl
dv/dt=14
dh/dt=?
when h=2/3 ft
3. The attempt at a solution
I tried doing it the straightforward way:
v=(1/2)bhl
dv/dt=(1/2)(5)(2/3)(dh/dt)(9)
14=(1/2)(5)(2/3)(dh/dt)(9)
dh/dt=14/15

This is wrong. Then I tried using similar triangles to get the new base to go along with the height of 8 in. For that I got 10/3 ft. So,
v=(1/2)bhl
dv/dt=(1/2)(10/3)(2/3)(dh/dt)(9)
14=(1/2)(10/3)(2/3)(dh/dt)(9)
dh/dt= 7/5

This is also incorrect. I really don't know why. Any help would be greatly appreciated!
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tiny-tim
#2
Mar6-10, 04:54 PM
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Hi hks118! Welcome to PF!

You need to write b as a function of h before you differentiate.
hks118
#3
Mar6-10, 05:49 PM
P: 19
Quote Quote by tiny-tim View Post
Hi hks118! Welcome to PF!

You need to write b as a function of h before you differentiate.
Thanks for the reply!

So b=5h correct? but if I plug in my h value, I get 10/3, which was wrong.

tiny-tim
#4
Mar7-10, 04:32 AM
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Isosceles Triangle related rates problem

(just got up )

What was your equation for dv/dt?
hks118
#5
Mar7-10, 11:41 AM
P: 19
Quote Quote by tiny-tim View Post
(just got up )

What was your equation for dv/dt?
Well, v=1/2bhl so I figured dv/dt=1/2bh(dh/dt)l. Or do I need db/dt and dl/dt as well?
tiny-tim
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Mar7-10, 02:44 PM
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Quote Quote by hks118 View Post
Well, v=1/2bhl so I figured dv/dt=1/2bh(dh/dt)l.
That wouldn't be right even if b was constant.
Try again.


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