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Ex 6.1,13 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2023 by Teachoo
Finding rate of change
Example 1
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 You are here
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
-
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
Ex 6.1, 13 A balloon, which always remains spherical, has a variable diameter 3/2 (2๐ฅ +1). Find the rate of change of its volume with respect to ๐ฅ.Let d be the diameter of the balloon Given that Diameter = d = 3/2 (2x + 1) Let r be the radius of the balloon r = ๐/2 = ๐/๐ (2x + 1) The balloon is a spherical Volume of the balloon = 4/3 ๐๐^3 We need to find rate of change of volume with respect to x i.e. ๐๐/๐๐ฅ Now, ๐๐/๐๐ฅ = ๐/๐๐ฅ (4/3 ๐๐^3 ) = 4๐/3 ร (๐๐^3)/๐๐ฅ = 4๐/3 ร ๐/๐๐ฅ (27/64 (2๐ฅ+1)^3 ) = 9๐/16 ร (๐(2๐ฅ + 1)^3)/๐๐ฅ = 9๐/16 ร 3(2x + 1)2 ร 2 = ๐๐๐ /๐ (๐๐+๐)^๐
