Question #26b83

3 Answers
Mar 30, 2017

(dV)/(dt) = 2sqrt(S/pi) (cm^3)/sec

Explanation:

The rate of increase of the area of the surface can be expressed as:

(dS)/(dt) = 8 (cm^2)/sec

We can find the rate of increase of the volume with the chain rule:

(dV)/(dt) = (dV)/(dS) * (dS)/(dt)

We must then express the volume as function of the surface.
As:

S=4pir^2 => r = sqrt(S/(4pi))

V = 4/3pir^3 = (4pi)/3 (sqrt(S/(4pi)))^3 = S^(3/2)/(6sqrtpi)

(dV)/(dS) = 1/(6sqrtpi) 3/2 S^(1/2) = 1/4 sqrt(S/pi)

So, finally:

(dV)/(dt) = (dV)/(dS) * (dS)/(dt) = 8*1/4 sqrt(S/pi) = 2sqrt(S/pi)

Mar 30, 2017

Rate of Change of volume is 4r \ cm^3s^(-1)

Explanation:

Let us set up the following variables:

{(V, "Volume of the sphere",cm^3), (S, "Surface Area of the sphere",cm^2), (t, "time",sec) :}

The Volume of the sphere is given by the standard formula:

V=4/3pir^3

Differentiating wrt r we get:

(dV)/(dr) = 4pir^2

The Surface Area of the sphere is given by the standard formula:

S = 4pir^2

Differentiating wrt r we get:

(dS)/(dr) = 8pir

We are given that:

(dS)/dt = 8 \ cm^2s^-1

By the chain rule:

(dV)/dt = (dV)/(dr) * (dr)/(dS) * (dS)/(dt)
" " = (4pir^2) * (1/(8pir)) * (8)
" " = 4r \ \ cm^3s^(-1)