Isosceles Triangle related rates problem

[hks118]

Homework Statement

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

Homework Equations

v=(1/2)bhl
dv/dt=14
dh/dt=?
when h=2/3 ft

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!

 

[tiny-tim]

Welcome to PF!

Hi hks118! Welcome to PF!

You need to write b as a function of h before you differentiate.

 

[hks118]

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]

(just got up :zzz: …)

What was your equation for dv/dt?

 

[hks118]

(just got up :zzz: …)

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]

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. ​