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

Transcript

Example 46 Find the equation of tangents to the curve y = cos (x + y), – 2πœ‹ ≀ x ≀ 2πœ‹ that are parallel to the line x + 2y = 0. Given curve is 𝑦 = cos (π‘₯+𝑦) , –2Ο€ ≀ π‘₯ ≀ 2Ο€ We need to find equation of tangent which is parallel to the line π‘₯ + 2𝑦 = 0 We know that slope of tangent is 𝑑𝑦/𝑑π‘₯ 𝑦 = cos (π‘₯+𝑦) Diff w.r.t. π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(π‘π‘œπ‘ (π‘₯ + 𝑦))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = –sin (π‘₯+𝑦) 𝑑(π‘₯ + 𝑦)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = – sin (π‘₯+𝑦) (𝑑π‘₯/𝑑π‘₯+𝑑𝑦/𝑑π‘₯) 𝑑𝑦/𝑑π‘₯ = – sin (π‘₯+𝑦) (1+𝑑𝑦/𝑑π‘₯) 𝑑𝑦/𝑑π‘₯ = – sin (π‘₯+𝑦) – sin(π‘₯+𝑦). 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ + sin (π‘₯+𝑦).𝑑𝑦/𝑑π‘₯ = – sin (π‘₯+𝑦) 𝑑𝑦/𝑑π‘₯ (1+𝑠𝑖𝑛(π‘₯+𝑦))=βˆ’π‘ π‘–π‘›(π‘₯+𝑦) 𝑑𝑦/𝑑π‘₯ = (βˆ’π‘ π‘–π‘›(π‘₯ + 𝑦))/(1 + 𝑠𝑖𝑛( π‘₯ + 𝑦) ) ∴ Slope of tangent is (βˆ’π‘ π‘–π‘›(π‘₯ + 𝑦))/(1 + 𝑠𝑖𝑛(π‘₯ + 𝑦) ) Given line is π‘₯ + 2𝑦 = 0 2𝑦 = –π‘₯ 𝑦 = (βˆ’π‘₯)/2 𝑦 = (( βˆ’1)/2)π‘₯+0 The above equation is of the form 𝑦= mπ‘₯ + c where m is slope ∴ Slope of line is (βˆ’1)/2 We know that If two lines are parallel than their slopes are equal Given line π‘₯ + 2𝑦 = 0 is parallel to tangent ∴ Slope of tangent = Slope of line (βˆ’π‘ π‘–π‘›(π‘₯ + 𝑦))/(1 + 𝑠𝑖𝑛(π‘₯ + 𝑦) )= (βˆ’1)/2 𝑠𝑖𝑛(π‘₯ + 𝑦)/(1 + 𝑠𝑖𝑛(π‘₯ + 𝑦) )= 1/2 2 sin(π‘₯+𝑦)=1+𝑠𝑖𝑛(π‘₯+𝑦) 2 sin (π‘₯+𝑦) – sin(π‘₯+𝑦)=1 sin (π‘₯+𝑦)=1 We know that sin πœ‹/2 = 1 sin(π‘₯+𝑦) = sin πœ‹/2 Hence general solution of π‘₯ + 𝑦 is π‘₯ + 𝑦 = nΟ€ + (βˆ’1)^n πœ‹/2 Now, Finding points through which tangents pass Given curve y = cos (π‘₯+𝑦) Putting value of x + y y = cos (π‘›πœ‹+(βˆ’1)^𝑛 πœ‹/2) y = 0 Putting y = 0 in x π‘₯ + 𝑦 = (π‘›πœ‹+(βˆ’1)^𝑛 πœ‹/2) π‘₯ + 0 = nΟ€ + (βˆ’1)^𝑛 πœ‹/2 π‘₯ = nΟ€ + (βˆ’1)^(𝑛 ) πœ‹/2 Since βˆ’2Ο€ ≀ π‘₯ ≀ 2Ο€ Thus, π‘₯ = (βˆ’3πœ‹)/2 & π‘₯ = πœ‹/2 ∴ Points are ((βˆ’3πœ‹)/2 , 0) & (πœ‹/2 , 0) Putting n = 0 π‘₯ = 0(πœ‹)+(βˆ’1)^0 πœ‹/2 π‘₯ = 0 + (πœ‹/2) π‘₯ = πœ‹/2 Putting n = –1 π‘₯ = –1(πœ‹)+(βˆ’1)^(βˆ’1) πœ‹/2 π‘₯ = βˆ’πœ‹βˆ’πœ‹/2 π‘₯ = (βˆ’2πœ‹ βˆ’ πœ‹)/2 π‘₯ = (βˆ’3πœ‹)/2 Finding equation of tangents We know that Equation of line at (π‘₯1 ,𝑦1) & having slope at π‘š is (π‘¦βˆ’π‘¦1)=π‘š(π‘₯βˆ’π‘₯1) We know that Equation of line at (π‘₯1 ,𝑦1) & having slope at π‘š is (π‘¦βˆ’π‘¦1)=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (πœ‹/2 , 0) & having slope (βˆ’1)/2 is (π‘¦βˆ’0)= (βˆ’1)/2 (π‘₯βˆ’πœ‹/2) 𝑦 = (βˆ’1)/2 ((2π‘₯ βˆ’ πœ‹)/2) 𝑦 = (βˆ’1)/4 (2π‘₯βˆ’πœ‹) 4y = –(2x – Ο€) " 4y = –2x + Ο€" 4𝑦 = –2π‘₯ – 3Ο€ 2x + 4y + 3Ο€ = 0 2x + 4y – Ο€ = 0 Hence Required Equation of tangent are 2x + 4y + 3Ο€ = 0 2x + 4y – Ο€ = 0

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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.