If a snowball melts so that its surface area decreases at a rate of 3 cm^2/min, find the rate at which the diameter decreases when the diameter is 8 cm.
Just about the same question as yours, all you have to do is change one number
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To find the rate at which the diameter decreases, we need to relate it to the rate at which the surface area decreases. We can do this using the formula for the surface area of a sphere:
Surface area of a sphere = 4πr^2,
where r is the radius of the sphere.
Given that the rate at which the surface area decreases is 3 cm^2/min, we can differentiate the surface area equation with respect to time to find the rate at which the radius changes:
d/dt(Surface area of a sphere) = d/dt(4πr^2).
The left-hand side represents the rate of change of the surface area, which is -3 cm^2/min (negative because the surface area is decreasing). The right-hand side represents the rate of change of the radius, which is what we're trying to find.
-3 cm^2/min = d/dt(4πr^2).
Now, we can solve for the rate at which the radius changes:
d/dt(4πr^2) = -3 cm^2/min.
Differentiating with respect to time, we get:
8πr(dr/dt) = -3 cm^2/min.
At this point, we have an equation in terms of the radius (r) and the rate at which the radius changes (dr/dt). We are given that the diameter is 8 cm, which means the radius (r) is half of that, so r = 4 cm.
Plugging r = 4 cm into the equation, we have:
8π(4)(dr/dt) = -3 cm^2/min.
Simplifying further, we get:
32π(dr/dt) = -3 cm^2/min.
Finally, we can solve for the rate at which the diameter decreases when the diameter is 8 cm:
dr/dt = (-3 cm^2/min) / (32π).
Calculating this expression gives us the rate at which the diameter decreases.