Assume the radius r of a sphere is expanding at a rate of 30 cm/min. the volume of a sphere is v = 4/3(pi)(r^3) and it's surface area is 4(pi)(r^2).determine the rate of change of the volume with respect to time at t = 2 min, assuming that r = 0 at t = 0.

Using implicit differentiation, it is found that the rate of change of the volume at time t = 2 is of 1,357,168 cm³/min.

The volume of the sphere is given by:

The rate of change is found applying implicit differentiation, so:

Radius expanding at a rate of 30 cm/min, thus .

Radius starts at 0 cm, thus, at 2 min, it will be of 60 cm, that is, .

Then:

The rate of change in the volume is of 1,357,168 cm³/min.

A similar problem is given at link

let's keep in mind that r = 0 at t = 0, namely the radius is 0 at 0 seconds, and we know the radius is expanding at 30 cm/min, so in t = 2, in 2 minutes the radius is then 30 + 30 = 60 cm wide.

step-by-step explanation:

1. practice, practice & more practice

2. review errors

3. master the key concepts

do not try to memorise the processes.

4. understand your doubts.

5. create a distraction free study environment

6. create a mathematical dictionary

7. apply maths to real world problems.