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If y = 1 - cos θ, x = 1 - sinθ then \(\frac{{dy}}{{dx}}\)at θ = \(\frac{\pi}{4}\)is
1. - 1
2. 1
3. 1/2
4. \(\frac{1}{\sqrt2}\)
5. None of these

1 Answer

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Correct Answer - Option 1 : - 1

CONCEPT:

  • \(\frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)
  • \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)


Differentiation of parametric functions:

If x = f(t), y = g(t), where t is a parameter, then \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{g'\left( t \right)}}{{f'\left( t \right)}}\)

CALCULATION:

Given: y = 1 - cosθ, x = 1 - sin θ

Here, we have to find \(\frac{{dy}}{{dx}}\) at θ = π/4

As we can see that, x and y are functions of θ

We also know that, if x = f(t), y = g(t), where t is a parameter, then \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{g'\left( t \right)}}{{f'\left( t \right)}}\)

Let's differentiate x and y w.r.t θ

⇒ \(\frac{{dy}}{{d\theta }} = \frac{{d\left( {1 - \cos \theta } \right)}}{{d\theta }} = \sin \theta \)       ---(1)

⇒ \(\frac{{dx}}{{d\theta }} = \frac{{d\left( {1 - \sin \theta } \right)}}{{d\theta }} = \; - \cos \theta \)       ---(2)

Now from (1) and (2) we can say that, \(\frac{{dy}}{{dx}} = \; - \tan \theta \)

⇒ \({\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {\theta = \frac{\pi }{4}} \right)}} = \; - \tan \frac{\pi }{4} = \; - 1\)

Hence, option A is the correct answer.

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