How Fast Is the Water Level Rising in a Trough as It Fills?

[ricky23i]

A trough is 8 ft long and its ends have the shape of isosceles triangles that are 2 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 11 ft3/min, how fast is the water level rising when the water is 9 inches deep?I got V=2h^2 (8)/2
V=8h^2
dv/dx=16h
11=16h

how do I get the height I am a confused and is everything else right? and can someone switch this to the homework thread just noticed

 

[HallsofIvy]

A trough is 8 ft long and its ends have the shape of isosceles triangles that are 2 ft across at the top and have a height of 1 ft.

The crucial point is that what ever the height of the water a cross section will have the same "shape". The base will be twice the height. Since the area of a triangle is "1/2 base times height", the are of a cross section will be "1/2 times twice the height times the height" or just $h^2$.

If the trough is being filled with water at a rate of 11 ft3/min, how fast is the water level rising when the water is 9 inches deep?


I got


V=2h^2 (8)/2
V=8h^2

Yes, this is correct.

dv/dx=16h

There is no "x" in the problem! What is true is that dv/dh= 16h.

11=16h

But dV/dh is not 11. You told that "the trough is being filled with water at a rate of 11 ft3/min". The denominator is in minutes- dv/dt= 11 cubic feet per minute where t is the time. What you need to do is to differentiate both sides of V= 8h^2 with respect to time, using the chain rule.

how do I get the height I am a confused and is everything else right? and can someone switch this to the homework thread just noticed

You don't get the height- that is given as 9 inches. You are asked "how fast is the water level rising" which is a rate of change: dh/dt.