Expert: Alon Mandes - 9/27/2010

Question
QUESTION: Dear Alon
To be honest, I have received two very different answers to this question, so need a third that hopefully matches one of them.
For any point on the surface of a sphere expanding at a given constant volumetric rate, what, at any given radius are the formulae for calculating (relative to the centre) its (1) instantaneous deceleration & (2) instantaneous velocity?  I am assuming this would lead to a value of velocity that is considerably greater than that of deceleration for any particular case (taking it that both would relate to the same units of time and distance – say seconds and metres).
Kind regards
Alan


ANSWER: Hello Alan,
1st of all let's write down the formula for volume : V=(4/3)πr³ .
In our case both V and r are function of time, thus
V(t)=(4/3)πr³(t). Now we get a relation between r and V :
r³(t)=3V(t)/4π . It's given that the sphere is expanding at a constant rate, let's call it α , this means that V'(t)=α which leads to V(t)=αt .
Now we plug this result into the expression of r(t), we get :
r³(t)=3αt/4π Or we can simply say r(t)=[3αt/4π]^(⅓) . Both forms
are the same !
Why the interest of r(t) ?
r(t) can be reviewed as the distance travelled by the point , and
since we know that the velocity and deceleration are the 1st & 2nd derivative of r(t), then r(t) is the key .
At a given raduis ro , the velocity and deceleration will be :
velocity(t)=dr/dt. In our expression r³(t)=3αt/4π , so
3r²(t)(dr/dt)=3α/4π . At r=ro we get dr/dt=r'=α/(4πro²).

deceleration(t)=d²r/dt².
From the expression : 3r²(t)(dr/dt)=3α/4π we can find r''(t) :
[ 3r²(t)r'(t) ]' = [ 3α/4π ]'
6r(t)r'(t)+3r²(t)r''(t)=0
2r'(t)+r(t)r''(t)=0
d²r/dt²=r''(t)=-2r'(t)/r(t)
At a given radius ro we get :
r''=-2r'(ro)/ro=-2α/(4πro²)/ro=-α/2πro

On the other hand , if we were interested in r' & r'' at t=to ,
then we will have to derive the 2nd expression "r(t)=[3αt/4π]^(⅓)".

Alon.


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

QUESTION: Dear Alom

Thanks for your rapid reply.
Can you please confirm (or explain otherwise) your answers may be summarised thus: -
(1)   Instantaneous deceleration = α/2πr, where α is the constant rate of volume increase.
(2)   Instantaneous velocity = α/4πr2.

If these interpretations are correct, your answer to (2) agrees with those from the other two volunteers, but that for (1) is very different and appears to be incorrect, as it means velocity would always be much less than deceleration, which cannot be feasible.

Kind regards
Alan


Answer
Hello Alan, yes there is amistake in calculating the deceleration
and for that I apologize ! here's the correction.
r'=velocity
r''=deceleration.
We know that V=(4/3)πr³ , thus (4/3)πr³=αt . Hence :
4πr²r'=α --> r'=α/4πr² (velocity)
We derive 2nd time, we get :
8πr(r')²+4πr²r''=0 --> r''=-[8πr(r')²]/[4πr²]=-2(r')²/r=
-2α²/[16π²r^5] .

So,

Velocity     = α/4πr²
Deceleration = -2α²/[16π²r^5]

Again, my apologies for the mistake .

Alon.