A spherical balloon is inflated so that its radius (r) increases at a rate of 2/r cm/sec. How fast is the volume of the balloon increasing when the radius is 4 cm?

Answer 1

#(dv)/dt = 32pi" "("cm"^3)/"s"#

Using the chain rule:

#(dv)/dt = (dv)/(dr)(dr)/dt#
given: #(dr)/dt = 2/r" ""cm"/"s"#

The volume of a sphere is:

#v = 4/3pir^3" cm"^3#
#(dv)/(dr) = 4pir^2" cm"^2#
#(dv)/dt = (4pir^2" cm"^2)(2/r" ""cm"/"s")#
#(dv)/dt = 8pir" "("cm"^3)/"s"" [1]"#
Evaluate equation [1] at #r = 4" cm"#
#(dv)/dt = 32pi" "("cm"^3)/"s"" [2]"#
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Answer 2

To find the rate at which the volume of the balloon is increasing when the radius is 4 cm, we use the formula for the volume of a sphere, V = (4/3)πr^3, and the given rate of change of the radius, dr/dt = 2/r cm/sec.

First, differentiate the volume formula with respect to time:

dV/dt = (4/3)π * 3r^2 * (dr/dt)

Now substitute the given values: r = 4 cm and dr/dt = 2/4 cm/sec.

dV/dt = (4/3)π * 3(4^2) * (2/4)

Calculate:

dV/dt = (4/3)π * 3(16) * (1/2)

dV/dt = 64π * (1/2)

dV/dt = 32π

So, when the radius is 4 cm, the volume of the balloon is increasing at a rate of 32π cubic centimeters per second.

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Answer from HIX Tutor

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.

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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