I need help with this

Question

A trough is 12 feet long and 3 feet across the top (see figure). 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? 

(b) If the water is rising at a rate of 3/8 inch per minute when h = 2.5, determine the rate at which water is being pumped into the trough. 

I am not sure which part I am getting mixed up on, please help.

TIA

Answer 1

1) You need an equation for the volume of water in terms of the height

2) You differentiate this equation and get dV/dt in terms of dh/dt

The volume will be the depth times the Area of the isosceles triangle cross-section. Noting that the base of the triangle is 1:1 with the height as all cross-sections are similar and the base goes from 0 to 3 and the height goes from 0 ro 3, we can write the volume equation:

V = 1/2 h²d where d is the depth of the trough. Taking the derivative to get the related rates:

dV/dt = hd dh/dt

In 1) You are given dV/dt, h, and d - solve for dh/dt.

in 2) You are given dh/dt, h, and d - solve for dV/dt (note h must be in feet)