a spherical ballon is being inflated at a constant rate of 25 cm^3/sec. At what rate in cm/sec is the radius of the ballon changing when the radius of the ballon is 2 cm? (VOLUME SPHERE FORMULA 4/3 pi r^3)

Well, well, well, it seems like we have a ballooning problem! Don't worry, I'll try to solve it without popping your bubble.

First, let's remind ourselves of the formula for the volume of a sphere: V = 4/3 * π * r^3, where V is the volume and r is the radius. We want to find dA/dt, the rate at which the radius is changing when the radius is 2 cm.

Now, to relate the volume and the radius, we need to differentiate both sides of the volume formula with respect to time (t). That gives us dV/dt = 4πr^2 * dr/dt. Since V is changing at a constant rate of 25 cm^3/sec, we have dV/dt = 25.

Now, when the radius is 2 cm, let's plug in the values we know: r = 2, dV/dt = 25. We can solve for dr/dt (the rate at which the radius is changing) as follows:

25 = 4π(2^2) * dr/dt
25 = 4π(4) * dr/dt
25 = 16π * dr/dt

dr/dt = 25 / (16π)

So, the rate at which the radius of the balloon is changing when the radius is 2 cm is approximately 25 / (16π) cm/sec.

But hey, remember to take everything I say with a grain of helium!

To find the rate at which the radius of the balloon is changing, we can use the formula for the volume of a sphere and differentiate it with respect to time.

Given:
- The rate of change of volume is constant, which is 25 cm^3/sec.
- The formula for the volume of a sphere is V = (4/3)πr^3.

We need to find dR/dt, the rate of change of the radius with respect to time.

Step 1: Differentiate the formula for the volume of a sphere with respect to time.

dV/dt = (4/3)(π)(3r^2)(dr/dt)

Step 2: Plug in the given rate of change of volume and the given radius value.

25 cm^3/sec = (4/3)(π)(3(2^2))(dr/dt)

Step 3: Solve for dr/dt, the rate at which the radius is changing.

dr/dt = (25 cm^3/sec) / [(4/3)(3(2^2))(π)]
dr/dt = (25 cm^3/sec) / [(4/3)(3(4))(π)]
dr/dt = (25 cm^3/sec) / [(4/3)(12)(π)]
dr/dt ≈ 25 cm^3/sec / (16π)

Therefore, when the radius of the balloon is 2 cm, the rate at which the radius is changing is approximately dr/dt ≈ 25 cm^3/sec / (16π) cm/sec.

To find the rate at which the radius of the balloon is changing, we can use the formula for the volume of a sphere and the given information about the rate of volume change.

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

Given that the balloon is being inflated at a constant rate of 25 cm^3/sec, we can express the rate of change of volume as dV/dt = 25 cm^3/sec.

We need to find the rate at which the radius of the balloon (dr/dt) is changing when the radius is 2 cm. To do this, we can differentiate the volume equation with respect to time, t.

dV/dt = d/dt[(4/3) * π * r^3]
dV/dt = (4/3) * π * d(r^3)/dt

Since V = (4/3) * π * r^3, we can substitute it into the differentiation equation:

25 = (4/3) * π * d(r^3)/dt

We can simplify this equation further:

25 = (4/3) * π * 3r^2 * (dr/dt)
25 = 4πr^2 * (dr/dt)

Now, substitute the given radius of 2 cm:

25 = 4π * (2)^2 * (dr/dt)
25 = 16π * (dr/dt)

Finally, we can solve for (dr/dt), i.e., the rate at which the radius is changing:

(dr/dt) = 25 / (16π)
(dr/dt) = 25 / (16 * 3.14)
(dr/dt) ≈ 0.498 cm/sec

Therefore, the rate at which the radius of the balloon is changing when the radius is 2 cm is approximately 0.498 cm/sec.

the surface area of the surface of the sphere, 4 pi r^2 times change in radius is the differential change in volume.

v = (4/3) pi r^3
dv/dt = 4 pi r^2 dr/dt

dr/dt = dv/dt /(4 pi r^2)