A spherical snowball is melting in such a way that it maintains its shape. The volume of the snowball is decreasing at a constant rate of 5 cubic inches per minute. At the instant when the radius of the snowball is decreasing at a rate of 3 inch per minute, what is the radius of the snowball, in inches? (The volume V of a sphere with radius r is V = Tr3.)
Question
A spherical snowball is melting in such a way that it maintains its shape. The volume of the snowball is decreasing at a constant rate of 5 cubic inches per minute. At the instant when the radius of the snowball is decreasing at a rate of 3 inch per minute, what is the radius of the snowball, in inches? (The volume V of a sphere with radius r is V = Tr3.)
Solution
The problem involves the concept of related rates in calculus. Here's how to solve it:
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We know that the volume V of a sphere is given by V = 4/3πr³.
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We're given that dV/dt = -5 cubic inches per minute (the negative sign indicates that the volume is decreasing).
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We're asked to find the radius r when dr/dt = -3 inches per minute.
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We can find an expression for dr/dt by differentiating both sides of the volume equation with respect to time t. Using the chain rule, we get dV/dt = 4πr² * dr/dt.
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We can substitute the given values into this equation to solve for r. So, -5 = 4πr² * -3.
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Simplifying this gives r² = 5 / (12π).
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Taking the square root of both sides gives r = sqrt(5 / (12π)).
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Therefore, the radius of the snowball at the instant when the radius is decreasing at a rate of 3 inches per minute is sqrt(5 / (12π)) inches.
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