If a snowball melts so that its surface area decreases at a rate of 4 cm2/min, find the rate at which the diameter decreases when the diameter is 9 cm.
SA = 4πr^2
d(SA)/dt = 8πr dr/dt
for the given data
4 = 8π(4/5) dr/dt
dr/dt = ....
d(diameter)/dt = 2 dr/dt
.1990
To find the rate at which the diameter decreases, we need to establish a relationship between the surface area of the snowball and its diameter.
The surface area of a sphere can be calculated using the formula: A = 4πr², where A represents the surface area and r represents the radius.
Since the diameter (d) is equal to twice the radius (r), we can rewrite the formula as: A = 4π(d/2)² = πd².
Now, we need to differentiate the surface area equation with respect to time (t) to find the rate of change of the surface area: dA/dt = d(πd²)/dt.
Using the chain rule, we can differentiate the equation as follows:
dA/dt = 2πd * dd/dt.
We are given that the surface area is decreasing at a rate of 4 cm²/min, so dA/dt = -4 cm²/min. We need to find the rate at which the diameter (d) is decreasing, or dd/dt.
Substituting the given values and rearranging the equation, we get:
-4 = 2π * 9 * dd/dt.
Simplifying further:
dd/dt = -4 / (18π).
Therefore, when the diameter is 9 cm, the rate at which the diameter decreases is approximately:
-4 / (18π) cm/min.