Thread: Instantaneous rate of change of a sphere

1. Instantaneous rate of change of a sphere

Hi, I need help finding the final answer of this, I know how to do it, but i'm not sure what the final answer should be:

The volume, V, of a sphere of radius r is given by V = f(r) = 4pi(r^3) / 3. Calculate the instantaneous rate of chnage of the volume, V, with respect to the change of the radius, r, at r = 15.1.

(doing this without derivatives)

We calculate the instantaneous rate of change of the volume, V, with the respect to change of the radius, r, at r = 15.1, as

lim r->15.1 ( f(r) - f(15.1) ) / (r - 15.1)
=
lim r->15.1 ( (4pi(r^3)/3) - (4pi(15.1^3)/3) ) / (r - 15.1)
=
4pi/3 * lim r->15.1 (r^3 - 15.1^3) / (r - 15.1)
= ___

I know to use the difference of cubes formula so I get:

4pi/3 * lim x->r (r² + 15.1r + 15.1²)

Thanks!

2. Originally Posted by coldfire
Hi, I need help finding the final answer of this, I know how to do it, but i'm not sure what the final answer should be:

The volume, V, of a sphere of radius r is given by V = f(r) = 4pi(r^3) / 3. Calculate the instantaneous rate of chnage of the volume, V, with respect to the change of the radius, r, at r = 15.1.

(doing this without derivatives)

We calculate the instantaneous rate of change of the volume, V, with the respect to change of the radius, r, at r = 15.1, as

lim r->15.1 ( f(r) - f(15.1) ) / (r - 15.1)
=
lim r->15.1 ( (4pi(r^3)/3) - (4pi(15.1^3)/3) ) / (r - 15.1)
=
4pi/3 * lim r->15.1 (r^3 - 15.1^3) / (r - 15.1)
= ___

I know to use the difference of cubes formula so I get:

4pi/3 * lim x->r (r² + 15.1r + 15.1²)

Thanks!
what happened to that red thing? it should be r->15.1 and do limit thing..

3. Yeah sorry my mistake, it should be:

4pi/3 * lim r->15.1 (r² + 15.1r + 15.1²)

Can anyone help me out?

4. from there, you can just substitute 15.1 to the remaining function since it is now defined there..