A spherical ballon is being inflated at the rate of 35cc/min. the rate of increase in the surfece area(un cm²/min.) of the ballon when its diameter is 14cm is:

1. 10 * under root 10
2. under root 10
3. 100
4. 10

V = (4/3)πr^3

dV/dt = 4π r^2 dr/dt
when r = 7 , dV/dt = 35
35 = 4π(49) dr/dt
dr/dt = 35/(196π)

A = 4πr^2
dA/dt = 8π r dr/dt
when r = 7, dr/dt = 35/(196π)

dA/dt = 8π)(7)(35/(196π) )
= 1960π/(196π)
= 10 cm^2/min

dA/dt = 8π(7)(35) = 1960π cm^2/min

To find the rate of increase in the surface area of the balloon, we need to use the formula for the surface area of a sphere. The surface area of a sphere is given by the formula: S = 4πr², where S is the surface area and r is the radius of the sphere.

Given that the diameter of the balloon is 14 cm, we can find the radius by dividing the diameter by 2:
radius (r) = 14 cm / 2 = 7 cm

Now, to find the rate of increase in the surface area, we need to differentiate the surface area formula with respect to time.

dS/dt = d/dt(4πr²)

Using the chain rule, we can differentiate the equation.

dS/dt = 8πr * dr/dt

Now, let's substitute the values into the equation. We are given that the balloon is being inflated at a rate of 35 cc/min, which is equivalent to 35 cm³/min. We can use this to find the rate at which the radius is increasing.

dr/dt = 35 cm³/min

Plugging in the values into the equation, we get:

dS/dt = 8π * 7 * 35

Simplifying further:

dS/dt = 1960π cm²/min

To determine the numerical value, we can approximate π as 3.14:

dS/dt ≈ 1960 * 3.14

dS/dt ≈ 6148.4 cm²/min

Therefore, the rate of increase in the surface area of the balloon when its diameter is 14 cm is approximately 6148.4 cm²/min.

None of the provided options matches this value exactly, but the closest approximation to 6148.4 is 100 cm²/min, which is option 3.