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

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Example 46 Find the equation of tangents to the curve y = cos (x + y), – 2𝜋 ≤ x ≤ 2 𝜋 that are parallel to the line x + 2y = 0. Given curve is 𝑦 = cos ﷐𝑥+𝑦﷯ , –2π ≤ 𝑥 ≤ 2π We need to find equation of tangent which is parallel to the line 𝑥 + 2𝑦 = 0 We know that slope of tangent is ﷐𝑑𝑦﷮𝑑𝑥﷯ 𝑦 = cos ﷐𝑥+𝑦﷯ Diff w.r.t. 𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐𝑑﷐𝑐𝑜𝑠﷐𝑥 + 𝑦﷯﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = –sin ﷐𝑥+𝑦﷯ ﷐𝑑﷐𝑥 + 𝑦﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ ﷐﷐𝑑𝑥﷮𝑑𝑥﷯+﷐𝑑𝑦﷮𝑑𝑥﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ ﷐1+﷐𝑑𝑦﷮𝑑𝑥﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ – sin﷐𝑥+𝑦﷯. ﷐𝑑𝑦﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ + sin ﷐𝑥+𝑦﷯.﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯﷐1+𝑠𝑖𝑛﷐𝑥+𝑦﷯﷯=−𝑠𝑖𝑛﷐𝑥+𝑦﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐−𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐ 𝑥 + 𝑦﷯﷯ ∴ Slope of tangent is ﷐−𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷯ Given line is 𝑥 + 2𝑦 = 0 2𝑦 = –𝑥 𝑦 = ﷐−𝑥﷮2﷯ 𝑦 = ﷐﷐ −1﷮2﷯﷯𝑥+0 The above equation is of the form 𝑦= m𝑥 + c where m is slope ⇒ Slope of line is ﷐−1﷮2﷯ We know that If two lines are parallel than their slopes are equal Given line 𝑥 + 2𝑦 = 0 is parallel to tangent ∴ Slope of tangent = Slope of line ﷐−𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷯= ﷐−1﷮2﷯ ﷐𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷯= ﷐1﷮2﷯ 2 sin﷐𝑥+𝑦﷯=1+𝑠𝑖𝑛﷐𝑥+𝑦﷯ 2 sin ﷐𝑥+𝑦﷯ – sin﷐𝑥+𝑦﷯=1 sin ﷐𝑥+𝑦﷯=1 We know that sin ﷐𝜋﷮2﷯ = 1 sin﷐𝑥+𝑦﷯ = sin ﷐𝜋﷮2﷯ Hence general solution of 𝑥 + 𝑦 is 𝑥 + 𝑦 = nπ + ﷐﷐−1﷯﷮n﷯ ﷐𝜋﷮2﷯ Now, Finding points through which tangents pass. Given curve y = cos ﷐𝑥+𝑦﷯ Putting value of x + y y = cos ﷐𝑛𝜋+﷐﷐−1﷯﷮𝑛﷯﷐𝜋﷮2﷯﷯ y = 0 Putting y = 0 in x 𝑥 + 𝑦 = ﷐𝑛𝜋+﷐﷐−1﷯﷮𝑛﷯﷐𝜋﷮2﷯﷯ 𝑥 + 0 = nπ + ﷐﷐−1﷯﷮𝑛﷯﷐𝜋﷮2﷯ 𝑥 = nπ + ﷐﷐−1﷯﷮𝑛 ﷯﷐𝜋﷮2﷯ Since 2π ≤ 𝑥 ≤ 2π Thus, 𝑥 = ﷐−3𝜋﷮2﷯ & 𝑥 = ﷐𝜋﷮2﷯ ∴ Points are ﷐﷐−3𝜋﷮2﷯ , 0﷯ & ﷐﷐𝜋﷮2﷯ , 0﷯ Finding equation of tangents Hence Required Equation of tangent are 2x + 4y + 3π = 0 2x + 4y – π = 0 Example 46 Find the equation of tangents to the curve y = cos (x + y), – 2𝜋 ≤ x ≤ 2 𝜋 that are parallel to the line x + 2y = 0. Given curve is 𝑦 = cos ﷐𝑥+𝑦﷯ , –2π ≤ 𝑥 ≤ 2π We need to find equation of tangent which is parallel to the line 𝑥 + 2𝑦 = 0 We know that slope of tangent is ﷐𝑑𝑦﷮𝑑𝑥﷯ 𝑦 = cos ﷐𝑥+𝑦﷯ Diff w.r.t. 𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐𝑑﷐𝑐𝑜𝑠﷐𝑥 + 𝑦﷯﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = –sin ﷐𝑥+𝑦﷯ ﷐𝑑﷐𝑥 + 𝑦﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ ﷐﷐𝑑𝑥﷮𝑑𝑥﷯+﷐𝑑𝑦﷮𝑑𝑥﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ ﷐1+﷐𝑑𝑦﷮𝑑𝑥﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ – sin﷐𝑥+𝑦﷯. ﷐𝑑𝑦﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ + sin ﷐𝑥+𝑦﷯.﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯﷐1+𝑠𝑖𝑛﷐𝑥+𝑦﷯﷯=−𝑠𝑖𝑛﷐𝑥+𝑦﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐−𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐ 𝑥 + 𝑦﷯﷯ ∴ Slope of tangent is ﷐−𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷯ Given line is 𝑥 + 2𝑦 = 0 2𝑦 = –𝑥 𝑦 = ﷐−𝑥﷮2﷯ 𝑦 = ﷐﷐ −1﷮2﷯﷯𝑥+0 The above equation is of the form 𝑦= m𝑥 + c where m is slope ⇒ Slope of line is ﷐−1﷮2﷯ We know that If two lines are parallel than their slopes are equal Given line 𝑥 + 2𝑦 = 0 is parallel to tangent ∴ Slope of tangent = Slope of line ﷐−𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷯= ﷐−1﷮2﷯ ﷐𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷯= ﷐1﷮2﷯ 2 sin﷐𝑥+𝑦﷯=1+𝑠𝑖𝑛﷐𝑥+𝑦﷯ 2 sin ﷐𝑥+𝑦﷯ – sin﷐𝑥+𝑦﷯=1 sin ﷐𝑥+𝑦﷯=1 We know that sin ﷐𝜋﷮2﷯ = 1 sin﷐𝑥+𝑦﷯ = sin ﷐𝜋﷮2﷯ Hence general solution of 𝑥 + 𝑦 is 𝑥 + 𝑦 = nπ + ﷐﷐−1﷯﷮n﷯ ﷐𝜋﷮2﷯ Now, Finding points through which tangents pass. Given curve y = cos ﷐𝑥+𝑦﷯ Putting value of x + y y = cos ﷐𝑛𝜋+﷐﷐−1﷯﷮𝑛﷯﷐𝜋﷮2﷯﷯ y = 0 Putting y = 0 in x 𝑥 + 𝑦 = ﷐𝑛𝜋+﷐﷐−1﷯﷮𝑛﷯﷐𝜋﷮2﷯﷯ 𝑥 + 0 = nπ + ﷐﷐−1﷯﷮𝑛﷯﷐𝜋﷮2﷯ 𝑥 = nπ + ﷐﷐−1﷯﷮𝑛 ﷯﷐𝜋﷮2﷯ Since 2π ≤ 𝑥 ≤ 2π Thus, 𝑥 = ﷐−3𝜋﷮2﷯ & 𝑥 = ﷐𝜋﷮2﷯ ∴ Points are ﷐﷐−3𝜋﷮2﷯ , 0﷯ & ﷐﷐𝜋﷮2﷯ , 0﷯ Finding equation of tangents Hence Required Equation of tangent are 2x + 4y + 3π = 0 2x + 4y – π = 0 Example 46 Find the equation of tangents to the curve y = cos (x + y), – 2𝜋 ≤ x ≤ 2 𝜋 that are parallel to the line x + 2y = 0. Given curve is 𝑦 = cos ﷐𝑥+𝑦﷯ , –2π ≤ 𝑥 ≤ 2π We need to find equation of tangent which is parallel to the line 𝑥 + 2𝑦 = 0 We know that slope of tangent is ﷐𝑑𝑦﷮𝑑𝑥﷯ 𝑦 = cos ﷐𝑥+𝑦﷯ Diff w.r.t. 𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐𝑑﷐𝑐𝑜𝑠﷐𝑥 + 𝑦﷯﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = –sin ﷐𝑥+𝑦﷯ ﷐𝑑﷐𝑥 + 𝑦﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ ﷐﷐𝑑𝑥﷮𝑑𝑥﷯+﷐𝑑𝑦﷮𝑑𝑥﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ ﷐1+﷐𝑑𝑦﷮𝑑𝑥﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ – sin﷐𝑥+𝑦﷯. ﷐𝑑𝑦﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ + sin ﷐𝑥+𝑦﷯.﷐𝑑𝑦﷮𝑑𝑥﷯ = – sin ﷐𝑥+𝑦﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯﷐1+𝑠𝑖𝑛﷐𝑥+𝑦﷯﷯=−𝑠𝑖𝑛﷐𝑥+𝑦﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐−𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐ 𝑥 + 𝑦﷯﷯ ∴ Slope of tangent is ﷐−𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷯ Given line is 𝑥 + 2𝑦 = 0 2𝑦 = –𝑥 𝑦 = ﷐−𝑥﷮2﷯ 𝑦 = ﷐﷐ −1﷮2﷯﷯𝑥+0 The above equation is of the form 𝑦= m𝑥 + c where m is slope ⇒ Slope of line is ﷐−1﷮2﷯ We know that If two lines are parallel than their slopes are equal Given line 𝑥 + 2𝑦 = 0 is parallel to tangent ∴ Slope of tangent = Slope of line ﷐−𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷯= ﷐−1﷮2﷯ ﷐𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷮1 + 𝑠𝑖𝑛﷐𝑥 + 𝑦﷯﷯= ﷐1﷮2﷯ 2 sin﷐𝑥+𝑦﷯=1+𝑠𝑖𝑛﷐𝑥+𝑦﷯ 2 sin ﷐𝑥+𝑦﷯ – sin﷐𝑥+𝑦﷯=1 sin ﷐𝑥+𝑦﷯=1 We know that sin ﷐𝜋﷮2﷯ = 1 sin﷐𝑥+𝑦﷯ = sin ﷐𝜋﷮2﷯ Hence general solution of 𝑥 + 𝑦 is 𝑥 + 𝑦 = nπ + ﷐﷐−1﷯﷮n﷯ ﷐𝜋﷮2﷯ Now, Finding points through which tangents pass. Given curve y = cos ﷐𝑥+𝑦﷯ Putting value of x + y y = cos ﷐𝑛𝜋+﷐﷐−1﷯﷮𝑛﷯﷐𝜋﷮2﷯﷯ y = 0 Putting y = 0 in x 𝑥 + 𝑦 = ﷐𝑛𝜋+﷐﷐−1﷯﷮𝑛﷯﷐𝜋﷮2﷯﷯ 𝑥 + 0 = nπ + ﷐﷐−1﷯﷮𝑛﷯﷐𝜋﷮2﷯ 𝑥 = nπ + ﷐﷐−1﷯﷮𝑛 ﷯﷐𝜋﷮2﷯ Since 2π ≤ 𝑥 ≤ 2π Thus, 𝑥 = ﷐−3𝜋﷮2﷯ & 𝑥 = ﷐𝜋﷮2﷯ ∴ Points are ﷐﷐−3𝜋﷮2﷯ , 0﷯ & ﷐﷐𝜋﷮2﷯ , 0﷯ Finding equation of tangents Hence Required Equation of tangent are 2x + 4y + 3π = 0 2x + 4y – π = 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.