A spherical snowball is placed in the sun. The snowball melts so that it's surface area
decreases at a rate of 2 cm2
/min. Find the rate at which the diameter decreases when the
diameter is 8 cm.
A = πd^2
dA/dt = 2πd dd/dt
Now plug in your numbers and solve for dd/dt
To find the rate at which the diameter decreases when the diameter is 8 cm, we need to use related rates.
Let's start by writing down what we know:
Given: The surface area of the snowball is decreasing at a rate of 2 cm^2/min.
We need to find: The rate at which the diameter decreases (dD/dt) when the diameter is 8 cm.
To solve this problem, we'll use the formula for the surface area of a sphere:
Surface Area (A) = 4πr^2
Where r is the radius of the sphere. Since we have the diameter, we can conclude that the radius (r) is half the diameter (r = D/2).
Differentiating both sides of the equation with respect to time (t) using the chain rule, we get:
dA/dt = 8πr * dr/dt
Now we can substitute the given rate of change of the surface area (dA/dt = -2 cm^2/min) and the given diameter (D = 8 cm) into the equation:
-2 = 8π(8/2) * dD/dt
Simplifying the equation:
-2 = 8π(4) * dD/dt
-2 = 32π * dD/dt
Now, we can solve for dD/dt, which represents the rate at which the diameter decreases:
dD/dt = -2 / (32π)
dD/dt = -1 / (16π) cm/min
So, the rate at which the diameter decreases when the diameter is 8 cm is -1 / (16π) cm/min.