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 47

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The cost function at American Gadget is $C (x) = x^{3} - 6 x^{2} + 15 x$ ( $x$ is thousands of units and $x > 0$ ). The production level at which average cost is minima 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 48

Let

f (x) = a x^{2} - b |x|

, where

a

and

b

are constant. Then at

x = 0

,

f (x)

has

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Gamma Question Bank for Engineering Mathematics>Chapter 25 - Application of Derivatives>Assignment>Q 49

The maximum value of

{(\frac{1}{x})}^{2 x^{2}}

is -

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Gamma Question Bank for Engineering Mathematics>Chapter 25 - Application of Derivatives>Assignment>Q 50

Let

x

be a number which exceeds it square by the greatest possible quantity. Then

x

is equal to

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Gamma Question Bank for Engineering Mathematics>Chapter 25 - Application of Derivatives>Assignment>Q 51

The point

(0, 3)

is nearest to the curve

x^{2} = 2 y

at

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Gamma Question Bank for Engineering Mathematics>Chapter 25 - Application of Derivatives>Assignment>Q 52

Let $f (x)$ be a function defined as $f (x) = \{\begin{cases} \sin (x^{2} - 3 x), & x \leq 0 \\ 6 x + 5 x^{2}, & x > 0 \end{cases} .$ Then, at $x = 0, f (x)$