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