# Example 1

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 1 Find the rate of change of the area of a circle per second with respect to its radius r when r = 5 cm. We have to find rate of change of area of circle with respect to radius i.e. we need to find 𝑑(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒)𝑑 (𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑑 𝑐𝑖𝑟𝑐𝑙𝑒) = 𝑑𝐴𝑑𝑟 We know that Area of circle = π r2 A = πr2 Finding 𝑑𝐴𝑑𝑟 𝑑𝐴𝑑𝑟 = 𝑑(𝜋𝑟2)𝑑𝑟 = π (𝑟2)𝑑𝑟 = π (2 . r2 –1) = 2π r For r = 5 cm 𝑑𝐴𝑑𝑟 = 2π (5) . 𝑐𝑚2𝑐𝑚 = 10π 𝒄𝒎𝟐𝒄𝒎

Example 1
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Ex 6.1,17

Example 5

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Example 6

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Example 42

Example 2

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Example 49

Example 3

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Misc. 19

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Example 43

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Example 4

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Misc 3

Example 44

Chapter 6 Class 12 Application of Derivatives

Concept wise

- Finding rate of change
- To show increasing/decreasing in whole domain
- To show increasing/decreasing in intervals
- Find intervals of increasing/decreasing
- Finding slope of tangent/normal
- Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
- Finding equation of tangent/normal when slope and curve are given
- Finding approximate value of numbers
- Finding approximate value of function
- Finding approximate value- Statement questions
- Finding minimum and maximum values from graph
- Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
- Absolute minima/maxima

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .