Question 1 (i) - Approximations (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

Last updated at December 16, 2024 by

Question 1 (i) - Find approximate value of āˆš25.3 (Using differentials) - Approximations (using Differentiation)

part 2 - Question 1 (i) - Approximations (using Differentiation) - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Question 1 (i) - Approximations (using Differentiation) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i) āˆš25.3Let y = āˆšš’™ Thus, āˆš(šŸšŸ“.šŸ‘) = y + āˆ†š’š Here, āˆ†š’š = š’…š’š/š’…š’™ ā–³x where x = 25 & ā–³x = 0.3 Since y = āˆšš‘„ š’…š’š/š’…š’™ = (š‘‘(āˆšš‘„))/š‘‘š‘„ = šŸ/(šŸāˆšš’™) Now, āˆ†š’š = š’…š’š/š’…š’™ ā–³x = 1/(2āˆšš‘„) ā–³x Putting x = 25 & ā–³x = 0.3 = 1/(2āˆš25) (0.3) = 1/(2 Ɨ 5) Ɨ 0.3 = 0.3/10 = 0.03 Therefore, āˆš25.3 = y + āˆ†š‘¦ Putting values āˆš25.3 =āˆš25+0.03 āˆš(šŸšŸ“. šŸ‘)=šŸ“. šŸŽšŸ‘ Hence, approximate value of āˆš25.3 is 5.03

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