a spherical ballon is inflated with gas at the rate 20cm^3 min. how fast is the radius of the ballon changing at the instant when the radius is 2cm

just so

To find the rate at which the radius of the balloon is changing, we can use the formula for the volume of a sphere, which is V = (4/3)πr^3, where V is the volume and r is the radius.

Given that the volume of the balloon is changing at a rate of 20 cm^3/min, we can express this as dV/dt = 20 cm^3/min.

We want to find how fast the radius, dr/dt, is changing when the radius is 2 cm. We can use implicit differentiation to determine the relationship between dV/dt and dr/dt.

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

dV/dt = (4/3)π(3r^2)(dr/dt)

Simplifying this equation, we have:

20 = (4/3)π(3r^2)(dr/dt)

Now, substitute the given radius, r = 2 cm, into the equation:

20 = (4/3)π(3(2^2))(dr/dt)

20 = (4/3)π(3(4))(dr/dt)

20 = (4/3)π(12)(dr/dt)

20 = 16π(dr/dt)

To solve for dr/dt, we can isolate it by dividing both sides by 16π:

20 / 16π = dr/dt

dr/dt ≈ 0.397 cm/min

Therefore, the radius of the balloon is changing at a rate of approximately 0.397 cm/min when the radius is 2 cm.

Time to start showing some of your work on these. They're all basically the same.

v = 4/3 πr^3
dv/dt = 4πr^2 dr/dt
...

We'll be happy to check your work on future problems.

okai now I have the idea I will do myself from now on and for this

20cm^3/min=4π(2cm)^2dr/dt