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Ex 6.3

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Ex 6.3, 17 - Find points on y = x3 at which slope of tangent

Ex 6.3,17 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,17 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.3,17 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Transcript

Ex 6.3, 17 Find the points on the curve 𝑦=𝑥3 at which the slope of the tangent is equal to the y-coordinate of the pointLet the Point be (ℎ , 𝑘) on the Curve 𝑦=𝑥3 Where Slope of tangent at (ℎ , 𝑘)=𝑦−𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 (ℎ, 𝑘) i.e. 〖𝑑𝑦/𝑑𝑥│〗_((ℎ, 𝑘) )=𝑘 Given 𝑦=𝑥^3 Differentiating w.r.t.𝑥 𝑑𝑦/𝑑𝑥=3𝑥^2 ∴ Slope of tangent at (ℎ , 𝑘) is 〖𝑑𝑦/𝑑𝑥│〗_((ℎ, 𝑘) )=3ℎ^2 From (1) 〖𝑑𝑦/𝑑𝑥│〗_((ℎ, 𝑘) )=𝑘 3ℎ^2=𝑘 Also Point (ℎ , 𝑘) is on the Curve 𝑦=𝑥^3 Point (ℎ , 𝑘) must Satisfy the Equation of Curve i.e. 𝑘=ℎ^3 Now our equations are 3ℎ^2=𝑘 …(1) & 𝑘=ℎ^3 …(2) Putting Value of 𝑘=3ℎ^2 in (3) 3ℎ^2=ℎ^3 ℎ^3−3ℎ^2=0 ℎ^2 (ℎ−3)=0 ℎ^2=0 ℎ=0 ℎ−3=0 ℎ=3 When 𝒉=𝟎 3ℎ^2=𝑘 3(0)=𝑘 𝑘=0 Hence, point is (0, 0) When 𝒉=𝟑 3ℎ^2=𝑘 3(3)^2=𝑘 𝑘=27 Hence, point is (3 , 27)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.