Question

If $x=a(\cos t+t\sin t)$ and $y=a(\sin t-t\cos t),$ then find $\frac{d^{2}y}{d{x}^2}.$

Answer 1

$\begin{array}{c}x=a\left(\cos t+t\sin t\right)andy=a\left(\sin t-t\cos t\right)\\ x=a\left(\cos t+t\sin t\right)\\ \frac{dx}{dt}=a\left[-\sin t+t\cos t+\sin t\right]\\ =-a\sin t+at\cos t+a\sin t\\ =at\cos t\\ \frac{d^{2}x}{d{t}^2}=a\left[t\left(-\sin t\right)+\cos t.1\right]\\ =\left[a\cos t-at\sin t\right]\\ y=a\left(\sin t-t\cos t\right)\\ \frac{dy}{dt}=a\left[\cos t-t\left(-\sin t\right)-\cos t.1\right]\\ =\left[a\cos t+at\sin t-\cos ta\right]\\ =at\sin t\\ \frac{d^{2}y}{d{t}^2}=a\left[t.\cos t+\sin t.1\right]\\ =\left[at\cos t+a\sin t\right]\\ =\left[at\cos t+a\sin t\right]\\ \frac{d^{2}y}{d{t}^2}=\frac{\frac{d^{2}y}{d{t}^2}\times \frac{dx}{dt}-\frac{d^{2}x}{d{t}^2}\times \frac{dy}{dt}}{(\frac{dx}{dt}{)}^3}=\frac{\left(at\cos t\right)(at\cos t+a\sin t)-(a\cos t-at\sin t)\left(at\sin t\right)}{(at\cos t{)}^3}=\frac{1}{at{\cos }^{3}t}.\end{array}$