Question

A trough is 8 feet long and has perpendicular cross section in the shape of an isosceles triangle (point down) with base 1 foot and height 2 feet. The trough is being filled with water at a rate of 1 cubic foot every 5 minutes. How fast is the water level rising whenthe water is 1/2 foot deep?

Answer 1

Dh/dt = 0.082 ft/min

Explanation:

As a perpendicular cross section of the trough is in the shape of an isosceles triangle the trough has a circular cone shape wit base of 1 feet and height h = 2 feet.

The volume of a circular cone is:

V(c) = 1/3 * π*r²*h

Then differentiating on both sides of the equation we get:

DV(c)/dt = 1/3* π*r² * Dh/dt (1)

We know that DV(c) / dt is 1 ft³ / 5 min or 1/5 ft³/min

and we are were asked how fast is the water rising when the water is 1/2 foot deep. We need to know what is the value of r at that moment

By proportion we know

r/h ( at the top of the cone 0,5/ 2) is equal to r/0.5 when water is 1/2 foot deep

Then r/h = 0,5/2 = r/0.5

r = (0,5)*( 0.5) / 2 ⇒ r = 0,125 ft

Then in equation (1) we got

(1/5) / 1/3* π*r² = Dh/dt

Dh/dt = 1/ 5*0.01635

Dh/dt = 0.082 ft/min