Ex 6.4, 1 (iv) - Find approximate value of (0.009)^1/3 - Ex 6.4

Ex 6.4, 1 (iv) - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.4, 1 (iv) - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.4, 1 (iv) - Chapter 6 Class 12 Application of Derivatives - Part 4

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Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (iv) ใ€–(0.009)ใ€—^(1/3)Let ๐‘ฆ=(๐‘ฅ)^(1/3) where ๐‘ฅ=0. 008 & โˆ†๐‘ฅ=0. 001 Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(ใ€–๐‘ฅ ใ€—^(1/3) )/๐‘‘๐‘ฅ=1/3 ๐‘ฅ^((โˆ’2)/3)=1/(3ใ€– ๐‘ฅใ€—^( 2/3) ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=1/(3ใ€– ๐‘ฅใ€—^( 2/3) ) ร—โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(3(0. 008)^( 2/3) ) ร—0. 001 โˆ†๐‘ฆ=(0. 001)/(3(8/1000)^(2/3) ) โˆ†๐‘ฆ=(0. 001)/(3(2/10)^( 3 ร— 2/3) ) โˆ†๐‘ฆ=(0. 001)/(3(2/10)^2 ) โˆ†๐‘ฆ=(0. 001)/(3 ร— 4/100) โˆ†๐‘ฆ=(0.001 ร— 100)/12 โˆ†๐‘ฆ=0. 008 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) โˆ†๐‘ฆ=ใ€–(๐‘ฅ+โˆ†๐‘ฅ) ใ€—^(1/3)โˆ’ใ€–๐‘ฅ ใ€—^(1/3) Putting Values 0. 008=(0. 008" " +0. 001)^( 1/3)โˆ’(0. 008" " )^( 1/3) 0. 008=(0. 009)^( 1/3)โˆ’(0. 008)^( 1/3) 0. 008=(0. 009)^( 1/3)โˆ’(8/1000)^( 1/3) 0. 008=(0. 009)^( 1/3)โˆ’(2/10)^( 3 ร— 1/3) 0. 008=(0. 009)^( 1/3)โˆ’(2/10) 0. 008=(0. 009)^( 1/3)โˆ’0. 2 0. 008+0. 2=(0. 009)^( 1/3) (0. 009)^( 1/3)=0. 208 Thus , Approximate Value of (0 . 009)^(1/3) is ๐ŸŽ. ๐Ÿ๐ŸŽ๐Ÿ–

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