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aliya0001 [1]
3 years ago
8

A spherical balloon is being inflated at a rate of 3 cubic inches per second. Determine the change in the rate of the radius.How

fast is the radius of the balloon changing at the instant the balloon’s diameter is 12 inches? Is the radius changing more rapidly when d=12 or when d=16 ? Why?
Mathematics
1 answer:
Novay_Z [31]3 years ago
5 0

Answer:

The rate rate of change of radius is \frac{1}{48\pi} inches per second when the diameter is 12 inches.

The radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Consider the provided information.

A spherical balloon is being inflated at a rate of 3 cubic inches per second.

The volume of sphere is V=\frac{4}{3}\pi r^3

Differentiate the above formula with respect to time.

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

Substitute the respective values in the above formula,

3=4\pi 6^2\frac{dr}{dt}

\frac{1}{48\pi}=\frac{dr}{dt}

The rate rate of change of radius is \frac{1}{48\pi} inches per second when the diameter is 12 inches.

When d=16

3=4\pi 8^2\frac{dr}{dt}

\frac{3}{256\pi}=\frac{dr}{dt}

Thus, the radius is changing more rapidly when the diameter is 12 inches.

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