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

Answer:

Rate of increase of radius = 0.0064 ft/sec

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

Step-by-step explanation:

Given that:

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

$\dfrac{dV}{dt} = 8 ft^3/min$

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:

$1.\ V = \dfrac{4}{3}\pi r^3\\2.\ A =4\pi r^2$

Now, differentiating the volume and area, we get:

$\dfrac{dV}{dt} = \dfrac{4}{3}\times 3 \pi r^2 \dfrac{dr}{dt}\\\Rightarrow \dfrac{dV}{dt} = 4 \pi r^2 \dfrac{dr}{dt}$

$\dfrac{dA}{dt} = 4\pi \times 2 r \dfrac{dr}{dt}\\\Rightarrow \dfrac{dA}{dt} = 8\pi r \dfrac{dr}{dt}$

$\dfrac{dV}{dt} = 8 = 4\pi r^2 \dfrac{dr}{dt}\\\Rightarrow 8 = 4\times 3.14 \times 10^2 \dfrac{dr}{dt}\\\Rightarrow 2 = 3.14 \times 10^2 \dfrac{dr}{dt}\\\Rightarrow \dfrac{dr}{dt} = 0.0064\ ft/sec$

$\dfrac{dA}{dt} = 8\times \pi \times r \dfrac{dr}{dt}\\\Rightarrow \dfrac{dA}{dt} = 8\times 3.14 \times 10 \times 0.0064\\\Rightarrow \dfrac{dA}{dt} = \bold{1.61 \ ft^2/sec}$