Question

The slope of the normal to the curve $x = a\left( {\theta - \sin \theta } \right)$, $y = a\left( {1 - \cos \theta } \right)$ at point $\theta = \dfrac{\pi }{2}$ is
A. 0.
B. 1.
C. -1.
D. $\dfrac{1}{{\sqrt 2 }}$.

Answer 1

To solve this question, we will use the concept of tangent and normal of application of derivatives. The slope of normal is the normal to a curve at $P = \left( {{x_1},{y_1}} \right)$ is a line perpendicular to the tangent at P and passing through P.
$\Rightarrow$ slope of the normal at P = $- \dfrac{1}{{{\text{slope of the tangent at P}}}} = - \dfrac{1}{{{{\left( {\dfrac{{dy}}{{dx}}} \right)}_P}}} = - {\left( {\dfrac{{dx}}{{dy}}} \right)_P}$.

Complete step-by-step answer:
Given that,
$x = a\left( {\theta - \sin \theta } \right)$ and $y = a\left( {1 - \cos \theta } \right)$ and $\theta = \dfrac{\pi }{2}$.
We have to find the slope of the normal of the given curve at the given point.
We know that,
$\Rightarrow$ slope of the normal at P = $- \dfrac{1}{{{\text{slope of the tangent at P}}}}$
So, first we have to find the slope of the tangent at the given point.
And we also know that,
Slope of tangent to the curve $y = f\left( x \right)$ at the point $\left( {{x_0},{y_0}} \right)$ is given by:
Slope of tangent = ${\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {{x_0},{y_0}} \right)}}\left( { = f'\left( x \right)} \right)$.
Here, we have x and y in terms of $\theta$.
Therefore, we will apply the chain rule.
Chain rule states that, let f be a real valued function which is a composite of two functions u and v, i.e. f = v o u.
Suppose $t = u\left( x \right)$ and if both $\dfrac{{dt}}{{dx}}$ and $\dfrac{{dv}}{{dt}}$ exist, we have
$\Rightarrow \dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}}.\dfrac{{dt}}{{dx}}$
So,
$x = a\left( {\theta - \sin \theta } \right)$
Differentiating both sides with respect to $\theta$, we will get
$\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left( {1 - \cos \theta } \right)$.
Similarly,
$y = a\left( {1 - \cos \theta } \right)$
Differentiating both sides with respect to $\theta$, we will get
$\Rightarrow \dfrac{{dy}}{{d\theta }} = a\sin \theta$.
Now,
Slope of tangent to the curve = $\dfrac{{dy}}{{dx}}$.
Applying the chain rule,
$\Rightarrow$ Slope of tangent to the curve = $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{d\theta }}.\dfrac{{d\theta }}{{dx}} = \dfrac{{dy}}{{d\theta }}.\dfrac{1}{{\left( {\dfrac{{dx}}{{d\theta }}} \right)}}$.
Putting the values, we will get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{a\sin \theta }}{{a\left( {1 - \cos \theta } \right)}}$.
Now, the slope of tangent at point $\theta = \dfrac{\pi }{2}$,
$\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\theta = \dfrac{\pi }{2}}} = \dfrac{{a\sin \dfrac{\pi }{2}}}{{a\left( {1 - \cos \dfrac{\pi }{2}} \right)}}$
$\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\theta = \dfrac{\pi }{2}}} = \dfrac{a}{a} \\\ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\theta = \dfrac{\pi }{2}}} = 1 \\\$
Therefore,
Slope of normal to the curve at point $\theta = \dfrac{\pi }{2}$,
$\Rightarrow$ slope of the normal = $- \dfrac{1}{{{\text{slope of the tangent }}}} = - 1$
Hence, the slope of normal of the given curve at the given point is 1.

So, the correct answer is “Option C”.

Note: Whenever we ask this type of question, we have to remember some basic points, tangents and normal and also, we have to remember the chain rule. First, we have to find the derivatives of both the given variables with respect to the given independent variable and then we will find the slope of the tangent of that curve at the given point. After that, by using the relationship of slope of tangent and slope of normal, we can easily find the slope of normal of that curve at the given point.