Question

A spherical balloon is inflating with helium at a rate of. How fast is the​ balloon's radius increasing at the instant the radius is ​ft? how fast is the surface area​ increasing?.

Answer 1

Using implicit differentiation, and considering that the inflation rate is of 200 cubic feet per minute and the radius is of 5 feet, it is found that:

• The radius is increasing at a rate of 0.6366 feet per minute.

• The surface area is changing at a rate of 80 feet squared per minute.

The volume of an sphere of radius r is given by:

$V = \frac{4\pi r^3}{3}$

Applying implicit differentiation, the rate of change is given by:

$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$

For this problem:

• Inflation rate of 200 cubic feet per minute, hence $\frac{dV}{dt} = 200$
• Radius is of 5 feet, hence $r = 5$

Then:

$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$

$200 = 4\pi (5)^2\frac{dr}{dt}$

$\frac{dr}{dt} = \frac{200}{100\pi}$

$\frac{dr}{dt} = 0.6366$

The radius is increasing at a rate of 0.6366 feet per minute.

The surface area is:

$S = 4\pi r^2$

Hence, the rate of change is:

$\frac{dS}{dt} = 8\pi r\frac{dr}{dt}$

Hence:

$\frac{dS}{dt} = 8\pi (5)(0.6366) = 80$

The surface area is changing at a rate of 80 feet squared per minute.

A similar problem is given at brainly.com/question/9543179