devarshi-dt-logo

Question:

If x = a sin2t(1+cos2t) and y = bcos2t(1-cos2t), show that at t = π/4, (dy/dx) = b/a.

Solution:

We have, x = a sin2t(1+cos2t)
Therefore, dx/dt = a.[2cos2t(1+cos2t) + sin2t(-2sin2t)] = 2a[cos2t + cos^2(2t) - sin^2(2t)] = 2a[cos2t + cos4t]
and y = bcos2t(1-cos2t)
then dy/dt = b[ -2sin2t(1-cos2t) + cos2t.2sin2t] = 2b[-sin2t + 2sin2tcos2t] = 2b[-sin2t + sin4t]
Therefore, dy/dx = (dy/dt)/(dx/dt) = 2b(-sin2t + sin4t)/2a(cos2t + cos4t)
Therefore, [dy/dx] at t = π/4 = b/a × (sin(π/2) - sin(π))/(cos(π/2) + cos(π)) = b/a × (1 - 0)/(0 -1) = -b/a