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September 22nd, 2009, 07:30 AM   #1
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Related Rates?????????

A trough is 12 feet long and 3 feet across the top . Its ends are isosceles triangles with altitudes of 3 feet.
(a) If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1.2 feet deep? (Round to three decimal places.)
V=6h^2
dv/dt= 12h(dh/dt)
2=12(1.2)(dh/dt)
dh/dt= .139

The answer I got was = .139


(b) If the water is rising at a rate of 3/8 inch per minute when h = 2.0 ft, determine the rate at which water is being pumped into the trough. (Round to three decimal places.)


this is what i dont get, where does the 3/8 go????
imcutenfresa is offline  
 
September 22nd, 2009, 08:20 AM   #2
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Where did a 3/8 come from?
stapel is offline  
September 22nd, 2009, 01:20 PM   #3
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Re: Related Rates?????????

well the 3/8 is part of the second part of the problem just that i dont know where it goes?
it says its rising at a rate of 3/8 do i multiply this by dh/dt?

this is the picture

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September 24th, 2009, 10:50 AM   #4
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Re: Related Rates?????????

i dont think the volume you wrote is right.
v=6h^2 is for a cube while your trough has an isosceles triangle.
So basically V= area of the triangle * height of the trough.
I hope I said it right.
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September 24th, 2009, 12:22 PM   #5
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Re: Related Rates?????????

v = 6h^2
dv/dh = 12h

dv/dt = 2

we need dh/dt = dh/dv * dv/dt

=> dh/dt = (1/12h) * 2

= 1/6h

At h = 1.2, dh/dt = 0.139


For b, we need to find dv/dt

we know, dh/dt = 3/8 = 0.375

dv/dh = 12h

Therefore, dv/dt = dv/dh * dh/dt

= 12h * 0.375

= 9h/2

At h = 2,

dv/dt = 9(2)/2

= 9

Also, 6h^2 is the correct volume
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