A spherical balloon is inflated so that its volume is increasing at the rate of 3.2 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.1 feet?

V=4/3 PI (D/2)^3=1/6 PI D^3

dV/dt=PI/6 *3D^2 * dD/dt
solve for d(diameter)/dt

To find out how rapidly the diameter of the balloon is increasing, we need to relate the volume of the balloon to its diameter.

The volume of a sphere is given by the formula V = (4/3)πr^3, where V is the volume and r is the radius. Since the diameter is given, we need to relate it to the radius.

The radius (r) of a sphere is half of its diameter (d), so r = d/2. Thus, we can rewrite the volume formula as V = (4/3)π(d/2)^3.

Next, we want to find the rate of change of the diameter (ddt) when the volume is increasing at a rate of 3.2 ft^3/min. We differentiate the volume equation with respect to time (t).

Differentiating both sides of the equation, we get:

dV/dt = d((4/3)π(d/2)^3)/dt

To calculate dV/dt, we can differentiate the equation with respect to t using the chain rule:

dV/dt = (4/3)π * 3(d/2)^2 * (1/2) * dd/dt

Simplifying the equation further:

dV/dt = (2/3)π * (d/2)^2 * dd/dt

Since we are given that dV/dt = 3.2 ft^3/min, and we want to find d(d)/dt when d = 1.1 ft, we can substitute these values into the equation:

3.2 = (2/3)π * (1.1/2)^2 * dd/dt

Now, we can solve for dd/dt:

dd/dt = 3.2 / [(2/3)π * (1.1/2)^2]

By calculating this expression, we can find the rate at which the diameter of the balloon is increasing when the diameter is 1.1 ft.