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Ex 6.3

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Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 13

Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 14
Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 15 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 16 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 17

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Transcript

Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (v) π‘₯=cos⁑𝑑, 𝑦=sin⁑𝑑 π‘Žπ‘‘ 𝑑= πœ‹/4At 𝒕= 𝝅/πŸ’ x = cos πœ‹/4 = 1/√2 y = sin πœ‹/4 = 1/√2 ∴ At 𝑑=πœ‹/4 , the point is (1/√2 " ," 1/√2) Finding 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=(𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Now, Hence, 𝑑𝑦/𝑑π‘₯=(𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) 𝒙=𝒄𝒐𝒔⁑𝒕 Differentiating w.r.t.𝑑 𝑑π‘₯/𝑑𝑑=𝑑(cos⁑𝑑 )/𝑑𝑑 𝑑π‘₯/𝑑𝑑=βˆ’sin⁑𝑑 π’š=π’”π’Šπ’β‘π’• Differentiating w.r.t. 𝑑 𝑑𝑦/𝑑𝑑=𝑑(sin⁑𝑑 )/𝑑𝑑 𝑑𝑦/𝑑𝑑=cos⁑𝑑 𝑑𝑦/𝑑π‘₯=cos⁑𝑑/(βˆ’sin⁑𝑑 ) 𝑑𝑦/𝑑π‘₯=βˆ’cot⁑𝑑 At 𝒕=𝝅/πŸ’ 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_(𝑑=πœ‹/4)=βˆ’π‘π‘œπ‘‘(πœ‹/4) =βˆ’1 ∴ Slope of tangent at (1/√2 " ," 1/√2) is – 1 We know that Slope of tangent Γ— Slope of Normal =βˆ’1 βˆ’1 Γ— Slope of Normal =βˆ’1 Slope of Normal =(βˆ’1)/(βˆ’1) Slope of Normal =1 Hence Slope of tangent is – 1 & Slope of Normal is 1 Finding equation of tangent & normal Now Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (1/√2 " ," 1/√2) & having Slope βˆ’1 is 𝑦 βˆ’1/√2 =βˆ’π‘₯+ 1/√2 π‘₯+𝑦 =1/√2+ 1/√2 π‘₯+𝑦 =2/√2 π‘₯+𝑦 =√2 𝒙+π’šβˆ’βˆšπŸ=𝟎 Equation of Normal at (1/√2 " ," 1/√2) & having Slope 1 is (π‘¦βˆ’1/√2)=1(π‘₯βˆ’1/√2) 𝑦 βˆ’1/√2 =π‘₯βˆ’ 1/√2 𝑦 =π‘₯βˆ’1/√2+ 1/√2 𝑦 =π‘₯ 𝒙 =π’š

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Davneet Singh

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.