Question #26b83

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Answer 1 · 2017-03-30T16:05:40

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

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:

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

Answer 2 · 2017-03-30T20:42:17

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

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:

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

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

Answer 3 · 2017-03-30T21:27:11