Aspherical balloon is inflated so that its volume is increasing at the rate of 3.9 ft3/min. how rapidly is the diameter of the balloon increasing when the diameter is 1.3 feet?

The volume of the sphere is obtained by the formula, 

                                          V = πD^3/6

Differentiating both sides in terms of time will give,

                                     dV/dt = (0.5πD^2)dD/dt

Given that the volume is changing at a rate of 3.9 ft^3/min and the diameter is 1.3 ft. dD/dt is obtained by simplifying the equation

                                    3.9 ft^3/min = (0.5π(1.3^2))dD/dt

Thus, dD/dt is approximately 1.469 ft/min.

The slope of the function is 0.5.

Bc equals mb that should be your answer