# How fast is the radius of the basketball increasing when the radius is 16 cm if air is being pumped into a basketball at a rate of 100 cm3/sec?

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To find the rate at which the radius of the basketball is increasing, we can use the formula relating the volume ( V ) of a sphere to its radius ( r ):

[ V = \frac{4}{3}\pi r^3 ]

Taking the derivative of both sides with respect to time ( t ), we get:

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

Given that the volume is increasing at a rate of ( \frac{dV}{dt} = 100 ) cm³/s when the radius ( r = 16 ) cm, we can plug in these values into the equation and solve for ( \frac{dr}{dt} ), the rate at which the radius is increasing:

[ 100 = 4\pi (16)^2 \frac{dr}{dt} ] [ \frac{dr}{dt} = \frac{100}{4\pi (16)^2} ] [ \frac{dr}{dt} = \frac{100}{4\pi \times 256} ] [ \frac{dr}{dt} = \frac{100}{1024\pi} ] [ \frac{dr}{dt} ≈ \frac{100}{3207} ] [ \frac{dr}{dt} ≈ 0.0312 , \text{cm/s} ]

Therefore, the radius of the basketball is increasing at a rate of approximately ( 0.0312 ) cm/s when the radius is 16 cm.

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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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