If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume.

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Solution

Let r be the radius of the sphere and ∆r be the error in measuring the radius. Then r = 9cm and ∆ r = 0.03 cm. Now, the volume V of the sphere is given by

straight V space equals space 4 over 3 πr cubed space space space space space space space or space space space dV over dr space equals space 4 πr squared

therefore space space space space dV space equals space open parentheses dV over dr close parentheses space increment straight r space equals space left parenthesis 4 πr squared right parenthesis space increment straight r space space space space space space space space space space space space space space equals space 4 straight pi left parenthesis 9 right parenthesis squared space left parenthesis 0.03 right parenthesis space equals space 9.72 space straight pi space cm cubed therefore space space space the space approximate space error space in space calculating space the space volume space is space 9.72 space straight pi space cm cubed.

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Application Of Derivatives

If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume.

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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