Question

If $x=a(k\sin t+\sin kt),y=a(k\cos t+\cos kt),$ then find $\frac{d^{2}y}{d{x}^2}$ in terms of t.

• A. $\frac{k+1}{4ak}\frac{1}{{\cos }^3\frac{k+1}{2}t\cos \frac{k+1}{2}t}.$
• B. $\frac{k+1}{4ak}\frac{1}{{\cos }^3\frac{k-1}{2}t\cos \frac{k+1}{2}t}.$
• C. $\frac{k+1}{4ak}\frac{1}{{\cos }^3\frac{k+1}{2}t\cos \frac{k-1}{2}t}.$
• D. $-\frac{k+1}{4ak}\frac{1}{{\cos }^3\frac{k+1}{2}t\cos \frac{k-1}{2}t}.$

Answer 1

Answer: (C)

$\frac{k+1}{4ak}\frac{1}{{\cos }^3\frac{k+1}{2}t\cos \frac{k-1}{2}t}.$

$x=a(k\sin t+\sin kt)$ and $y=a(k\cos t+\cos kt),$
Differentiating both equation w.r.t $t$ we get,
$\frac{dx}{dt}=ak(\cos t+\cos kt)$ and $\frac{dy}{dx}=-ak(\sin t+\sin kt),$
$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\frac{\sin t+\sin kt}{\cos t+\cos kt}=-\tan \frac{k+1}{2}t$
$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(-\tan \frac{k+1}{2}t)$
$=\frac{d}{dt}(-\tan \frac{k+1}{2}t)\frac{dt}{dx}$
$=\frac{\frac{k+1}{2}{\sec }^2\frac{k+1}{2}t}{ak(\cos t+\cos kt)}$
$=\frac{k+1}{4ak}\frac{1}{{\cos }^3\frac{k+1}{2}t\cos \frac{k-1}{2}t}.$
$[\because \cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}]$