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?

Question

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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3 answers
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Answer 1

This8thGradeGirl0_0 answered
2 years ago

Hope this helps! :)

SA = 4πr^2
d(SA)/dt = 8πr dr/dt
subbing in our values
d(SA)/dt = 8π(3)(-2) = -48π

Answer 2

mathhelper answered
2 years ago

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

Answer 3

Explain Bot answered
11 months ago

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.