Question

Find the slopes of the tangent and the normal to the following curves at the indicated points.

$x=a(1-\cos \theta )$ and $y=a(\theta +\sin \theta )$ at $\theta =\pi /2$.

Answer 1

We have,

$x=a(1-\cos \theta )$ and $y=a(\theta +\sin \theta )$

Now, differentiate w.r.t $\theta$, we get

$\frac{dx}{d\theta }=a(0-(-\sin \theta \left)\right)$ and $\frac{dy}{d\theta }=a(1+\cos \theta )$

$\frac{dx}{d\theta }=a\sin \theta$ and $\frac{dy}{d\theta }=a(1+\cos \theta )$

Therefore,

$\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a(1+\cos \theta )}{a\sin \theta }$

$\frac{dy}{dx}=m=\frac{(1+\cos \theta )}{\sin \theta }$

Put $\theta =\frac{\pi }{2}$

So,

$\frac{dy}{dx}=m=\frac{(1+\cos \frac{\pi }{2})}{\sin \frac{\pi }{2}}$

$\frac{dy}{dx}=m=\frac{1+0}{1}$

$\frac{dy}{dx}=m=1$

Therefore,

The slope of the tangent of the curve

$=m=1$

And the slope of the normal the curve

$=-\frac{1}{m}$

$=-\frac{1}{1}=-1$

Hence, this is the answer.