A spherical balloon is being filled with air in such a way that its radius is increasing at a rate of 2 centimeters per second. At what rate is the volume of the balloon changing at the instant when its surface has an area of 4 pi or 12.56637061 square centimeters?

Volume = V = (4/3) pi r^3

dV/dt = 4 pi r^2 dr/dt
When the area 4 pi r^2 = 4 pi, the value of r is 1.

At that instant,
dV/dt = 4 pi *(1 cm^2)*2 cm/s
= 8 pi cm^3/s

A. Spherical balloon filled with helium at 1 atm just lift a 2kg lead what is its radius

To find the rate at which the volume of the balloon is changing at the instant when its surface has an area of 4π square centimeters, we need to find the rate of change of the volume with respect to time.

Let's denote the radius of the balloon at any given time as r(t), and the volume of the balloon at that time as V(t).

We are given that the radius of the balloon is increasing at a rate of 2 centimeters per second, which means dr/dt = 2 cm/s.

We also know that the surface area of the balloon is given by 4πr^2. We are given that the surface area is 4π square centimeters. So, we can write:

4π = 4πr^2

Dividing both sides by 4π gives us:

r^2 = 1

Taking the square root of both sides gives us:

r = 1 cm

So, at the instant when the surface area is 4π square centimeters, the radius of the balloon is 1 cm.

To find the rate of change of the volume with respect to time, we need to find dV/dt. The volume of a sphere is given by V = (4/3)πr^3. So, we can write:

V = (4/3)π(1^3)
V = (4/3)π cubic centimeters

Differentiating both sides with respect to time gives us:

dV/dt = d/dt [(4/3)π]
dV/dt = 0

Therefore, at the instant when the surface area is 4π square centimeters, the volume of the balloon is not changing, or dV/dt = 0 cubic centimeters per second.

To find the rate at which the volume of the balloon is changing, we need to differentiate the volume formula with respect to time.

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

Differentiating both sides of the equation with respect to time (t), we get:

dV/dt = d/dt [(4/3)πr^3]

Using the chain rule, the derivative of r^3 with respect to t is:

d(r^3)/dt = 3r^2 (dr/dt)

Thus, we end up with:

dV/dt = (4/3)π * 3r^2 * (dr/dt)

Simplifying further, we have:

dV/dt = 4πr^2 * (dr/dt)

We are given that the rate of change of the radius (dr/dt) is 2 cm/s. We need to find the rate of change of the volume (dV/dt) when the surface area of the balloon is 4π or 12.56637061 cm^2.

The surface area of a sphere can be calculated using the formula A = 4πr^2, where A is the surface area and r is the radius.

Given A = 4π, we can solve for the radius:

4π = 4πr^2

Dividing both sides by 4π, we get:

r^2 = 1

Taking the square root of both sides, we find:

r = 1 cm

Substituting the given values of r and dr/dt into our equation for dV/dt:

dV/dt = 4π * (1 cm)^2 * (2 cm/s)

Simplifying, we have:

dV/dt = 4π * 2 cm^3/s

Finally, we can compute the value of dV/dt:

dV/dt = 8π cm^3/s

Therefore, at the instant when the surface area of the balloon is 4π or 12.56637061 cm^2, the volume of the balloon is changing at a rate of 8π cm^3/s.