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by linear approximation, do you mean \(V = \frac{4}{3} \pi r^3\), \(\Delta V = \frac{dV}{dr}\Delta r\) ?
Would I derive the first one?
yes differentiate (4/3) pi r^3 which is the formula for the volume of the sphere
4pir^2
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And then what
you know the value of r and delta r - Just plug them in and calculate
36pi
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correct to within 0.01 inch in maths speak is: \(\Delta r = \pm 0.01\) you got \(\frac{dV}{dr} = 4 \pi r^2\) yourself, so finish the job
36pi is correct for the value of 4 pi r^2 now you plug is the +/- 0.01
* plug in
.36
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