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anonymous

  • one year ago

The radius of a sphere is claimed to be 3.0 inches, correct to within 0.01 inch. Use linear approximation to estimate the resulting error, measured in cubic inches, in the volume of the sphere. ± 0.000001 ± 0.36π ± 0.036π ± 0.06

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  1. anonymous
    • one year ago
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    where did you get dy = 4πx^2*dx

  2. mathmate
    • one year ago
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    Volume of a sphere of radius r is \(V=\frac{4}{3}\pi r^3 \) \(dV/dr = \frac{4}{3}\pi (3)r2 = 4\pi r^2\), or \(dV = \frac{4}{3}\pi (3)r2~dr = 4\pi r^2~dr\) where dr=0.01" and r=3". If you think of the situation physically, 4 pi r^2 is the surface area of the sphere. Multiply that by the error thickness will give you the error in volume.

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