Question 7 - Examples - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Question 7 Find the equation of tangent to the curve given by x = a sin3 t , y = b cos3 t at a point where t = ๐/2 . The curve is given as x = a sin3t , y = b cos3t Slope of the tangent = ๐๐ฆ/๐๐ฅ Here, ๐ ๐/๐ ๐ = (๐ ๐/๐ ๐)/(๐ ๐/๐ ๐) ๐ ๐/๐ ๐ = (๐(๐ cos^3โกใ๐ก)ใ)/๐๐ก = โ3b cos^2 ๐ก sinโก๐ก ๐ ๐/๐ ๐ = (๐(๐ sin^3โกใ๐ก)ใ)/๐๐ก = 3a sin^2โก๐ก cosโก๐ก Hence, ๐๐ฆ/๐๐ฅ = (dy/dt)/(๐๐ฅ/dt) = (โ3๐๐๐๐ ^2 ๐ก sinโก๐ก)/(3๐ sin^2โกใ๐ก cosโก๐ก ใ ) = (โ๐ ๐๐๐โก๐)/(๐ ๐๐๐โก๐ ) Now, Slope of the tangent at "t = " ๐/2 is ๐ ๐/๐ ๐ = (โ๐ ใcos ใโกใ๐/2ใ)/(๐ ใsin ใโกใ๐/2ใ ) = (โ๐(0))/(๐(1)) = 0 To find Equation of tangent, we need to find point (x, y) Putting t = ๐/2 in equation of x and y ๐ฅ = ๐ sin3 (๐/2) ๐=๐ ๐ฆ = b cos3 (๐/2) y = 0 Hence, point is (a, 0) Now, Equation of tangent at point (๐, 0) and with slope 0 is y โ 0 = 0 (x โ ๐) y = 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo