RaySix

Post

Rate of Surface Area Increase of Inflating Balloon

Created by @nathanedwards

at November 4th 2023, 10:26:47 pm.

AP_Calculus_AB-Questions

#geometry

#calculus

#mathematics

Question:

A spherical balloon is being inflated at a constant rate of 3 cubic centimeters per second. At the instant when the radius of the balloon is 4 centimeters, find the rate at which the surface area of the balloon is increasing.

Answer:

To find the rate at which the surface area of the balloon is increasing, we need to relate the surface area of the sphere to its radius.

The surface area of a sphere is given by the formula:

A = 4\pi r^2

Where:

A is the surface area of the sphere in square units, and
r is the radius of the sphere in units

Given that the balloon is being inflated at a constant rate of 3 cubic centimeters per second, this means that the volume of the balloon is increasing at a rate of 3 cubic centimeters per second.

The volume of a sphere is given by the formula:

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

Let's differentiate both sides of the equation with respect to time (t):

\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)

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

The left side represents the rate of change of volume (3 cubic centimeters per second), so we have:

3 = 4\pi r^2 \frac{dr}{dt}

Now, we are given that the radius of the balloon is 4 centimeters, so substituting this value into the equation, we get:

3 = 4\pi (4^2) \frac{dr}{dt}

Simplifying the equation:

3 = 4\pi (16) \frac{dr}{dt}

\frac{dr}{dt} = \frac{3}{64\pi}

Therefore, the rate at which the surface area of the balloon is increasing when the radius of the balloon is 4 centimeters is $\frac{3}{64\pi}$ cubic centimeters per second.