Related Rates: If a snowball melts so that its surface area decreases at a rate of 6 cm2/min, find the rate at which the diameter decreases when the diameter is 11 cm.

for a sphere,

a = πd^2
da/dt = 2π dd/dt
-6 = 2π*11 dd/dt

So, dd/dt = -3/(11π) cm/min

To find the rate at which the diameter decreases, we can use related rates. We know that the surface area of the snowball decreases at a rate of 6 cm^2/min. We need to find the rate of change of the diameter when the diameter is 11 cm.

Let's start by relating the surface area of a sphere to its diameter. The formula for the surface area of a sphere is given by:

A = 4πr^2,

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

Since the diameter (d) of a sphere is twice the radius (r), we can write:

d = 2r.

To find the rate of change of the diameter (dd/dt), we differentiate both sides of the equation with respect to time (t):

dd/dt = 2(dr/dt).

Now, we need to relate the rate of change of the surface area (dA/dt) to the rate of change of the radius (dr/dt). We are given that dA/dt = -6 cm^2/min (the negative sign indicates a decrease), and we need to find dr/dt when d = 11 cm.

To do this, we use the chain rule from calculus. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). In this case, we have:

A = 4πr^2,
dA/dt = 8πr * dr/dt.

Now we can substitute the given values and solve for dr/dt:

-6 = 8π(11) * dr/dt.

Simplifying the equation gives:

-6 = 88π * dr/dt.

Dividing both sides by 88π:

dr/dt = -6 / (88π) ≈ -0.0022 cm/min.

Therefore, when the diameter is 11 cm, the rate at which the diameter decreases is approximately -0.0022 cm/min.