Question

If x = f(t) cos t − f' (t) sin t and y = f(t) sin t + f'(t) cos t, then ${\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2=$

(a) f(t) − f''(t)
(b) {f(t) − f'' (t)}²
(c) {f(t) + f''(t)}²
(d) none of these

Answer 1

(c){f(t) + f''(t)}²

Here,
$$\begin{gathered} x=f\left(t\right)\cos t-f^{\text{'}}\left(t\right)\sin t\mathrm{and}y=f\left(t\right)\sin t+f^{\text{'}}\left(t\right)\cos t \\ \Rightarrow \frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=f^{\text{'}}\left(t\right)\cos t-f\left(t\right)\sin t-f^{\text{'}\text{'}}\left(t\right)\sin t-f^{\text{'}}\left(t\right)\cos t\mathrm{and}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}=f^{\text{'}}\left(t\right)\sin t+f\left(t\right)\cos t+f^{\text{'}\text{'}}\left(t\right)\cos t-f^{\text{'}}\left(t\right)\sin t \\ \Rightarrow \frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=-f\left(t\right)\sin t-f^{\text{'}\text{'}}\left(t\right)\sin t\mathrm{and}\frac{\mathit{d}y}{\mathit{d}t}=f\left(t\right)\cos t+f^{\text{'}\text{'}}\left(t\right)\cos t \\ \mathrm{Thus}, \\ {\left(\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}\right)}^2+{\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}\right)}^2={\left\{-f\left(t\right)\sin t-f^{\text{'}\text{'}}\left(t\right)\sin t\right\}}^2+{\left\{f\left(t\right)\cos t+f^{\text{'}\text{'}}\left(t\right)\cos t\right\}}^2 \\ ={\left\{f\left(t\right)\sin t+f^{\text{'}\text{'}}\left(t\right)\sin t\right\}}^2+{\left\{f\left(t\right)\cos t+f^{\text{'}\text{'}}\left(t\right)\cos t\right\}}^2 \\ ={\sin }^{2}t{\left\{f\left(t\right)+f^{\text{'}\text{'}}\left(t\right)\right\}}^2+{\cos }^{2}t{\left\{f\left(t\right)+f^{\text{'}\text{'}}\left(t\right)\right\}}^2 \\ ={\left\{f\left(t\right)+f^{\text{'}\text{'}}\left(t\right)\right\}}^2\left({\sin }^{2}t+{\cos }^{2}t\right) \\ ={\left\{f\left(t\right)+f^{\text{'}\text{'}}\left(t\right)\right\}}^2 \end{gathered}$$