Question

If x = a sin t − b cos t, y = a cos t + b sin t, prove that $\frac{d^{2}y}{d{x}^2}=-\frac{x^2+y^2}{y^3}$.

Answer 1

Here,
$$\begin{gathered} x=a\sin t-b\cos t\mathrm{and}y=a\cos t+b\sin t \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}t,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=a\cos t+b\sin t\mathrm{and}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}=-a\sin t+b\cos t \\ \therefore \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{-a\sin t+b\cos t}{a\cos t+b\sin t} \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}\mathit{y}}{\mathit{d}{\mathit{x}}^{\mathit{2}}}=\left[\frac{-{\left(a\cos t+b\sin t\right)}^2-{\left(a\sin t-b\cos t\right)}^2}{{\left(a\cos t+b\sin t\right)}^2}\right]\frac{dt}{dx} \\ =\left[\frac{-{\left(a\cos t+b\sin t\right)}^2-{\left(a\sin t-b\cos t\right)}^2}{{\left(a\cos t+b\sin t\right)}^2}\right]\frac{1}{a\cos t+b\sin t} \\ =\frac{-\left(y^2+x^2\right)}{y^3} \end{gathered}$$