September 22nd, 2009, 07:30 AM  #1 
Member Joined: Sep 2009 Posts: 51 Thanks: 0  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???? 
September 22nd, 2009, 08:20 AM  #2 
Senior Member Joined: Mar 2009 Posts: 318 Thanks: 0 
Where did a 3/8 come from? 
September 22nd, 2009, 01:20 PM  #3 
Member Joined: Sep 2009 Posts: 51 Thanks: 0  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 
September 24th, 2009, 10:50 AM  #4 
Newbie Joined: Sep 2009 Posts: 1 Thanks: 0  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. 
September 24th, 2009, 12:22 PM  #5 
Newbie Joined: Sep 2009 Posts: 27 Thanks: 0  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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