The radius of a spherical snowball is decreasing at a rate of 2 inches per hour. How fast is the snowball melting in terms of its volume when its diameter is 8 inches?
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SA = 4πr^2
d(SA)/dt = 8πr dr/dt
subbing in our values
d(SA)/dt = 8π(3)(-2) = -48π
It asked for the rate of change in terms of its VOLUME.
V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
= 4π (16)(-2)
= -128π cubic inches
To find the rate at which the snowball is melting in terms of its volume, we need to use the formula for the volume of a sphere:
V = (4/3)πr³
where V is the volume of the sphere and r is the radius.
Given that the rate of change of the radius is -2 inches per hour, we can differentiate the volume formula with respect to time (t) using the chain rule:
dV/dt = dV/dr * dr/dt
To find dV/dr, we differentiate the volume formula:
dV/dr = 4πr²
Substituting the given diameter of 8 inches, we can find the corresponding radius:
diameter = 2 * radius
8 = 2 * r
r = 4 inches
Now, we can calculate dV/dr at r = 4 inches:
dV/dr = 4π(4)²
dV/dr = 4π(16)
dV/dr = 64π
Next, we need to find dr/dt, the rate of change of radius with respect to time, which is given as -2 inches per hour.
Now we can substitute the values into the first equation:
dV/dt = dV/dr * dr/dt
dV/dt = 64π * (-2)
Finally, we can simplify the equation to find the rate at which the snowball is melting in terms of its volume:
dV/dt = -128π inches³ per hour
Therefore, the snowball is melting at a rate of -128π inches³ per hour when its diameter is 8 inches.