Can somebody help me with this Calculus related rates problem?

Question

A trough has ends shaped like isosceles triangles, and has a length of 9 m, width 3 m, and height 4 m. Water is being pumped into the trough at a rate of 3 m^3/min. At what rate does the height of the water change when the water is 2 m deep?

Answer 1

You need an expression for V related to height and width:

V = 1/2 HWL

You have a constraint that the W/H= 3/4, so W = 3/4H and the equation becomes

V = 9/2 * 3/4 H² = 27/8 * H² (L is 9 m)

Now, take the derivative wrt time:

dV/dt = 27/8 * (2H) dH/dt = 27/4 * H *dH/dt

You are given dV/dt, H ; Find dH/dt