Ex 6.4, 1 - Using differentials, find approximate value of - Ex 6.4

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i) โˆš25.3 Let y = โˆš๐‘ฅ Let x = 25 & โ–ณ x = 0.3 Since y = โˆš๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘(โˆš๐‘ฅ))/๐‘‘๐‘ฅ = 1/(2โˆš๐‘ฅ) Now, โˆ†๐‘ฆ = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โ–ณx = 1/(2โˆš๐‘ฅ) (0.3) = 1/(2โˆš25) (0.3) = 1/(2 ร— 5) ร— 0.3 = 0.3/10 = 0.03 Also, โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) Putting values โˆ†๐‘ฆ=โˆš(๐‘ฅ+โˆ†๐‘ฅ)โˆ’โˆš๐‘ฅ 0. 03=โˆš(25+0. 3)โˆ’โˆš25 0. 03=โˆš(25. 3)โˆ’5 0. 03+5=โˆš(25. 3) โˆš(25. 3)=5. 03 Hence, approximate value of โˆš25.3 is 5.03 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (ii) โˆš49.5 Let y = โˆš๐‘ฅ Let x = 49 & โ–ณ x = 0.5 Since y = โˆš๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘(โˆš๐‘ฅ))/๐‘‘๐‘ฅ = 1/(2โˆš๐‘ฅ) Now, โˆ†๐‘ฆ = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โ–ณx = 1/(2โˆš๐‘ฅ) (0.5) = 1/(2โˆš49) (0.5) = 1/(2 ร— 7) ร— 0.5 = 0.5/14 = 0.036 Also, โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) Putting values โˆ†๐‘ฆ=โˆš(๐‘ฅ+โˆ†๐‘ฅ)โˆ’โˆš๐‘ฅ 0. 036=โˆš(49+0. 5)โˆ’โˆš49 0. 036=โˆš49.5โˆ’7 0. 036+7=โˆš49.5 โˆš49.5=7. 036 Hence, approximate value of โˆš49.5 is 7.036 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (iii) โˆš0.6 Let y = โˆš๐‘ฅ Let x = 0.64 & โ–ณx = โ€“0.04 Since y = โˆš๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘(โˆš๐‘ฅ))/๐‘‘๐‘ฅ = 1/(2โˆš๐‘ฅ) Now, โˆ†๐‘ฆ = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โ–ณx = 1/(2โˆš0.64) (โ€“0.04) = 1/(2โˆš0.64) (โ€“0.04) = 1/(2 ร— 0.8) ร— (โ€“0.04) = (โˆ’0.04)/1.6 = โ€“0.025 Also, โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) Putting values โˆ†๐‘ฆ=โˆš(๐‘ฅ+โˆ†๐‘ฅ)โˆ’โˆš๐‘ฅ โˆ’0. 025=โˆš(0.64โˆ’0.04)โˆ’โˆš0.64 โˆ’0.025=โˆš0.60โˆ’0.8 0.8 โ€“ 0. 025=โˆš0.60 โˆš0.60=0.775 Hence, approximate value of โˆš0.60 is 0.775 Ex 6.4,1 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(0. 2)^2 ) โˆ†๐‘ฆ=(0. 001)/(0. 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– Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (v) ใ€–(0.999)ใ€—^(1/10) Let ๐‘ฆ=ใ€–๐‘ฅ ใ€—^(1/10) where ๐‘ฅ=1 , โˆ†๐‘ฅ=โˆ’0. 001 Now, ๐‘ฆ=๐‘ฅ^( 1/10) Differentiating w.r.t. ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 1/10) )/๐‘‘๐‘ฅ=1/10 ๐‘ฅ^((โˆ’9)/10)=1/(10ใ€– ๐‘ฅใ€—^(9/10) )=1/(4(๐‘ฅ)^( 3/4) ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ= 1/(10ใ€– ๐‘ฅใ€—^(9/10) ) โˆ†๐‘ฅ โˆ†๐‘ฆ= 1/(10 (1)^(9/10) ) ร— (โˆ’0. 001) โˆ†๐‘ฆ= 1/10 ร—(โˆ’0. 001) โˆ†๐‘ฆ=โˆ’0. 0001 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€—^(1/10)โˆ’๐‘ฅ^( 1/10) Putting Values โˆ†๐‘ฆ=(1+(โˆ’0. 001))^(1/10)โˆ’(1)^(1/10) โˆ’0. 0001=(0. 999)^(1/10)โˆ’1 โˆ’0. 0001+1=(0. 999)^(1/10) 0. 9999=(0. 999)^(1/10) Thus, the Approximate Value of (0. 999)^(1/10) is ๐ŸŽ. ๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ— Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (vi) ใ€–(15)ใ€—^(1/4) Let ๐‘ฆ=ใ€–๐‘ฅ ใ€—^(1/4) where ๐‘ฅ=16 , โˆ†๐‘ฅ=โˆ’1 Now, ๐‘ฆ=๐‘ฅ^( 1/4) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 1/4) )/๐‘‘๐‘ฅ=1/4 ๐‘ฅ^( (1 โˆ’ 4)/4 )=1/4 ๐‘ฅ^( (โˆ’3)/( 4) ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=1/(4(๐‘ฅ)^( 3/4) ) โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(4(16)^( 3/4) ) . (โˆ’1) โˆ†๐‘ฆ=1/(4(2^4 )^( 3/4) ) (โˆ’1) โˆ†๐‘ฆ=(โˆ’1)/(4 ร— 2^3 ) โˆ†๐‘ฆ=(โˆ’1)/(4 ร— 8) โˆ†๐‘ฆ=(โˆ’1)/32 โˆ†๐‘ฆ=โˆ’0. 03125 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) โˆ†๐‘ฆ=(๐‘ฅ+โˆ†๐‘ฅ)^(1/4)โˆ’(๐‘ฅ)^(1/4) Putting Values โˆ’0. 03125=(16+(โˆ’1))^( 1/4)โˆ’ใ€–(16) ใ€—^(1/4) โˆ’0. 03125=(16โˆ’1)^( 1/4)โˆ’ใ€–(2) ใ€—^( 4 ร— 1/4) โˆ’0. 03125=(15)^( 1/4)โˆ’2 โˆ’0. 03125+2=(15)^( 1/4) โˆ’0. 03125+2=(15)^( 1/4) 1. 96875=(15)^( 1/4) Thus, Approximate Value of (15)^( 1/4) is ๐Ÿ. ๐Ÿ—๐Ÿ”๐Ÿ–๐Ÿ•๐Ÿ“ Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (vii) ใ€–(26)ใ€—^(1/3) Let ๐‘ฆ=(๐‘ฅ)^(1/3) where ๐‘ฅ=27 & โˆ†๐‘ฅ=โˆ’1 Now, ๐‘ฆ=ใ€–๐‘ฅ ใ€—^(1/3) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(ใ€–๐‘ฅ ใ€—^(1/3) )/๐‘‘๐‘ฅ=1/3 ใ€–๐‘ฅ ใ€—^(1/3 โˆ’ 1) =1/3 ใ€–๐‘ฅ ใ€—^((โˆ’ 2)/( 3) )=1/(3ใ€–๐‘ฅ ใ€—^(2/( 3) ) ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=1/(3ใ€–๐‘ฅ ใ€—^(2/( 3) ) ) โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(3(27)^( 2/3 ) )ร— (โˆ’1) โˆ†๐‘ฆ=1/(3(3^3 )^( 2/3 ) )ร— (โˆ’1) โˆ†๐‘ฆ=(โˆ’1)/(3ใ€– ร— 3ใ€—^(2 ) ) โˆ†๐‘ฆ=(โˆ’1)/(3 ร— 9 ) โˆ†๐‘ฆ=(โˆ’1)/27 โˆ†๐‘ฆ=โˆ’0. 037037 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’(๐‘ฅ)^( 1/3) Putting Values โˆ’0. 037037=ใ€–(27+(โˆ’1)) ใ€—^(1/3)โˆ’(27)^( 1/3) โˆ’0. 037037=ใ€–(26) ใ€—^(1/3)โˆ’(3)^( 3 ร— 1/3) โˆ’0. 037037=ใ€–(26) ใ€—^(1/3)โˆ’3 โˆ’0. 037037+3=ใ€–(26) ใ€—^(1/3) 2. 9629=ใ€–(26) ใ€—^(1/3) Thus, Approximate Value of (26)^(1/3) is ๐Ÿ. ๐Ÿ—๐Ÿ”๐Ÿ๐Ÿ— Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (viii) ใ€–(255)ใ€—^(1/4) Let ๐‘ฆ=(๐‘ฅ)^(1/4) where ๐‘ฅ=256 & โˆ†๐‘ฅ=โˆ’1 Now, ๐‘ฆ=(๐‘ฅ)^(1/4) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(ใ€–๐‘ฅ ใ€—^(1/4) )/๐‘‘๐‘ฅ=1/4 ๐‘ฅ^( 1/4 โˆ’ 1) =1/4 ๐‘ฅ^( (โˆ’3)/( 4))=1/(4 ๐‘ฅ^( 3/( 4)) ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=1/(4ใ€– ๐‘ฅใ€—^( 3/( 4)) ) ร—โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(4 (256)^(3/4) ) ร— (โˆ’1) โˆ†๐‘ฆ=(โˆ’ 1)/(4 (4^4 )^(3/4) ) โˆ†๐‘ฆ=(โˆ’ 1)/(4 ร— 4^3 ) โˆ†๐‘ฆ=(โˆ’ 1)/(4 ร— 64) โˆ†๐‘ฆ=(โˆ’ 1)/256 โˆ†๐‘ฆ=โˆ’0. 0039 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€—^(1/4)โˆ’๐‘ฅ^( 1/4) Putting Values โˆ’0. 0039=ใ€–(256+(โˆ’1)) ใ€—^(1/4)โˆ’(256)^( 1/4) โˆ’0. 0039=ใ€–(255) ใ€—^(1/4)โˆ’(4^4 )^( 1/4) โˆ’0. 0039=ใ€–(255) ใ€—^(1/4)โˆ’4 โˆ’0. 0039=ใ€–(255) ใ€—^(1/4)โˆ’4 โˆ’0. 0039+4=ใ€–(255) ใ€—^(1/4) 3. 9961=ใ€–(255) ใ€—^(1/4) Thus, the Approximate Values of ใ€–(255) ใ€—^(1/4) is ๐Ÿ‘. ๐Ÿ—๐Ÿ—๐Ÿ”๐Ÿ Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (ix) ใ€–(82)ใ€—^(1/4) Let ๐‘ฆ=(๐‘ฅ)^( 1/4) where ๐‘ฅ=81 โˆ†๐‘ฅ=1 Now, ๐‘ฆ=(๐‘ฅ)^( 1/4) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 1/4) )/๐‘‘๐‘ฅ=1/4 ใ€–๐‘ฅ ใ€—^(1/4 โˆ’ 1)=1/4 ใ€–๐‘ฅ ใ€—^((โˆ’3)/( 4))=1/(4ใ€–๐‘ฅ ใ€—^(3/4) ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=1/(4ใ€– ๐‘ฅ ใ€—^(3/4) ) โˆ†๐‘ฆ Putting Values โˆ†๐‘ฆ=1/(4(81)^( 3/4) ) ร— (1) โˆ†๐‘ฆ=1/(4 ร—3^(4 ร— 3/4 ) ) ร— (1) โˆ†๐‘ฆ=1/(4 ร— 3^3 ) โˆ†๐‘ฆ=1/(4 ร—27) โˆ†๐‘ฆ=1/108 โˆ†๐‘ฆ=0. 009 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) So, โˆ†๐‘ฆ=(๐‘ฅ+โˆ†๐‘ฅ)^( 1/4) โˆ’(๐‘ฅ)^( 1/4) Putting Values 0. 009=(81+1)^( 1/4)โˆ’ใ€–(81) ใ€—^(1/4) 0. 009=(82)^( 1/4)โˆ’(3)^(4 ร— 1/4 ) 0. 009=(82)^( 1/4)โˆ’3 0. 009+3=(82)^( 1/4) 3. 009=(82)^( 1/4) Thus, the Approximate Value of (82)^( 1/4) is ๐Ÿ‘. ๐ŸŽ๐ŸŽ๐Ÿ— Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (x) ใ€–(401)ใ€—^(1/2) Let ๐‘ฆ=๐‘ฅ^( 1/2) where ๐‘ฅ=400 & โˆ†๐‘ฅ=1 Now, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=1/(2โˆš๐‘ฅ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=๐‘‘๐‘ฆ/(2โˆš๐‘ฅ) โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(2โˆš400) ร—(1) โˆ†๐‘ฆ=1/(2โˆš400) ร—(1) โˆ†๐‘ฆ=1/(2โˆš(ใ€–20ใ€—^2 )) โˆ†๐‘ฆ=1/(2 ร— 20) โˆ†๐‘ฆ=1/40 โˆ†๐‘ฆ=0. 025 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) So, โˆ†๐‘ฆ= (๐‘ฅ+โˆ†๐‘ฅ)^( 1/2)โˆ’(๐‘ฅ)^( 1/2) Putting Values 0. 025=(400+1)^( 1/(2 ))โˆ’ใ€–(400) ใ€—^(1/2) 0. 025=(401)^( 1/2)โˆ’(20)^( 2 ร— 1/2) 0. 025=(401)^( 1/2)โˆ’20 0. 025+20=(401)^( 1/2) 20. 025=(401)^( 1/2) (401)^( 1/2)=20. 025 Thus, the Approximate Value of (401)^( 1/2) is ๐Ÿ๐ŸŽ. ๐ŸŽ๐Ÿ๐Ÿ“ Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xi) ใ€–(0.0037)ใ€—^(1/2) Let y = โˆš๐‘ฅ Let x = 0.0036 & โ–ณ x = 0.0001 Since y = โˆš๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘(โˆš๐‘ฅ))/๐‘‘๐‘ฅ = 1/(2โˆš๐‘ฅ) Now, โˆ†๐‘ฆ = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โ–ณx = 1/(2โˆš๐‘ฅ) (0.0001) = 1/(2โˆš0.0036) (0.0001) = 1/(2 ร— 0.06) ร— (0.0001) = 0.0001/0.12 = 1/1200 = 0.000833 Also, โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) So, โˆ†๐‘ฆ=(๐‘ฅ+โˆ†๐‘ฅ)^( 1/2)โˆ’(๐‘ฅ)^( 1/2) Putting Values 0. 000833=(0. 0036+0. 0001)^( 1/2)โˆ’(0. 0036)^( 1/2) 0. 000833=(0. 0037)^( 1/2)โˆ’(36/10000)^(1/2) 0. 000833=(0. 0037)^( 1/2)โˆ’(6/100)^( 2 ร— 1/2) 0. 000833=(0. 0037)^( 1/2)โˆ’(0. 06)^(2 ร— 1/2 ) 0. 000833=(0. 0037)^( 1/2)โˆ’(0. 06) 0. 000833+0. 06=(0. 0037)^( 1/2) 0. 060833=(0. 0037)^( 1/2) (0. 0037)^( 1/2)=0. 060833 Thus, the Approximate Value of (0. 0037)^( 1/2)=๐ŸŽ. ๐ŸŽ๐Ÿ”๐ŸŽ๐Ÿ–๐Ÿ‘๐Ÿ‘ Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xii) ใ€–(26.57)ใ€—^(1/3) Let ๐‘ฆ=๐‘ฅ^( 1/3) where ๐‘ฅ=27 & โˆ† ๐‘ฅ=โˆ’0. 43 Now, ๐‘ฆ=๐‘ฅ^( 1/3) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 1/3) )/๐‘‘๐‘ฅ=1/3 ๐‘ฅ^( 1/3 โˆ’ 1)=1/3 ๐‘ฅ^( (โˆ’ 2)/( 3))=1/(3 ๐‘ฅ^( 2/( 3)) ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=1/(3 ๐‘ฅ^( 2/( 3)) ) ร— โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(3 (27)^( 2/( 3)) ) ร— (โˆ’0. 43) โˆ†๐‘ฆ=(โˆ’0. 43)/(3 ร— 3^(3 ร— 2/( 3)) ) โˆ†๐‘ฆ=(โˆ’0. 43)/(3 ร— 3^2 ) โˆ†๐‘ฆ=(โˆ’0. 43)/(3 ร— 9) โˆ†๐‘ฆ=(โˆ’0. 43)/27 โˆ†๐‘ฆ=โˆ’0. 015926 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) So, โˆ†๐‘ฆ=ใ€–(๐‘ฅ+โˆ†๐‘ฅ) ใ€—^(1/3)โˆ’๐‘ฅ^( 1/3) Putting Values โˆ’0. 015926=(27+(โˆ’0. 43))^( 1/3)โˆ’(27)^( 1/3) โˆ’0. 015926=(26. 57)^( 1/3)โˆ’(27)^( 3 ร— 1/3) โˆ’0. 015926=(26. 57)^( 1/3)โˆ’3 โˆ’0. 015926+3=(26. 53)^( 1/3) 2. 984=(26. 53)^( 1/3) Thus, the Approximate Values of (26. 53)^( 1/3) is ๐Ÿ. ๐Ÿ—๐Ÿ–๐Ÿ’ Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xiii)ใ€–(81.5)ใ€—^(1/4) Let ๐‘ฆ=(๐‘ฅ)^( 1/4) where ๐‘ฅ=81 & โˆ†๐‘ฅ=0. 5 Now, ๐‘ฆ=๐‘ฅ^( 1/4) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 1/4) )/๐‘‘๐‘ฅ=1/4 ๐‘ฅ^( 1/4 โˆ’ 1)=1/4 ๐‘ฅ^( (โˆ’ 3)/( 4)) =1/(4๐‘ฅ^(3/4) ) โˆ† ๐‘ฅ Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=๐‘‘๐‘ฆ/(4๐‘ฅ^(3/4) ) โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(4(81)^( 3/4) ) ร— (0. 5) โˆ†๐‘ฆ=(0. 5)/(4 ร— (3^( 4) )^( 3/4) ) โˆ†๐‘ฆ=(0. 5)/(4 ร— 3^( 3) ) โˆ†๐‘ฆ=(0. 5)/(4 ร— 27) โˆ†๐‘ฆ=(0. 5)/108 โˆ†๐‘ฆ=0. 0046 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) So, โˆ†๐‘ฆ=(๐‘ฅ+โˆ†๐‘ฅ)^( 1/4)โˆ’๐‘ฅ^( 1/4) Putting Values 0. 0046=(81+0. 5)^( 1/4)โˆ’(81)^( 1/4) 0. 0046=(81. 5)^( 1/4)โˆ’(3^4 )^( 1/4) 0. 0046=(81. 5)^( 1/4)โˆ’3 0. 0046+3=(81. 5)^( 1/4) 3. 0046=(81. 5)^( 1/4) Thus, Approximate Value of (81. 5)^( 1/4) is ๐Ÿ‘. ๐ŸŽ๐ŸŽ๐Ÿ’๐Ÿ” Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xiv) ใ€–(3.968)ใ€—^(3/2) Let ๐‘ฆ=๐‘ฅ^( 3/2) where ๐‘ฅ=4 & โˆ†๐‘ฅ=โˆ’0. 032 Now, ๐‘ฆ=๐‘ฅ^( 3/2) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 3/2) )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3/2 ใ€–๐‘ฅ ใ€—^(1/2) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=3/2 ๐‘ฅ^( 1/2) โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=3/2 (4)^( 1/2) . (โˆ’0. 032) โˆ†๐‘ฆ=3/2 (2^2 )^( 1/2) . (โˆ’0. 032) โˆ†๐‘ฆ=3/2 ร—2 ร— (โˆ’0. 032) โˆ†๐‘ฆ=3 ร— (โˆ’0. 032) โˆ†๐‘ฆ=โˆ’0. 096 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) So, โˆ†๐‘ฆ=(๐‘ฅ+โˆ†๐‘ฅ)^( 3/2)โˆ’๐‘ฅ^( 3/2) Putting Values โˆ’0. 096=(4+(โˆ’0. 032))^( 3/2)โˆ’(4)^( 3/2) โˆ’0. 096=(4โˆ’0. 032)^( 3/2)โˆ’ใ€–(2)^2ใ€—^( ร— 3/2) โˆ’0. 096=(3. 968)^( 3/2)โˆ’2^3 โˆ’0. 096=(3. 968)^( 3/2)โˆ’8 โˆ’0. 096+8=(3. 968)^( 3/2) 7. 904=(3. 968)^( 3/2) (3. 968)^( 3/2)=7.904 Thus, Approximate Values of (3. 968)^( 3/2) is ๐Ÿ•. ๐Ÿ—๐ŸŽ๐Ÿ’ Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xv) ใ€–(32.15)ใ€—^(1/5) Let ๐‘ฆ=(๐‘ฅ)^( 1/5) where ๐‘ฅ=32 & โˆ†๐‘ฅ=0. 15 Now, ๐‘ฆ=(๐‘ฅ)^( 1/5) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 1/5) )/๐‘‘๐‘ฅ=1/5 ๐‘ฅ^( 1/5 โˆ’ 1) =1/5 ๐‘ฅ^( (โˆ’ 4)/( 5) )=1/(5ใ€– ๐‘ฅใ€—^( 4/5 ) ) Let ๐‘ฆ=(๐‘ฅ)^( 1/5) where ๐‘ฅ=32 & โˆ†๐‘ฅ=0. 15 Now, ๐‘ฆ=(๐‘ฅ)^( 1/5) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 1/5) )/๐‘‘๐‘ฅ=1/5 ๐‘ฅ^( 1/5 โˆ’ 1) =1/5 ๐‘ฅ^( (โˆ’ 4)/( 5) )=1/(5ใ€– ๐‘ฅใ€—^( 4/5 ) ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=1/(5 ใ€–๐‘ฅ ใ€—^(4/5) ) โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(5(32)^( 4/5) ) ร— (0. 15) โˆ†๐‘ฆ=1/(5(2)^( 5 ร— 4/5) ) ร— (0. 15) โˆ†๐‘ฆ=1/(5(2)^( 4) ) ร— (0. 15) โˆ†๐‘ฆ=1/(5(2)^( 4) ) ร— (0. 15) โˆ†๐‘ฆ=1/(5 ร—16) ร—0. 15 โˆ†๐‘ฆ=(0. 15" " )/80 โˆ†๐‘ฆ=0. 00187 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) So, โˆ†๐‘ฆ=(๐‘ฅ+โˆ†๐‘ฅ)^( 1/5)โˆ’ใ€–๐‘ฅ ใ€—^(1/5) Putting Values 0. 00187=(32+0. 15)^( 1/5)โˆ’(32)^( 1/5) 0. 00187=(32. 15)^( 1/5)โˆ’(2)^(5 ร— 1/5 ) 0. 00187=(32. 15)^( 1/5)โˆ’2 0. 00187+2=(32. 15)^( 1/5) 2. 00187=(32. 15)^( 1/5) (32. 15)^( 1/5)=2. 00187 Hence Approximate Values of (32. 15)^( 1/5) is ๐Ÿ. ๐ŸŽ๐ŸŽ๐Ÿ๐Ÿ–๐Ÿ•

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.