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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
If $x = {e^t}$ $\sin t,y = {e^t}\cos t,\,\,t$ is a parameter, then $\frac{{{d^2}y}}{{d{x^2}}}$ at $(1, 1)$ is equal to
  • ✓
    $ - \frac{1}{2}$
  • B
    $ - \frac{1}{4}$
  • C
    $0$
  • D
    $\frac{1}{2}$
✓

Answer

Correct option: A.
$ - \frac{1}{2}$
a
(a) At point $(1, 1)$, $1 = {e^t}\sin t,\,1 = {e^t}\cos t$
$ \Rightarrow \tan t = 1 \Rightarrow t = \pi /4$.
Now, $\frac{{dy}}{{dt}} = {e^t}(\cos t - \sin t)$and $\frac{{dx}}{{dt}} = {e^t}(\sin t + \cos t)$
==> $\frac{{dy}}{{dx}} = \frac{{\cos t - \sin t}}{{\cos t + \sin t}}$
Now,  $\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dt}}\left( {\frac{{\cos t - \sin t}}{{\cos t + \sin t}}} \right)\,\frac{{dt}}{{dx}}$
$\left[ {\frac{{(\cos t + \sin t)\,( - \sin t - \cos t) - (\cos t - \sin t)( - \sin t + \cos t)}}{{{{(\cos t + \sin t)}^2}}}} \right]\,\frac{{dt}}{{dx}}$
= $\frac{{ - 2}}{{{{(\cos t + \sin t)}^2}}}\,.\,\frac{1}{{{e^t}(\sin t + \cos t)}}$
= $\frac{{ - 2}}{{({e^t}\cos t + {e^t}\sin t)}}\,.\,\frac{1}{{{{(\cos t + \sin t)}^2}}}$
=$\frac{{ - 2}}{{x + y}}\,.\,\frac{1}{{{{(\cos t + \sin t)}^2}}}$= $\frac{{ - 2}}{{1 + 1}}\,.\,\frac{1}{{{{\left( {\cos \frac{\pi }{4} + \sin \frac{\pi }{4}} \right)}^2}}} = \frac{{ - 1}}{2}$.

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