हिंदी

If X = F(T) Cos T − F' (T) Sin T and Y = F(T) Sin T + F'(T) Cos T, Then ( D X D T ) 2 + ( D Y D T ) 2 = (A) F(T) − F''(T) - Mathematics

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प्रश्न

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

MCQ
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उत्तर

(c){f(t) + f''(t)}2

Here,

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

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

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

\[\text { Thus }, \]

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

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

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

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

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

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अध्याय 12: Higher Order Derivatives - Exercise 12.3 [पृष्ठ २४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 12 Higher Order Derivatives
Exercise 12.3 | Q 21 | पृष्ठ २४

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