the question located in the description box.

Question

Shaun orders the construction of an inverted circular cone [an once cream cone pointing downward, V = 1/3 pie r^2 h] with base radius 2m and height 4m. if spice melange is being pumped into the tank at a rate of 2m^3/min, what is the rate at which the spice level is rising when the spice is 3m deep.

Answer 1

The volume of the frustum is V = 1/3 π(R²H - r²h).

 

R = base radius

H = cone height

 

r = radius at spice level

h = distance from spice level to apex.

 

y = H - h for the level height of the spice.

 

h/H = r/R from similar triangles.

 

 

H - y = rH/R from the last two equations.

 

r = R(H - y)/H from the last equation.

 

 

We want V in terms of y:

 

V = 1/3 π(R²H - R²(H - y)³/H²)

 

 

dV/dt = πR²(H - y)²/H² dy/dt

 

dy/dt = dV/dt H²/[πR²(H - y)²]

 

= (2)(4²)/[π(2²)(4 - 3)²] = 8/π meter/min