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Example 46 - Find equation of tangents to y = cos (x + y)

Example 46 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 46 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 46 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 46 - Chapter 6 Class 12 Application of Derivatives - Part 5 Example 46 - Chapter 6 Class 12 Application of Derivatives - Part 6 Example 46 - Chapter 6 Class 12 Application of Derivatives - Part 7 Example 46 - Chapter 6 Class 12 Application of Derivatives - Part 8

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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) 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. 𝒙 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) π’…π’š/𝒅𝒙 = (βˆ’π’”π’Šπ’(𝒙 + π’š))/(𝟏 + π’”π’Šπ’( 𝒙 + π’š) ) ∴ 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 π’š = (( βˆ’πŸ)/𝟐)𝒙+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 Since line is parallel to tangent ∴ Slope of tangent = Slope of line (βˆ’π’”π’Šπ’(𝒙 + π’š))/(𝟏 + π’”π’Šπ’(𝒙 + π’š) )= (βˆ’πŸ)/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 (𝒙+π’š)=𝟏 Since sin πœ‹/2 = 1 sin(π‘₯+𝑦) = sin 𝝅/𝟐 Hence, (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’πŸ‘π…)/𝟐 , 𝟎) & (𝝅/𝟐 , 𝟎) Putting n = 0 π‘₯ = 0(πœ‹)+(βˆ’1)^0 πœ‹/2 π‘₯ = 0 + (πœ‹/2) 𝒙 = 𝝅/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 𝒙 = (βˆ’πŸ‘π…)/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) Equation of tangent at ((βˆ’πŸ‘π…)/𝟐 , 𝟎) & having slope (βˆ’πŸ)/𝟐 is (π‘¦βˆ’0) = (βˆ’1)/2 (π‘₯βˆ’((βˆ’3πœ‹)/2)) y = (βˆ’1)/2 (π‘₯+3πœ‹/2) y = (βˆ’1)/2 ((2π‘₯ + 3πœ‹)/2) 2x + 4y + 3Ο€ = 0 Equation of tangent at (𝝅/𝟐 , 𝟎) & having slope (βˆ’πŸ)/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€) 2x + 4y βˆ’ Ο€ = 0 Hence Required Equation of tangent are 2x + 4y + 3Ο€ = 0 2x + 4y – Ο€ = 0

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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.