Question

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

 

विकल्प

•  f(t) − f''(t)

• {f(t) − f'' (t)}²

• {f(t) + f''(t)}²

• none of these

Answer 1

(c){f(t) + f''(t)}²

Here,

\[x = f\left( t \right)\cos t - f^{'} \left( t \right) \sin t \text { and y } = f\left( t \right) \sin t + f^{'} \left( t \right)\cos t\]

\[ \Rightarrow \frac{d x}{d t} = f^{'} \left( t \right)\cos t - f\left( t \right)\sin t - f^{''} \left( t \right)\sin t - f^{'} \left( t \right)\cos t \text { and } \frac{d y}{d t} = f^{'} \left( t \right) \sin t + f\left( t \right)\cos t + f^{''} \left( t \right)\cos t - f^{'} \left( t \right) \sin t\]

\[ \Rightarrow \frac{d x}{d t} = - f\left( t \right)\sin t - f^{''} \left( t \right)\sin t \text { and } \frac{d y}{d t} = f\left( t \right)\cos t + f^{''} \left( t \right)\cos t\]

\[\text { Thus }, \]

\[ \left( \frac{d x}{d t} \right)^2 + \left( \frac{d y}{d t} \right)^2 = \left\{ - f\left( t \right)\sin t - f^{''} \left( t \right)\sin t \right\}^2 + \left\{ f\left( t \right)\cos t + f^{''} \left( t \right)\cos t \right\}^2 \]

\[ = \left\{ f\left( t \right)\sin t + f^{''} \left( t \right)\sin t \right\}^2 + \left\{ f\left( t \right)\cos t + f^{''} \left( t \right)\cos t \right\}^2 \]

\[ = \sin^2 t \left\{ f\left( t \right) + f^{''} \left( t \right) \right\}^2 + \cos^2 t \left\{ f\left( t \right) + f^{''} \left( t \right) \right\}^2 \]

\[ = \left\{ f\left( t \right) + f^{''} \left( t \right) \right\}^2 \left( \sin^2 t + \cos^2 t \right)\]

\[ = \left\{ f\left( t \right) + f^{''} \left( t \right) \right\}^2\]