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.

Consider the radius of the sphere be R and ∆ R be the error in measuring the radius. Given R = 9 cm and ∆ R = 0 . 03 cm . Now, the volume V of the sphere is given by V = 4 3 π R 3 . . . ( 1 ) Now, differentiating equation ( 1 ) , with respect to R , we get, d V d R = 4 π R 2 We have d V = d V d R ∆ R = 4 π R 2 ∆ R = 4 π 9 2 0 . 03 = 9 . 72 π cm 3 Thus, the approximate error in calculating the volume is 9 . 72 π cm 3 .

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Gamma Question Bank for Engineering Mathematics>Chapter 25 - Application of Derivatives>Assignment>Q 34

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The difference between the greatest and least values of the function $f (x) = \sin 2 x - x$ on $[- \frac{π}{2}, \frac{π}{2}]$ is

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Important Questions on Application of Derivatives

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Gamma Question Bank for Engineering Mathematics>Chapter 25 - Application of Derivatives>Assignment>Q 35

Difference between the greatest and the least values of the function

f (x) = x (\ln x - 2)

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[1, e^{2}]

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Gamma Question Bank for Engineering Mathematics>Chapter 25 - Application of Derivatives>Assignment>Q 36

The sum of lengths of the hypotenuse and another side of a right angled triangle is given. The area of the triangle will be maximum, if the angle between them is

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Gamma Question Bank for Engineering Mathematics>Chapter 25 - Application of Derivatives>Assignment>Q 37

Divide