Question

7) A spherical balloon is inflated so that its radius increases at a rate of 2 cm/sec. How fastis the volume of the balloon increasing when the radius is 3 cm?4(Use V =for the volume of a sphere)3A) 7270 cm/sec B) 791 cm/sec C) 70 cm/sec D) 8210 cm/sec

Answer 1

The formula for the volume of a sphere is given by:

`V=(4)/(3)\pi r^3`

Since we are asked the rate of change of the volume with repect to time, we take the derivative on both sides, taking into account the chain rule:

`(dV)/(dt)=(d((4)/(3)\pi r^3))/(dt)`

taking out the constants:

`(dV)/(dt)=(4)/(3)\pi(d(r^3))/(dx)`

Now we derivate, using the chain rule, that is:

`(df(g(x)))/(dx)=f^(\prime)(x)g^(\prime)(x)`

Applying the rule:

`(dV)/(dt)=(4)/(3)\pi(3r^2)(dr)/(dt)`

Simplifying:

`(dV)/(dt)=4\pi(r^2)(dr)/(dt)`

We have the following known values:

Replacing we get:

`(dV)/(dt)=4\pi(3)^2(2)`

Solving we get: