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

Transcript

Misc 3 The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ? Let x be the equal sides isosceles triangle with fixed base b. i.e. AB = AC = ๐‘ฅ & BC = b Given that side of triangle decreasing when ๐‘ฅ = b i.e. ๐‘‘๐‘ฅ/๐‘‘๐‘ก= 3 cm/sec. We need to find how fast area is decreasing when ๐‘ฅ = b i.e. ๐‘‘๐ด/๐‘‘๐‘ก when ๐‘ฅ = b Finding Area Draw a perpendicular AD to BC โ‡’ i.e. AD โŠฅ BC In isosceles triangle, perpendicular from vertex to the side bisects the side i.e. D is the mid point of BC BD = DC โˆด BD = DC = ๐‘/2 In โˆ† ๐ด๐ท๐ต Using Pythagoras theorem (๐ด๐ต)^2=(๐ด๐ท)^2+(๐ต๐ท)^2 (๐‘ฅ)^2=(๐ด๐ท)^2+ (๐‘/2)^2 ๐‘ฅ2 โ€“ (๐‘/2)^2=(๐ด๐ท)^2 (๐ด๐ท)^2 = ๐‘ฅ2 โ€“ (๐‘/2)^2 ๐ด๐ท=โˆš(๐‘ฅ2โˆ’(๐‘/2)^2 ) We know that Area of isosceles triangle = 1/2 ร— Base ร— Height A = 1/2 ร— b ร— โˆš(๐‘ฅ2โˆ’(๐‘/2)^2 ) A = 1/2 ร— b ร— /2 โˆš(๐‘ฅ2โˆ’๐‘^2/4) We need ๐‘‘๐ด/๐‘‘๐‘ก Diff w.r.t.x ๐‘‘๐ด/๐‘‘๐‘ก= 1/2 ๐‘ . ๐‘‘(โˆš(๐‘ฅ^2 โˆ’ ๐‘^2/4))/๐‘‘๐‘ก ๐‘‘๐ด/๐‘‘๐‘ก= 1/2 ๐‘ [1/(2โˆš(๐‘ฅ2 โˆ’ ๐‘^2/4)) ร— ๐‘‘(๐‘ฅ^2 โˆ’ ๐‘^2/4)/๐‘‘๐‘ก] ๐‘‘๐ด/๐‘‘๐‘ก= 1/2 ๐‘ [1/(2โˆš(๐‘ฅ2 โˆ’ ๐‘^2/4)) ร—(๐‘‘(๐‘ฅ2)/๐‘‘๐‘กโˆ’0)] ๐‘‘๐ด/๐‘‘๐‘ก= 1/2 ๐‘ [1/(2โˆš(๐‘ฅ2 โˆ’ ๐‘^2/4)) ร—(๐‘‘(๐‘ฅ2)/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฅ/๐‘‘๐‘ก)] ๐‘‘๐ด/๐‘‘๐‘ก= 1/2 ๐‘ [1/(2โˆš(๐‘ฅ2 โˆ’ ๐‘^2/4)) ร— 2๐‘ฅ ร— ๐‘‘๐‘ฅ/๐‘‘๐‘ก] ๐‘‘๐ด/๐‘‘๐‘ก= 1/2 ๐‘ [1/(2โˆš(๐‘ฅ2 โˆ’ ๐‘^2/4)) ร— 2๐‘ฅ ร— 3] ๐‘‘๐ด/๐‘‘๐‘ก= ๐‘/(4โˆš(๐‘ฅ2 โˆ’ ๐‘^2/4)) ร— 6๐‘ฅ Finding ๐‘‘๐ด/๐‘‘๐‘ก At ๐‘ฅ = b โ”œ ๐‘‘๐ด/๐‘‘๐‘กโ”ค|_(๐‘ฅ = ๐‘)=(6๐‘^2)/(4โˆš(๐‘^2 โˆ’ ๐‘^2/4))= (6๐‘^2)/(4โˆš((4๐‘^2 โˆ’ ๐‘^2)/4))= (6๐‘^2)/(4โˆš((3๐‘^2)/4))= (6๐‘^2)/((4โˆš3 ๐‘)/2)= (6๐‘^2)/(2โˆš3 ๐‘) = 3๐‘/โˆš3=๐‘โˆš3 Since dimension of area is cm2 s โ‡’ ๐‘‘๐ด/๐‘‘๐‘ก at ๐‘ฅ = b is ๐’ƒโˆš๐Ÿ‘ cm2/s

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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.