If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.2.46πcm38.62πcm39.72πcm37.6πcm3
Question
If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.2.46πcm38.62πcm39.72πcm37.6πcm3
Solution
The volume of a sphere is given by the formula V = 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 with respect to r is dV/dr = 4πr².
Substituting r = 9 cm and dr = 0.03 cm into the equation gives:
dV = 4π(9 cm)² * 0.03 cm = 9.72π cm³.
So, the approximating error in calculating its volume is 9.72π cm³.
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