Question

Find $\frac{dy}{dx}$, if x = 2 cos t + cos 2t, y = 2 sin t - sin 2t, at t = π/4

Answer 1

I hope it is correct

Answer 2

Answer:

$\dfrac {dy} {dx} = \dfrac {-1}{1 + \sqrt {2}}$

Step-by-step explanation:

We need to find dy/dx

given:

x = 2 cos t + cos 2t

$\dfrac {dx} {dt} = -2sint - 2sin2t\\\\\dfrac {dx} {dt} = -2(sint + sin2t)...(1)$

y = 2 sin t - sin 2t

$\dfrac {dy} {dt} = 2cost - 2cos2t\\\\\dfrac {dx} {dt} = -2(cos2t - cost)...(2)$

Now, let us find dy/dx

divide eqution (1) by (2)

$\dfrac { \dfrac {dx} {dt} }{\dfrac {dy} {dt}}= \dfrac {-2(cos2t - cost)}{-2(sint + sin2t)}\\\\\dfrac {dy} {dx} = \dfrac {cos2t - cost}{sint + sin2t}$

Now, put the vale of t = $\pi/4$ = 45°

$\dfrac {dy} {dx} = \dfrac {cos2\times 45^o  - cos45^o}{sin45^o + sin2\times 45^o}$

$\dfrac {dy} {dx} = \dfrac {cos90^o  - cos45^o}{sin45^o + sin90^o}$

$\dfrac {dy} {dx} = \dfrac {0 - \dfrac{1}{\sqrt{2}}} {{\dfrac1{\sqrt{2}}+ 1}}\\\dfrac {dy} {dx} = \dfrac { \dfrac{-1}{\sqrt{2}}} {{\dfrac{1+\sqrt{2}}{ \sqrt{2}}}}\\\dfrac {dy} {dx} = \dfrac {-1}{1 + \sqrt {2}}$