Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12




Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Ex 6.3, 23 Prove that the curves ๐ฅ=๐ฆ2 & ๐ฅ๐ฆ=๐ cut at right angles if 8๐2 = 1 We need to show that the curves cut at right angles Two Curve intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other First we Calculate the point of intersection of Curve (1) & (2) ๐ฅ=๐ฆ2 ๐ฅ๐ฆ=๐ Putting ๐ฅ=๐ฆ2 in (2) ๐ฅ๐ฆ=๐ ๐ฆ^2 ร ๐ฆ=๐ ๐ฆ^3=๐ ๐ฆ=๐^(1/3) Putting Value of ๐ฆ=๐^(1/3) in (1) ๐ฅ=(๐^(1/3) )^2 ๐ฅ=๐^(2/3) Thus , Point of intersection of Curve is (๐^(๐/๐) ,๐^(๐/๐) ) We know that Slope of tangent to the Curve is ๐๐ฆ/๐๐ฅ For ๐=๐^๐ Differentiating w.r.t.๐ฅ ๐๐ฅ/๐๐ฅ=๐(๐ฆ^2 )/๐๐ฅ 1=๐(๐ฆ^2 )/๐๐ฅ ร ๐๐ฆ/๐๐ฆ 1=๐(๐ฆ^2 )/๐๐ฆ ร ๐๐ฆ/๐๐ฅ 1=2๐ฆ ร๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ=1/2๐ฆ Slope of tangent at (๐^(2/3) , ๐^(1/3) ) is ใ๐๐ฆ/๐๐ฅโใ_((๐^(2/3) , ๐^(1/3) ) )=1/2(๐^(1/3) ) =1/(2 ๐^(1/3) ) For ๐๐=๐ Differentiating w.r.t ๐(๐ฅ๐ฆ)/๐๐ฅ=๐(๐)/๐๐ฅ ๐(๐ฅ๐ฆ)/๐๐ฅ=0 ๐(๐ฅ)/๐๐ฅ ร๐ฆ+๐๐ฆ/๐๐ฅ ร๐ฅ=0 ๐ฆ+๐๐ฆ/๐๐ฅ ๐ฅ=0 ๐๐ฆ/๐๐ฅ=(โ๐ฆ)/๐ฅ Slope of tangent at (๐^(2/3) , ๐^(1/3) ) is ใ๐๐ฆ/๐๐ฅโใ_((๐^(2/3) , ๐^(1/3) ) )=(โ๐^(1/3))/๐^(2/3) =โใ ๐ใ^(1/3 โ 2/3) =โใ ๐ใ^((โ 1)/( 3) )=(โ1)/๐^(1/3) We need to show that Curves cut at right Angle i.e. tangents of their Curves are perpendicular to each other . Now, (Slope of tangent to the Curve ๐ฅ=๐ฆ^2) ร (Slope of tangent to the Curve ๐ฅ๐ฆ=๐) =โ1 1/(2 ๐^( 1/3) ) ร (โ1)/๐^( 1/3) =โ1 We know that if two lines are perpendicular then Product of their Slopes = โ1 1/(2 ใ๐ ใ^(1/3) ร๐^(1/3) )=โ1 1/(2 ๐^( 1/3 + 1/3) )=1 1/(2 ๐^( 2/3) )=1 1=2๐^( 2/3) 2๐^( 2/3)=1 ๐^( 2/3)=1/2 (๐^( 2/3) )^3=(1/2)^3 ๐^2=1/8 ใ๐๐ใ^๐=๐ Hence Proved
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