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Example 1 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2023 by Teachoo
Finding rate of change
Example 1
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Ex 6.1, 1
Ex 6.1,17 (MCQ)
Example 5
Ex 6.1,15 Important
Example 6
Ex 6.1,16
Ex 6.1, 18 (MCQ) Important
Example 2
Ex 6.1,2
Example 35
Example 3
Ex 6.1,5 Important
Ex 6.1,3
Ex 6.1,6
Ex 6.1,12
Ex 6.1,13 Important
Misc 16 (MCQ)
Ex 6.1,14
Example 31 Important
Ex 6.1,4 Important
Example 30 Important
Example 4 Important
Ex 6.1,7
Ex 6.1,8
Ex 6.1,9
Ex 6.1,11 Important
Misc 2 Important
Ex 6.1,10 Important
Example 32 Important
Chapter 6 Class 12 Application of Derivatives
Concept wise
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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
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 (π (π¨πππ ππ ππππππ))/(π (πππ πππ ππ ππππππ)) = π π¨/π π We know that Area of circle = Ο r2 A = Οr2 Finding π π¨/π π ππ΄/ππ = (π(ππ2))/ππ = Ο ((π2))/ππ = Ο (2r) = 2Οr For r = 5 cm ππ΄/ππ = 2Ο (5) = 10Ο cmx/s
