��Example: 1�Gas is escaping from a spherical balloon at the rate of 2 ft3 / min .How fast is the surface area shrinking when the radius is 12 ft ? ��

given that V'=2ft^3/min;r=12ft,V=(4pi r^3)/3,S=4pi r^2.

dV/dr=4pir^2; dS/dr=8pi r
dS/dt=dS/dr x V'/(dV/dr)
dS/dt=(8pi r x 2)/4pi r^2=1/3
dS/dt=1/3 ft^2/min

To find how fast the surface area is shrinking, we can use the formula for the rate of change in the surface area of a sphere with respect to time.

The formula is given by:

dS/dt = -4πr^2(dr/dt)

Where:
- dS/dt is the rate at which the surface area is changing,
- r is the radius of the sphere, and
- dr/dt is the rate at which the radius is changing.

In this example:
- The rate at which gas is escaping from the balloon is given as 2 ft^3/min.
- The radius of the balloon is 12 ft.

We need to find dS/dt when r = 12 ft.

First, let's find dr/dt when the volume of gas is escaping at a rate of 2 ft^3/min:

The volume V of a sphere is given by:
V = (4/3)πr^3

Differentiating with respect to time:
dV/dt = (dV/dr)(dr/dt)

Since dV/dt is given as 2 ft^3/min, we can solve for dr/dt:
2 = (4/3)π(12^2)(dr/dt)

Simplifying the equation:
2 = 576π(dr/dt)
dr/dt = 2 / (576π)
dr/dt ≈ 0.00108 ft/min

Now that we have the rate at which the radius is changing, let's find dS/dt:

Using the formula:
dS/dt = -4πr^2(dr/dt)

Substituting the values:
dS/dt = -4π(12^2)(0.00108)
dS/dt ≈ -2.066 ft^2/min

Therefore, the surface area is shrinking at a rate of approximately 2.066 ft^2/min when the radius is 12 ft.

To find the rate at which the surface area is shrinking, we need to use the formula for the surface area of a sphere, which is given by:

A = 4πr^2

Where A is the surface area of the sphere and r is the radius.

We are given that the radius is 12 ft and the gas is escaping at a rate of 2 ft^3/min. We need to find the rate at which the surface area is shrinking, so we need to find dA/dt, the derivative of the surface area with respect to time.

To do this, we can use implicit differentiation. Let's assume that A is a function of time, so we can write:

A(t) = 4πr(t)^2

Differentiating both sides of this equation with respect to time, we get:

dA/dt = 8πr(t) * dr(t)/dt

Now we need to find dr(t)/dt, the rate at which the radius is changing with respect to time. We are given that the rate at which the gas is escaping is 2 ft^3/min, which can also be written as dv(t)/dt, where v(t) is the volume of the balloon. Since the volume of a sphere is given by:

V = (4/3)πr^3

We can rewrite this in terms of r(t):

r(t) = (3V(t)/4π)^(1/3)

Now, differentiating both sides of this equation with respect to time, we get:

dr(t)/dt = (1/3) * (3V(t)/4π)^(-2/3) * dV(t)/dt

Since dV(t)/dt is the rate at which the gas is escaping, which is given as 2 ft^3/min, we can substitute this value into the equation:

dr(t)/dt = (1/3) * (3V(t)/4π)^(-2/3) * 2

Finally, substituting this value for dr(t)/dt back into the equation for dA/dt:

dA/dt = 8πr(t) * (1/3) * (3V(t)/4π)^(-2/3) * 2

Simplifying this expression, we can substitute the given values of r(t) = 12 ft and dV(t)/dt = 2 ft^3/min:

dA/dt = 8π * 12 * (1/3) * (3V(t)/4π)^(-2/3) * 2

Now we just need to calculate the value of (3V(t)/4π)^(-2/3) using the volume of the balloon:

V(t) = (4/3)πr(t)^3 = (4/3)π(12)^3 = 2304π

(3V(t)/4π)^(-2/3) = (3(2304π)/(4π))^(-2/3) = 18^-2/3

Simplifying further, we get:

18^-2/3 = 1/(18^2) = 1/324

Substituting this value back into the equation for dA/dt, we have:

dA/dt = 8π * 12 * (1/3) * (1/324) * 2

Simplifying this expression gives us the final answer for the rate at which the surface area is shrinking when the radius is 12 ft.