A spherical balloon is being inflated in such a way that its radius increases at a rate of 3 cm/min. If the volume of the balloon is 0 at time 0, at what rate is the volume increasing after 5 minutes?

my answer is 45 cm/min.

is this right. thanks

way off.

First, volume is cm^3

v = 4/3 pi r^3
dv/dt = 4 pi r^2 dr/dt
at t=5, r=15 so
dv/dt = 4 pi * 225 * 3 = 2700 pi cm^3/min

How did you come up with 45? Way small, and I'd surely expect a pi to appear somewhere in any problem involving a sphere.

Thank you!

My fault lied in plugging in directly. Thank you for showing me the steps.

Well, let's see if we can calculate it together!

The formula for the volume of a sphere is V = (4/3)πr³, where V is the volume and r is the radius.

We know that the radius is increasing at a rate of 3 cm/min. So, after 5 minutes, the radius will have increased by 15 cm, since 3 cm/min * 5 min = 15 cm.

Now, let's substitute this new radius into the volume formula:

V = (4/3)π(15)³
V = (4/3)π(3375)
V ≈ 4500π cm³

To find the rate at which the volume is increasing after 5 minutes, we need to calculate the derivative dV/dt:

dV/dt = 4πr²(dr/dt)

Using the given rate of change for the radius (dr/dt = 3 cm/min), we can calculate:

dV/dt = 4π(15)²(3)
dV/dt = 540π cm³/min

So, the rate at which the volume is increasing after 5 minutes is approximately 540π cm³/min, which is roughly 1696 cm³/min.

Well, that's quite an increase! I guess that's one balloon you wouldn't want to let go of during a party!

To solve this problem, we need to use the formula for the volume of a sphere and apply the concept of rates of change.

The formula for the volume of a sphere is given by V = (4/3)πr^3, where V is the volume and r is the radius of the sphere.

We know that the radius of the balloon is increasing at a rate of 3 cm/min. We want to find the rate at which the volume is increasing after 5 minutes.

To find the rate at which the volume is changing, we need to take the derivative of the volume function with respect to time. Since the radius is changing with respect to time, we can use the chain rule to find the rate of change of the volume.

Let's start by finding the rate at which the volume is changing, which can be denoted as dV/dt.

dV/dt = dV/dr * dr/dt

dV/dr is the derivative of the volume function with respect to the radius, which is given by dV/dr = 4πr^2.

dr/dt is the rate at which the radius is changing, which is 3 cm/min, as given in the problem.

Substituting these values into the equation, we have:

dV/dt = 4πr^2 * 3

Now, we need to find the radius r after 5 minutes. Since the radius is increasing at a constant rate of 3 cm/min, after 5 minutes, the radius will be 3 * 5 = 15 cm.

Now, substitute the value of r into the equation:

dV/dt = 4π(15^2)*3 = 4π(225)*3 = 2700π cm^3/min

So, the rate at which the volume is increasing after 5 minutes is 2700π cm^3/min, which is approximately 8477 cm^3/min.

Therefore, your answer of 45 cm/min is not correct. The correct answer is approximately 8477 cm^3/min.