Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 21 feet high?

dx/dt =  0,029 ft

Step-by-step explanation:

Let call "x " diameter  of the base of the cone  then  h (height ) is equal to x

V  = ( 1/3)* A(b) *h

A(b)  = π*(x/2)²

Then

Volume of the cone is

V = ( 1/3)*π*(x/2)²*x

V =  ( 1/12)*π*x³

Then:

dV/dt  =   ( 1/12)*π* 3*x²*dx/dt     ⇒   dV/dt  = ( 1/4)*π*x²*dx/dt

dV/dt  = (1/4)*π*(21)²*dx/dt

As  dV/dt =  10 ft³

10*4   =  441*π* dx/dt     ⇒   dx/dt  =  40/ 1384,74

dx/dt =  0,029 ft

step-by-step explanation:

answer: -10

step-by-step explanation:

s(t) = 1 - 10t

s(x) = 1 - 10x

s(x + h) = 1 - 10(x + h)

            = 1 - 10x - 10h

s(x + h) - s(x) = 1 - 10x - 10h - (1 - 10x)

                      = 1 - 10x - 10h - 1 + 10x

                      = -10h

= -10