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A sphere is to be designed with a radius of 72 in.Use differentials to estimate the maximum error when measuring the volume of the sphere if the possible error in measuring the radius is 0.5 in. (Hint: The formula for the volume of a sphere is V left parenthesis r right parenthesis equals 4 over 3 straight pi r cubed.)A. 452.39 space i n cubedB. 16 comma 286.02 space i n cubedC. 32 comma 572.03 space i n cubedD. 65 comma 144.07 space i n cubed

Question

A sphere is to be designed with a radius of 72 in.Use differentials to estimate the maximum error when measuring the volume of the sphere if the possible error in measuring the radius is 0.5 in. (Hint: The formula for the volume of a sphere is V left parenthesis r right parenthesis equals 4 over 3 straight pi r cubed.)A. 452.39 space i n cubedB. 16 comma 286.02 space i n cubedC. 32 comma 572.03 space i n cubedD. 65 comma 144.07 space i n cubed

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Solution

The formula for the volume of a sphere is V(r) = 4/3πr³.

We can find the differential dV, which represents the error in the volume, by taking the derivative of the volume with respect to r (the radius), and then multiplying by the error in the radius (dr).

The derivative of V(r) = 4/3πr³ with respect to r is dV/dr = 4πr².

If the possible error in measuring the radius is 0.5 in (dr = 0.5 in), then the maximum error in the volume (dV) is given by:

dV = (dV/dr) * dr = 4πr² * dr = 4π(72 in)² * 0.5 in = 32,572.03 in³.

So, the maximum error when measuring the volume of the sphere is approximately 32,572.03 cubic inches.

Therefore, the correct answer is C. 32,572.03 in³.

This problem has been solved

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