A trough is 12 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top

Question

A trough is 12 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft³/min, how fast is the water level rising when the water is 7 inches deep?

In feet, we know that

h =

I can't figure this out

Answer 1

Start by sketching the trough. Note that the depth is 1 ft and max width is 3 ft.

Given any depth of water what would the subsequent width of water in the trough?

In this case the width of water is three times the height so w = 3h (Remember similar triangles)

The formula for Volume is:

Area of triangle * length

= 1/2 b*h *l

= 1/2 * 3h * h * 12

V = 18 h²

The next step is to take the derivative with respect to time of both side.

d/dt (V) = d/dt (18h²)

dV/dt = 36 h * (dh/dt) You might be wondering about the dh/dt; if so keep reading.

If we were deriving with respect to h then we would get 36 h * dh/dh dh/dh = 1

We do this because of the chain rule. Derivative of outside times derivative of inside.

Because the height depends on time we must include the dh/dt.

Remember dh/dt is just a number dictating how quickly the water level rises at this moment.

The rest is substitution and algebra.

dV/dt = 36 h * (dh/dt)

14 = 36 * 7/12 * dh/dt Remember to keep everything in feet. 7 in = 7/12 ft

2/3 ft/min = dh/dt

or

dh/dt = 8 in/min