Example 20 - Find equation of tangent x = a sin3 t , y = b cos3 t

Example 20 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 20 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 20 - Chapter 6 Class 12 Application of Derivatives - Part 4

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

Transcript

Example 20 Find the equation of tangent to the curve given by x = a sin3 t , y = b cos3 t at a point where t = πœ‹/2 . The curve is given as x = a sin3t , y = b cos3t Slope of the tangent = 𝑑𝑦/𝑑π‘₯ Here, π’…π’š/𝒅𝒙 = (π’…π’š/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) π’…π’š/𝒅𝒕 = (𝑑(𝑏 cos^3⁑〖𝑑)γ€—)/𝑑𝑑 = βˆ’3b cos^2 𝑑 sin⁑𝑑 𝒅𝒙/𝒅𝒕 = (𝑑(π‘Ž sin^3⁑〖𝑑)γ€—)/𝑑𝑑 = 3a sin^2⁑𝑑 cos⁑𝑑 Hence, 𝑑𝑦/𝑑π‘₯ = (dy/dt)/(𝑑π‘₯/dt) = (βˆ’3π‘π‘π‘œπ‘ ^2 𝑑 sin⁑𝑑)/(3π‘Ž sin^2⁑〖𝑑 cos⁑𝑑 γ€— ) = (βˆ’π’ƒ 𝒄𝒐𝒔⁑𝒕)/(𝒂 π’”π’Šπ’β‘π’• ) Now, Slope of the tangent at "t = " πœ‹/2 is π’…π’š/𝒅𝒙 = (βˆ’π‘ γ€–cos γ€—β‘γ€–πœ‹/2γ€—)/(π‘Ž γ€–sin γ€—β‘γ€–πœ‹/2γ€— ) = (βˆ’π‘(0))/(π‘Ž(1)) = 0 To find Equation of tangent, we need to find point (x, y) Putting t = πœ‹/2 in equation of x and y π‘₯ = π‘Ž sin3 (πœ‹/2) 𝒙=𝒂 𝑦 = b cos3 (πœ‹/2) y = 0 Hence, point is (a, 0) Now, Equation of tangent at point (π‘Ž, 0) and with slope 0 is y βˆ’ 0 = 0 (x βˆ’ π‘Ž) y = 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 10 years. He provides courses for Maths and Science at Teachoo.