Check sibling questions

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


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 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.