# The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 40 mm?

or

The formula for a solid sphere's volume is:

When you take the derivative of both sides with respect to time...

...remember the Chain Rule for implicit differentiation. The general format for this is:

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.

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To find the rate at which the volume of the sphere is increasing when the diameter is 40 mm, we'll first find the rate of change of the volume with respect to time using the formula for the volume of a sphere, ( V = \frac{4}{3}\pi r^3 ), where ( r ) is the radius.

Given that the radius is increasing at a rate of ( \frac{dr}{dt} = 4 ) mm/s, we can find ( \frac{dV}{dt} ) using the chain rule:

[ \frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} ]

Substituting ( V = \frac{4}{3}\pi r^3 ):

[ \frac{dV}{dt} = 4\pi r^2 \times 4 ]

Given that the diameter is 40 mm, the radius ( r = \frac{d}{2} = \frac{40}{2} = 20 ) mm.

[ \frac{dV}{dt} = 4\pi (20)^2 \times 4 ]

[ \frac{dV}{dt} = 3200\pi ]

So, the volume of the sphere is increasing at a rate of ( 3200\pi ) mm³/s when the diameter is 40 mm.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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