Question

The volume of a gas-filled spherical balloon increases 6 cubic inches for each degree (Celsius) increase in temperature. If the temperature increases at a constant rate of 2 degrees per minute, then at what rate is the radius of the balloon changing at the instant when the volume is 36π cubic inches?

A. 1/π inches per minute
B. 1/2π inches per minute
C. 1/3π inches per minute
D. 1/6π inches per minute
E. 2/3π inches per minute

Answer 1

The rate at which the radius of the balloon is changing at the instant when the volume is 36π cubic inches is 1/3π inches per minute.

Change in volume of the sphere with temperature

The change in volume of the sphere with temperature is calculated as follows;

$V = \frac{4}{3} \pi r^3$

Given;

$\frac{dV}{dT} = 6 \ in^3/^0C\\\\\frac{dT}{dt} = 2 \ ^0C/\min$

Radius of the sphere when the volume of 36π cubic inches

$\frac{4}{3} \pi r^3 = 36 \pi\\\\r^3 = 27\\\\r^3 = 3^3\\\\r = 3 \ in$

Change in volume of the sphere with radius

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

Rate at which the radius of the balloon is changing

$\frac{dV}{dT} \times \frac{dT}{dt} = \frac{dV}{dt} \\\\\frac{dV}{dt} = 6\ (in^3/^0C) \ \times 2\ (^0C/\min) = 12 \ \frac{in^3}{\min} \\\\\frac{1}{\frac{dV}{dr} } \times \frac{dV}{dt} = \frac{dr}{dt} \\\\\frac{dr}{dt} = \frac{1}{4\pi r^2} \times 12\\\\\frac{dr}{dt} = \frac{12}{4\pi r^2} \\\\Recall, \ r = 3 \ in\\\\\frac{dr}{dt} = \frac{12}{4\pi (3)^2} \\\\\frac{dr}{dt} = \frac{12}{36\pi }\\\\\frac{dr}{dt} = \frac{1}{3\pi} \ \frac{in}{\min}$

Thus, the rate at which the radius of the balloon is changing at the instant when the volume is 36π cubic inches is 1/3π inches per minute.

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