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Expert: - 9/14/2010


Question
QUESTION: Dear Scott
For any point on the surface of a sphere expanding at a constant volumetric rate, what are the formulae for calculating at any instant relative to the centre its (1) deceleration & (2) speed; given that rate of expansion and the radius?
Kind Regards

ANSWER: The volume of a sphere is (4*pi*r^3)/3.  The surface area is the derivative with respect to r.  That is, the surface is 4*pi*r^2.
The rate of expansion would be the surface area, and this is constant.

Therefore we can say that 4*pi*r^2(dt/dt) = K.  This  means that dr/dt decreases as r increases.

The deceleratioin would be the negative of the acceleration.

Is this what you're after?


---------- FOLLOW-UP ----------

QUESTION: Thanks for that Scott, but I also need an answer to the question (2), which is what is the instantaneous speed of any point relative to the sphere's centre?  I imagine this to be much greater than the deceleration.

Answer
The problem states that dV/dt = K.
It is known that dV/dt can be expanded to dV/dt = (dV/dr)(dr/dt).

Since V(t)=4*pi*r^3(t)/3, then dV/dr=4*pi*r^2(t).

Putting dV/dt, dV/dr into the equation dV/dt = (dV/dr)(dr/dt)
gives K=4*pi*r^2(t)*(dr/dt).

This says that r'(t) = K/(4*pi*r^2(t)).

To find the change in speed, take the derivative of this with respect to t.  This gives
r"(t) = [-2K/(4*pi*r^3(t))]r'(t)
     = [-K/(2*pi*r^3(t))]r'(t).

A few lines ago, r'(t) was given as K/(4*pi*r^2(t)).
Putting this in to the equation for r"(t) gives
r"(t) = [-K/(2*pi*r^3(t))][K/(4*pi*r^2(t))].

This reduces to -K^2/[8*pi^2*r^5(t)].

I'm not sure this is right, but take a look at it and point out any errors.

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