# Isosceles Triangle related rates problem

by hks118
Tags: isosceles, rates, triangle
 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!
 Sci Advisor HW Helper Thanks P: 26,148 Hi hks118! Welcome to PF! You need to write b as a function of h before you differentiate.
P: 19
 Quote by tiny-tim Hi hks118! Welcome to PF! You need to write b as a function of h before you differentiate.

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

 Sci Advisor HW Helper Thanks P: 26,148 Isosceles Triangle related rates problem (just got up …) What was your equation for dv/dt?
P: 19
 Quote by tiny-tim (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?