If a snowball melts so that its surface area decreases at a rate of 9 cm^2/min, find the rate at which the diameter decreases when the diameter is 11 cm.

We know that the surface area of a sphere is given by the formula:

A = 4\pi r^2

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

Given that the snowball is melting and the surface area is decreasing at a rate of 9 cm^2/min , we can find the rate at which the radius is decreasing by differentiating the formula for surface area with respect to time:

\frac{dA}{dt} = 8\pi r \frac{dr}{dt}

Given that \frac{dA}{dt} = -9 cm^2/min (since the surface area is decreasing), and we are looking for the rate at which the diameter (which is twice the radius) is decreasing when the diameter is 11 cm, we need to express the rate of change in terms of the diameter.

Since diameter D = 2r , we can write:

r = \frac{1}{2} D

Differentiating with respect to time:

\frac{dr}{dt} = \frac{1}{2} \frac{dD}{dt}

Substitute this relationship into the formula for the rate of change of surface area:

8\pi r \frac{dr}{dt} = -9

8\pi \left(\frac{1}{2} D\right) \left(\frac{1}{2} \frac{dD}{dt}\right) = -9

2\pi D \frac{dD}{dt} = -9

\frac{dD}{dt} = -\frac{9}{2\pi D}

When the diameter D = 11 ,

\frac{dD}{dt} = -\frac{9}{2\pi \times 11}

\frac{dD}{dt} = -\frac{9}{22\pi}

\boxed{\frac{dD}{dt} = -\frac{9}{22\pi} \, cm/min}