Posted by **Reggie** on Tuesday, December 4, 2012 at 10:28pm.

A trough is triangular in cross section, an isosceles triangle with sides of 12 inches, and a top of 10 inches. The trough is 40 inches long. How fast is the depth changing if you are pumping one cubic foot per minute into the trough?

I am having a hard time getting it in terms of h.

- Calculus -
**Reiny**, Wednesday, December 5, 2012 at 12:03am
Make a sketch of the cross-section of the isosceles triangle. Draw it some water level, letting the height be h and letting the width of the water level be x

by similar triangles

x/h = 10/12

12x = 10h

x = 5h/6

volume of water = 40(area of triangle)

= 40(1/2)(x)h

= 20(5h/6)(h) = 50/3 h^3

V = (50/3) h^3

dV/dt = 50 h^2 dh/dt

have to watch our units, 1 cubic foot = 12^3 or 1728 cubic inches

1728 = 50 h^2 dh/dt

dh/dt = 1728/(50h^2) = 864/25 inches/minute

- Calculus -
**Reggie**, Wednesday, December 5, 2012 at 12:08am
wow I was headed that way didn't watch units Thank You very much

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