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Isosceles Triangle related rates problem 
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#1
Mar610, 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! 


#2
Mar610, 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. 


#3
Mar610, 05:49 PM

P: 19

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


#4
Mar710, 04:32 AM

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Isosceles Triangle related rates problem
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What was your equation for dv/dt? 


#5
Mar710, 11:41 AM

P: 19




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