Question

The volume of a spherical balloon is increasing at the constant rate of 8 cubic feet per minute. How fast is the radius incresing when the radius is exactly 10 feet? How fast is the surface area increasing at that time?

Answer 1

Rate of increase of radius = 0.0064 ft/sec

Rate of increase of surface area = 1.61
`ft^2/sec`

Explanation:

Given that:

Rate of change of volume of a spherical balloon = 8 cubic feet per minute

`(dV)/(dt) = 8 ft^3/min`

Radius,
`r` = 10 feet

To find:

The rate of change of radius at this moment and rate of change of surface area at this moment?

Solution:

First of all, let us have a look at the formula:

`1.\ V = (4)/(3)\pi r^3\\2.\ A =4\pi r^2`

Now, differentiating the volume and area, we get:

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

`(dA)/(dt) = 4\pi * 2 r (dr)/(dt)\\\Rightarrow (dA)/(dt) = 8\pi r (dr)/(dt)`

`(dV)/(dt) = 8 = 4\pi r^2 (dr)/(dt)\\\Rightarrow 8 = 4* 3.14 * 10^2 (dr)/(dt)\\\Rightarrow 2 = 3.14 * 10^2 (dr)/(dt)\\\Rightarrow (dr)/(dt) = 0.0064\ ft/sec`

`(dA)/(dt) = 8* \pi * r (dr)/(dt)\\\Rightarrow (dA)/(dt) = 8* 3.14 * 10 * 0.0064\\\Rightarrow (dA)/(dt) = \bold{1.61 \ ft^2/sec}`