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

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

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.