Can the Volume of a Sphere be Calculated Using its Surface Area and Radius?

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SUMMARY

The volume of a sphere can be calculated using its surface area and radius through the equations v = (4/3)πr³ for volume and x = 4πr² for surface area. Differentiating these equations with respect to time t yields dv/dt = 4πr²(dr/dt) and dx/dt = 8πr(dr/dt). By comparing these results, it is established that dv/dt = (1/2)r(dx/dt), confirming the relationship between the rates of change of volume and surface area with respect to time.

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wolfsprint
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The volume of a sphere is = v cm^3, it's radius is = r cm, it's surface area is = x cm^2, prove that : dv/dt = 1/2 r.dx/dt
 
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Express the volume and the surface area through r(t) and differentiate them.
 
I tried and I couldn't, can you please show the work?
 
wolfsprint said:
I tried and I couldn't, can you please show the work?
Well, start by showing your work in expressing the volume and the surface area through the radius. You should not do it yourself; you should just find the formulas in a textbook or online. Then what exactly is your difficulty in differentiating? You need to use the chain rule because the volume is a function of r and r is a function of t.
 
The only equation i managed to find is v=4/3 "pi" r^3 which is he volume of the sphere.
So if we say that the surface area = x ,then
V=1/3 r . X
But then when i differentiate it it turns out like this ,
Dv/dt = 1/3 dr/dt . Dx/dt , and that prove needs dv/dt = 1/2 r . Dx/dt
 
First, do not mix uppercase and lowercase letters, such as X and x. In mathematics, they often denote different entities.

wolfsprint said:
The only equation i managed to find is v=4/3 "pi" r^3 which is he volume of the sphere.
So if we say that the surface area = x ,then
V=1/3 r . X
Since you found the relationship between v and x, you must know that $x = 4\pi r^2$, so this is the second equation that you have. You can also find this in Wikipedia.

wolfsprint said:
But then when i differentiate it it turns out like this ,
Dv/dt = 1/3 dr/dt . Dx/dt , and that prove needs dv/dt = 1/2 r . Dx/dt
In post #2, I recommended expressing both v and x only through r and then differentiating them as compositions v(r(t)) and x(r(t)). Here you expressed v through x and r.
 
I still don't understand , what do you by "expressing"?
 
wolfsprint said:
I still don't understand , what do you by "expressing"?
By expressing $v$ and $x$ through $r$ I mean finding formulas containing constants and $r$ only that give the values of $v$ and $x$. These formulas are $v=\frac{4}{3}\pi r^3$ and $x=4\pi r^2$. Then we have $\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$. Similarly, find $\frac{dx}{dt}$ (keep in mind that $r$ is a function of $t$, i.e., $x(t)$ is a composite function $x(r(t))$, just like $v(r(t))$) and compare the results.
 
Thanks a lot but i still didnt get what you are trying to tell me , I've got all these equations on the paper and i tried to substitute each in the original but it just won't get the right answer! the fact that dr/dt exists is not right :(:confused:
 
  • #10
So, we established that

\[\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}\tag1\]

Next, since $x=4\pi r^2$, we have

\[\frac{dx}{dt}=8\pi r\frac{dr}{dt}\tag2\]

Comparing the right-hand sides of (1) and (2), we see that $\frac{dv}{dt}=\frac{1}{2}r\frac{dx}{dt}$.
 
  • #11
Using the given variables, we have:

The volume of the sphere is:

(1) $\displaystyle v=\frac{4}{3}\pi r^3$

The surface area is:

(2) $\displaystyle x=4\pi r^2$

We are asked to show:

$\displaystyle \frac{dv}{dt}=\frac{1}{2}r\frac{dx}{dt}$

Now, differentiating (1) and (2) with respect to time t, solving both for $\displaystyle \frac{dr}{dt}$ and equating, what do we find?
 
  • #12
Thanks a lot! i can't believe i missed that
 

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