Question

A high-altitude spherical weather balloon expands as it rises due to the drop in atmospheric pressure. Suppose that the radius r increases at the rate of 0.03 inches per second and that r=48 inches at time t=0. Determine a function that models the volume V of the balloon at time t and find the volume at t=300 seconds.

Answer 1

Considering a linear function for the radius, we have that:

• The volume of the sphere is modeled by $V(t) = \frac{4\pi(0.03t + 48)^3}{3}$.

• After 300 seconds, it is of 775,735 in³.

What is a linear function?

A linear function is modeled by:

$y = mx + b$

In which:

• m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.

• b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value.

What is the volume of a sphere?

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

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

In this problem, radius r increases at the rate of 0.03 inches per second and that r=48, hence it is represented by the following linear function:

$r(t) = 0.03t + 48$

Thus, the volume of the sphere is given by:

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

$V(t) = \frac{4\pi(0.03t + 48)^3}{3}$

After 300 seconds, the volume in cubic inches is given by:

$V(300) = \frac{4\pi(0.03(300) + 48)^3}{3} = 775735$

More can be learned about linear functions at https://brainly.com/question/24808124