If $x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1-\cos 2t)$, find the values of $\dfrac{dy}{dx}$ at $t=\dfrac{\pi}{4}$ and $t=\dfrac{\pi}{3}$.

$\dfrac{b}{a}$ at $t=\pi/4$; $\dfrac{\sqrt3\,b}{a}$ at $t=\pi/3$

Differentiation — CBSE Class 12 · 2016 Board Question with Solution

x = a sin 2 t (1 + cos 2 t)

and

y = b cos 2 t (1 - cos 2 t)

, find the values of

\frac{d y}{d x}

t = \frac{π}{4}

and

t = \frac{π}{3}

Answer: $\frac{b}{a}$ at $t = π /4$ ; $\frac{3 b}{a}$ at $t = π /3$

As in the reduced forms,

x = a sin 2 t + \frac{a}{2} sin 4 t

and

y = b cos 2 t - \frac{b}{2} - \frac{b}{2} cos 4 t

, so

\frac{d x}{d t} = 2 a (cos 2 t + cos 4 t), \frac{d y}{d t} = 2 b (sin 4 t - sin 2 t)

Hence:

\frac{d y}{d x} = \frac{b ( sin 4 t - sin 2 t )}{a ( cos 4 t + cos 2 t )}

t = \frac{π}{4}

sin 4 t = 0, sin 2 t = 1, cos 4 t = - 1, cos 2 t = 0 \Rightarrow \frac{d y}{d x} = \frac{b ( - 1 )}{a ( - 1 )} = \frac{b}{a}

.
At

t = \frac{π}{3}

sin 4 t = - \frac{3}{2}, sin 2 t = \frac{3}{2}, cos 4 t = - \frac{1}{2}, cos 2 t = - \frac{1}{2} \Rightarrow \frac{d y}{d x} = \frac{b ( - 3 )}{a ( - 1 )} = \frac{3 b}{a}

\frac{d y}{d x}_{t = π /4} = \frac{b}{a}, \frac{d y}{d x}_{t = π /3} = \frac{3 b}{a}

CBSE Class 12 Board · 2016 · Subjective