Atrough 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?

Dh/dt  = 0.082 ft/min

Step-by-step 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

ab will be equal to