Ex 6.4,1 - Chapter 6 Class 12 Application of Derivatives

Transcript

Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i) ﷮25.3﷯ Let y = ﷮𝑥﷯ Let x = 25 & △ x = 0.3 Since y = ﷮𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑( ﷮𝑥﷯)﷮𝑑𝑥﷯ = 1﷮2 ﷮𝑥﷯﷯ Now, ∆𝑦 = 𝑑𝑦﷮𝑑𝑥﷯ △x = 1﷮2 ﷮𝑥﷯﷯ (0.3) = 1﷮2 ﷮25﷯﷯ (0.3) = 1﷮2 × 5﷯ × 0.3 = 0.3﷮10﷯ = 0.03 Also, ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ Putting values ∆𝑦= ﷮𝑥+∆𝑥﷯− ﷮𝑥﷯ 0. 03= ﷮25+0. 3﷯− ﷮25﷯ 0. 03= ﷮25. 3﷯−5 0. 03+5= ﷮25. 3﷯ ﷮25. 3﷯=5. 03 Hence, approximate value of ﷮25.3﷯ 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) ﷮49.5﷯ Let y = ﷮𝑥﷯ Let x = 49 & △ x = 0.5 Since y = ﷮𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑( ﷮𝑥﷯)﷮𝑑𝑥﷯ = 1﷮2 ﷮𝑥﷯﷯ Now, ∆𝑦 = 𝑑𝑦﷮𝑑𝑥﷯ △x = 1﷮2 ﷮𝑥﷯﷯ (0.5) = 1﷮2 ﷮49﷯﷯ (0.5) = 1﷮2 × 7﷯ × 0.5 = 0.5﷮14﷯ = 0.036 Also, ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ Putting values ∆𝑦= ﷮𝑥+∆𝑥﷯− ﷮𝑥﷯ 0. 036= ﷮49+0. 5﷯− ﷮49﷯ 0. 036= ﷮49.5﷯−7 0. 036+7= ﷮49.5﷯ ﷮49.5﷯=7. 036 Hence, approximate value of ﷮49.5﷯ 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) ﷮0.6﷯ Let y = ﷮𝑥﷯ Let x = 0.64 & △x = –0.04 Since y = ﷮𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑( ﷮𝑥﷯)﷮𝑑𝑥﷯ = 1﷮2 ﷮𝑥﷯﷯ Now, ∆𝑦 = 𝑑𝑦﷮𝑑𝑥﷯ △x = 1﷮2 ﷮0.64﷯﷯ (–0.04) = 1﷮2 ﷮0.64﷯﷯ (–0.04) = 1﷮2 × 0.8﷯ × (–0.04) = −0.04﷮1.6﷯ = –0.025 Also, ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ Putting values ∆𝑦= ﷮𝑥+∆𝑥﷯− ﷮𝑥﷯ −0. 025= ﷮0.64−0.04﷯− ﷮0.64﷯ −0.025= ﷮0.60﷯−0.8 0.8 – 0. 025= ﷮0.60﷯ ﷮0.60﷯=0.775 Hence, approximate value of ﷮0.60﷯ 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) (0.009)﷮ 1﷮3﷯﷯ Let 𝑦= 𝑥﷯﷮ 1﷮3﷯﷯ where 𝑥=0. 008 & ∆𝑥=0. 001 Differentiating w.r.t.𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥 ﷮ 1﷮3﷯﷯﷯﷮𝑑𝑥﷯= 1﷮3﷯ 𝑥﷮ −2﷮3﷯﷯= 1﷮3 𝑥﷮ 2﷮3﷯﷯﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦= 1﷮3 𝑥﷮ 2﷮3﷯﷯﷯ ×∆𝑥 Putting Values ∆𝑦= 1﷮3 0. 008﷯﷮ 2﷮3﷯﷯﷯ ×0. 001 ∆𝑦= 0. 001﷮3 8﷮1000﷯﷯﷮ 2﷮3﷯﷯﷯ ∆𝑦= 0. 001﷮3 2﷮10﷯﷯﷮ 3 × 2﷮3﷯﷯﷯ ∆𝑦= 0. 001﷮3 2﷮10﷯﷯﷮2﷯﷯ ∆𝑦= 0. 001﷮3 0. 2﷯﷮2﷯﷯ ∆𝑦= 0. 001﷮0. 12﷯ ∆𝑦=0. 008 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ ∆𝑦= 𝑥+∆𝑥﷯ ﷮ 1﷮3﷯﷯− 𝑥 ﷮ 1﷮3﷯﷯ Putting Values 0. 008= 0. 008 +0. 001﷯﷮ 1﷮3﷯﷯− 0. 008 ﷯﷮ 1﷮3﷯﷯ 0. 008= 0. 009﷯﷮ 1﷮3﷯﷯− 0. 008 ﷯﷮ 1﷮3﷯﷯ 0. 008= 0. 009﷯﷮ 1﷮3﷯﷯− 8﷮1000﷯﷯﷮ 1﷮3﷯﷯ 0. 008= 0. 009﷯﷮ 1﷮3﷯﷯− 2﷮10﷯﷯﷮ 3 × 1﷮3﷯﷯ 0. 008= 0. 009﷯﷮ 1﷮3﷯﷯− 2﷮10﷯﷯ 0. 008= 0. 009﷯﷮ 1﷮3﷯﷯−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 𝟎. 𝟐𝟎𝟖 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (v) (0.999)﷮ 1﷮10﷯﷯ Let 𝑦= 𝑥 ﷮ 1﷮10﷯﷯ where 𝑥=1 , ∆𝑥=−0. 001 Now, 𝑦= 𝑥﷮ 1﷮10﷯﷯ Differentiating w.r.t. 𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥﷮ 1﷮10﷯﷯﷯﷮𝑑𝑥﷯= 1﷮10﷯ 𝑥﷮ −9﷮10﷯﷯= 1﷮10 𝑥﷮ 9﷮10﷯﷯﷯= 1﷮4 𝑥﷯﷮ 3﷮4﷯﷯﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 Putting Values ∆𝑦= 1﷮10 𝑥﷮ 9﷮10﷯﷯﷯ ∆𝑥 ∆𝑦= 1﷮10 1﷯﷮ 9﷮10﷯﷯﷯ × −0. 001﷯ ∆𝑦= 1﷮10﷯ × −0. 001﷯ ∆𝑦=−0. 0001 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ ∆𝑦= 𝑥+∆𝑥﷯ ﷮ 1﷮10﷯﷯− 𝑥﷮ 1﷮10﷯﷯ Putting Values ∆𝑦= 1+ −0. 001﷯﷯﷮ 1﷮10﷯﷯− 1﷯﷮ 1﷮10﷯﷯ −0. 0001= 0. 999﷯﷮ 1﷮10﷯﷯−1 −0. 0001+1= 0. 999﷯﷮ 1﷮10﷯﷯ 0. 9999= 0. 999﷯﷮ 1﷮10﷯﷯ Thus, the Approximate Value of 0. 999﷯﷮ 1﷮10﷯﷯ is 𝟎. 𝟗𝟗𝟗𝟗 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (vi) (15)﷮ 1﷮4﷯﷯ Let 𝑦= 𝑥 ﷮ 1﷮4﷯﷯ where 𝑥=16 , ∆𝑥=−1 Now, 𝑦= 𝑥﷮ 1﷮4﷯﷯ Differentiating w.r.t.𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥﷮ 1﷮4﷯﷯﷯﷮𝑑𝑥﷯= 1﷮4﷯ 𝑥﷮ 1 − 4﷮4﷯ ﷯= 1﷮4﷯ 𝑥﷮ −3﷮ 4﷯ ﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦= 1﷮4 𝑥﷯﷮ 3﷮4﷯﷯﷯ ∆𝑥 Putting Values ∆𝑦= 1﷮4 16﷯﷮ 3﷮4﷯﷯﷯ . −1﷯ ∆𝑦= 1﷮4 2﷮4﷯﷯﷮ 3﷮4﷯﷯﷯ −1﷯ ∆𝑦= −1﷮4 × 2﷮3﷯﷯ ∆𝑦= −1﷮4 × 8﷯ ∆𝑦= −1﷮32﷯ ∆𝑦=−0. 03125 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ ∆𝑦= 𝑥+∆𝑥﷯﷮ 1﷮4﷯﷯− 𝑥﷯﷮ 1﷮4﷯﷯ Putting Values −0. 03125= 16+ −1﷯﷯﷮ 1﷮4﷯﷯− 16﷯ ﷮ 1﷮4﷯﷯ −0. 03125= 16−1﷯﷮ 1﷮4﷯﷯− 2﷯ ﷮ 4 × 1﷮4﷯﷯ −0. 03125= 15﷯﷮ 1﷮4﷯﷯−2 −0. 03125+2= 15﷯﷮ 1﷮4﷯﷯ −0. 03125+2= 15﷯﷮ 1﷮4﷯﷯ 1. 969= 15﷯﷮ 1﷮4﷯﷯ Thus, Approximate Value of 15﷯﷮ 1﷮4﷯﷯ is 𝟏. 𝟗𝟔𝟗 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (vii) (26)﷮ 1﷮3﷯﷯ Let 𝑦= 𝑥﷯﷮ 1﷮3﷯﷯ where 𝑥=27 & ∆𝑥=−1 Now, 𝑦= 𝑥 ﷮ 1﷮3﷯﷯ Differentiating w.r.t.𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥 ﷮ 1﷮3﷯﷯﷯﷮𝑑𝑥﷯= 1﷮3﷯ 𝑥 ﷮ 1﷮3﷯ − 1﷯ = 1﷮3﷯ 𝑥 ﷮ − 2﷮ 3﷯ ﷯= 1﷮3 𝑥 ﷮ 2﷮ 3﷯ ﷯﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯∆𝑥 ∆𝑦= 1﷮3 𝑥 ﷮ 2﷮ 3﷯ ﷯﷯∆𝑥 Putting Values ∆𝑦= 1﷮3 27﷯﷮ 2﷮3﷯ ﷯ ﷯× −1﷯ ∆𝑦= 1﷮3 3﷮3﷯﷯﷮ 2﷮3﷯ ﷯ ﷯× −1﷯ ∆𝑦= −1﷮3 × 3﷮2 ﷯ ﷯ ∆𝑦= −1﷮3 × 9 ﷯ ∆𝑦= −1﷮27﷯ ∆𝑦=−0. 037037 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ ∆𝑦= 𝑥+∆𝑥﷯ ﷮ 1﷮3﷯﷯− 𝑥﷯﷮ 1﷮3﷯﷯ Putting Values −0. 037037= 27+ −1﷯﷯ ﷮ 1﷮3﷯﷯− 27﷯﷮ 1﷮3﷯﷯ −0. 037037= 26﷯ ﷮ 1﷮3﷯﷯− 3﷯﷮ 3 × 1﷮3﷯﷯ −0. 037037= 26﷯ ﷮ 1﷮3﷯﷯−3 −0. 037037+3= 26﷯ ﷮ 1﷮3﷯﷯ 2. 9629= 26﷯ ﷮ 1﷮3﷯﷯ Thus, Approximate Value of 26﷯﷮ 1﷮3﷯﷯ is 𝟐. 𝟗𝟔𝟐𝟗 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (viii) (255)﷮ 1﷮4﷯﷯ Let 𝑦= 𝑥﷯﷮ 1﷮4﷯﷯ where 𝑥=256 & ∆𝑥=−1 Now, 𝑦= 𝑥﷯﷮ 1﷮4﷯﷯ Differentiating w.r.t.𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥 ﷮ 1﷮4﷯﷯﷯﷮𝑑𝑥﷯= 1﷮4﷯ 𝑥﷮ 1﷮4﷯ − 1﷯ = 1﷮4﷯ 𝑥﷮ −3﷮ 4﷯﷯= 1﷮4 𝑥﷮ 3﷮ 4﷯﷯﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦= 1﷮4 𝑥﷮ 3﷮ 4﷯﷯﷯ ×∆𝑥 Putting Values ∆𝑦= 1﷮4 256﷯﷮ 3﷮4﷯﷯﷯ × −1﷯ ∆𝑦= − 1﷮4 4﷮4﷯﷯﷮ 3﷮4﷯﷯﷯ ∆𝑦= − 1﷮4 × 4﷮3﷯﷯ ∆𝑦= − 1﷮4 × 64﷯ ∆𝑦= − 1﷮256﷯ ∆𝑦=−0. 0039 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ ∆𝑦= 𝑥+∆𝑥﷯ ﷮ 1﷮4﷯﷯− 𝑥﷮ 1﷮4﷯﷯ Putting Values −0. 0039= 256+ −1﷯﷯ ﷮ 1﷮4﷯﷯− 256﷯﷮ 1﷮4﷯﷯ −0. 0039= 255﷯ ﷮ 1﷮4﷯﷯− 4﷮4﷯﷯﷮ 1﷮4﷯﷯ −0. 0039= 255﷯ ﷮ 1﷮4﷯﷯−4 −0. 0039= 255﷯ ﷮ 1﷮4﷯﷯−4 −0. 0039+4= 255﷯ ﷮ 1﷮4﷯﷯ 3. 9961= 255﷯ ﷮ 1﷮4﷯﷯ Thus, the Approximate Values of 255﷯ ﷮ 1﷮4﷯﷯ is 𝟑. 𝟗𝟗𝟔𝟏 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (ix) (82)﷮ 1﷮4﷯﷯ Let 𝑦= 𝑥﷯﷮ 1﷮4﷯﷯ where 𝑥=81 ∆𝑥=1 Now, 𝑦= 𝑥﷯﷮ 1﷮4﷯﷯ Differentiating w.r.t.𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥﷮ 1﷮4﷯﷯﷯﷮𝑑𝑥﷯= 1﷮4﷯ 𝑥 ﷮ 1﷮4﷯ − 1﷯= 1﷮4﷯ 𝑥 ﷮ −3﷮ 4﷯﷯= 1﷮4 𝑥 ﷮ 3﷮4﷯﷯﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦= 1﷮4 𝑥 ﷮ 3﷮4﷯﷯﷯ ∆𝑦 Putting Values ∆𝑦= 1﷮4 81﷯﷮ 3﷮4﷯﷯﷯ × 1﷯ ∆𝑦= 1﷮4 × 3﷮4 × 3﷮4﷯ ﷯﷯ × 1﷯ ∆𝑦= 1﷮4 × 3﷮3﷯﷯ ∆𝑦= 1﷮4 ×27﷯ ∆𝑦= 1﷮108﷯ ∆𝑦=0. 009 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ So, ∆𝑦= 𝑥+∆𝑥﷯﷮ 1﷮4﷯﷯ − 𝑥﷯﷮ 1﷮4﷯﷯ Putting Values 0. 009= 81+1﷯﷮ 1﷮4﷯﷯− 81﷯ ﷮ 1﷮4﷯﷯ 0. 009= 82﷯﷮ 1﷮4﷯﷯− 3﷯﷮4 × 1﷮4﷯ ﷯ 0. 009= 82﷯﷮ 1﷮4﷯﷯−3 0. 009+3= 82﷯﷮ 1﷮4﷯﷯ 3. 009= 82﷯﷮ 1﷮4﷯﷯ Thus, the Approximate Value of 82﷯﷮ 1﷮4﷯﷯ is 𝟑. 𝟎𝟎𝟗 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (x) (401)﷮ 1﷮2﷯﷯ Let 𝑦= 𝑥﷮ 1﷮2﷯﷯ where 𝑥=400 & ∆𝑥=1 Now, 𝑑𝑦﷮𝑑𝑥﷯= 1﷮2 ﷮𝑥﷯﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦= 𝑑𝑦﷮2 ﷮𝑥﷯﷯ ∆𝑥 Putting Values ∆𝑦= 1﷮2 ﷮400﷯﷯ × 1﷯ ∆𝑦= 1﷮2 ﷮400﷯﷯ × 1﷯ ∆𝑦= 1﷮2 ﷮ 20﷮2﷯﷯﷯ ∆𝑦= 1﷮2 × 20﷯ ∆𝑦= 1﷮40﷯ ∆𝑦=0. 025 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ So, ∆𝑦= 𝑥+∆𝑥﷯﷮ 1﷮2﷯﷯− 𝑥﷯﷮ 1﷮2﷯﷯ Putting Values 0. 025= 400+1﷯﷮ 1﷮2 ﷯﷯− 400﷯ ﷮ 1﷮2﷯﷯ 0. 025= 401﷯﷮ 1﷮2﷯﷯− 20﷯﷮ 2 × 1﷮2﷯﷯ 0. 025= 401﷯﷮ 1﷮2﷯﷯−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 𝟐𝟎. 𝟎𝟐𝟓 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xi) (0.0037)﷮ 1﷮2﷯﷯ Let y = ﷮𝑥﷯ Let x = 0.0036 & △ x = 0.0001 Since y = ﷮𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑( ﷮𝑥﷯)﷮𝑑𝑥﷯ = 1﷮2 ﷮𝑥﷯﷯ Now, ∆𝑦 = 𝑑𝑦﷮𝑑𝑥﷯ △x = 1﷮2 ﷮𝑥﷯﷯ (0.0001) = 1﷮2 ﷮0.0036﷯﷯ (0.0001) = 1﷮2 × 0.06﷯ × (0.0001) = 0.0001﷮0.12﷯ = 1﷮1200﷯ = 0.000833 Also, ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ So, ∆𝑦= 𝑥+∆𝑥﷯﷮ 1﷮2﷯﷯− 𝑥﷯﷮ 1﷮2﷯﷯ Putting Values 0. 000833= 0. 0036+0. 0001﷯﷮ 1﷮2﷯﷯− 0. 0036﷯﷮ 1﷮2﷯﷯ 0. 000833= 0. 0037﷯﷮ 1﷮2﷯﷯− 36﷮10000﷯﷯﷮ 1﷮2﷯﷯ 0. 000833= 0. 0037﷯﷮ 1﷮2﷯﷯− 6﷮100﷯﷯﷮ 2 × 1﷮2﷯﷯ 0. 000833= 0. 0037﷯﷮ 1﷮2﷯﷯− 0. 06﷯﷮2 × 1﷮2﷯ ﷯ 0. 000833= 0. 0037﷯﷮ 1﷮2﷯﷯− 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﷯﷯=𝟎. 𝟎𝟔𝟎𝟖𝟑𝟑 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xii) (26.57)﷮ 1﷮3﷯﷯ Let 𝑦= 𝑥﷮ 1﷮3﷯﷯ where 𝑥=27 & ∆ 𝑥=−0. 43 Now, 𝑦= 𝑥﷮ 1﷮3﷯﷯ Differentiating w.r.t.𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥﷮ 1﷮3﷯﷯﷯﷮𝑑𝑥﷯= 1﷮3﷯ 𝑥﷮ 1﷮3﷯ − 1﷯= 1﷮3﷯ 𝑥﷮ − 2﷮ 3﷯﷯= 1﷮3 𝑥﷮ 2﷮ 3﷯﷯﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦= 1﷮3 𝑥﷮ 2﷮ 3﷯﷯﷯ × ∆𝑥 Putting Values ∆𝑦= 1﷮3 27﷯﷮ 2﷮ 3﷯﷯﷯ × −0. 43﷯ ∆𝑦= −0. 43﷮3 × 3﷮3 × 2﷮ 3﷯﷯﷯ ∆𝑦= −0. 43﷮3 × 3﷮2﷯﷯ ∆𝑦= −0. 43﷮3 × 9﷯ ∆𝑦= −0. 43﷮27﷯ ∆𝑦=−0. 015926 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ So, ∆𝑦= 𝑥+∆𝑥﷯ ﷮ 1﷮3﷯﷯− 𝑥﷮ 1﷮3﷯﷯ Putting Values −0. 015926= 27+ −0. 43﷯﷯﷮ 1﷮3﷯﷯− 27﷯﷮ 1﷮3﷯﷯ −0. 015926= 26. 57﷯﷮ 1﷮3﷯﷯− 27﷯﷮ 3 × 1﷮3﷯﷯ −0. 015926= 26. 57﷯﷮ 1﷮3﷯﷯−3 −0. 015926+3= 26. 53﷯﷮ 1﷮3﷯﷯ 2. 9843= 26. 53﷯﷮ 1﷮3﷯﷯ Thus, the Approximate Values of 26. 53﷯﷮ 1﷮3﷯﷯ is 𝟐. 𝟗𝟖𝟒𝟑 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xiii) (81.5)﷮ 1﷮4﷯﷯ Let 𝑦= 𝑥﷯﷮ 1﷮4﷯﷯ where 𝑥=81 & ∆𝑥=0. 5 Now, 𝑦= 𝑥﷮ 1﷮4﷯﷯ Differentiating w.r.t.𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥﷮ 1﷮4﷯﷯﷯﷮𝑑𝑥﷯= 1﷮4﷯ 𝑥﷮ 1﷮4﷯ − 1﷯= 1﷮4﷯ 𝑥﷮ − 3﷮ 4﷯﷯ = 1﷮4 𝑥﷮ 3﷮4﷯﷯﷯ ∆ 𝑥 Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦= 𝑑𝑦﷮4 𝑥﷮ 3﷮4﷯﷯﷯ ∆𝑥 Putting Values ∆𝑦= 1﷮4 81﷯﷮ 3﷮4﷯﷯﷯ × 0. 5﷯ ∆𝑦= 0. 5﷮4 × 3﷮ 4﷯﷯﷮ 3﷮4﷯﷯﷯ ∆𝑦= 0. 5﷮4 × 3﷮ 3﷯﷯ ∆𝑦= 0. 5﷮4 × 27﷯ ∆𝑦= 0. 5﷮108﷯ ∆𝑦=0. 0046 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ So, ∆𝑦= 𝑥+∆𝑥﷯﷮ 1﷮4﷯﷯− 𝑥﷮ 1﷮4﷯﷯ Putting Values 0. 0046= 81+0. 5﷯﷮ 1﷮4﷯﷯− 81﷯﷮ 1﷮4﷯﷯ 0. 0046= 81. 5﷯﷮ 1﷮4﷯﷯− 3﷮4﷯﷯﷮ 1﷮4﷯﷯ 0. 0046= 81. 5﷯﷮ 1﷮4﷯﷯−3 0. 0046+3= 81. 5﷯﷮ 1﷮4﷯﷯ 3. 0046= 81. 5﷯﷮ 1﷮4﷯﷯ Thus, Approximate Value of 81. 5﷯﷮ 1﷮4﷯﷯ is 𝟑. 𝟎𝟎𝟒𝟔 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xiv) (3.968)﷮ 3﷮2﷯﷯ Let 𝑦= 𝑥﷮ 3﷮2﷯﷯ where 𝑥=4 & ∆𝑥=−0. 032 Now, 𝑦= 𝑥﷮ 3﷮2﷯﷯ Differentiating w.r.t.𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥﷮ 3﷮2﷯﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 3﷮2﷯ 𝑥 ﷮ 1﷮2﷯﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦= 3﷮2﷯ 𝑥﷮ 1﷮2﷯﷯ ∆𝑥 Putting Values ∆𝑦= 3﷮2﷯ 4﷯﷮ 1﷮2﷯﷯ . −0. 032﷯ ∆𝑦= 3﷮2﷯ 2﷮2﷯﷯﷮ 1﷮2﷯﷯ . −0. 032﷯ ∆𝑦= 3﷮2﷯ ×2 × −0. 032﷯ ∆𝑦=3 × −0. 032﷯ ∆𝑦=−0. 096 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ So, ∆𝑦= 𝑥+∆𝑥﷯﷮ 3﷮2﷯﷯− 𝑥﷮ 3﷮2﷯﷯ Putting Values −0. 096= 4+ −0. 032﷯﷯﷮ 3﷮2﷯﷯− 4﷯﷮ 3﷮2﷯﷯ −0. 096= 4−0. 032﷯﷮ 3﷮2﷯﷯− 2﷯﷮2﷯﷮ × 3﷮2﷯﷯ −0. 096= 3. 968﷯﷮ 3﷮2﷯﷯− 2﷮3﷯ −0. 096= 3. 968﷯﷮ 3﷮2﷯﷯−8 −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 𝟕. 𝟗𝟎𝟒 Ex 6.4,1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xv) (32.15)﷮ 1﷮5﷯﷯ Let 𝑦= 𝑥﷯﷮ 1﷮5﷯﷯ where 𝑥=32 & ∆𝑥=0. 15 Now, 𝑦= 𝑥﷯﷮ 1﷮5﷯﷯ Differentiating w.r.t.𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥﷮ 1﷮5﷯﷯﷯﷮𝑑𝑥﷯= 1﷮5﷯ 𝑥﷮ 1﷮5﷯ − 1﷯ = 1﷮5﷯ 𝑥﷮ − 4﷮ 5﷯ ﷯= 1﷮5 𝑥﷮ 4﷮5﷯ ﷯﷯ Using ∆𝑦= 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦= 1﷮5 𝑥 ﷮ 4﷮5﷯﷯﷯ ∆𝑥 Putting Values ∆𝑦= 1﷮5 32﷯﷮ 4﷮5﷯﷯﷯ × 0. 15﷯ ∆𝑦= 1﷮5 2﷯﷮ 5 × 4﷮5﷯﷯﷯ × 0. 15﷯ ∆𝑦= 1﷮5 2﷯﷮ 4﷯﷯ × 0. 15﷯ ∆𝑦= 1﷮5 2﷯﷮ 4﷯﷯ × 0. 15﷯ ∆𝑦= 1﷮5 ×16﷯ ×0. 15 ∆𝑦= 0. 15 ﷮80﷯ ∆𝑦=0. 00187 We know that ∆𝑦=𝑓 𝑥+∆𝑥﷯−𝑓 𝑥﷯ So, ∆𝑦= 𝑥+∆𝑥﷯﷮ 1﷮5﷯﷯− 𝑥 ﷮ 1﷮5﷯﷯ Putting Values 0. 00187= 32+0. 15﷯﷮ 1﷮5﷯﷯− 32﷯﷮ 1﷮5﷯﷯ 0. 00187= 32. 15﷯﷮ 1﷮5﷯﷯− 2﷯﷮5 × 1﷮5﷯ ﷯ 0. 00187= 32. 15﷯﷮ 1﷮5﷯﷯−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 𝟐. 𝟎𝟎𝟏𝟖𝟕