Question-17
Find the angle between the tangents to the curve \(\displaystyle y=\frac{\pi \sin x}{x}\) at \(\displaystyle x=\pi\) and \(\displaystyle x=-\pi\).
The derivative at a point is the slope of the tangent at that point.
Let \(\displaystyle f( x) =\frac{\pi \sin x}{x}\). The slopes of the tangents to the curve at the two given points are \(\displaystyle f^{\prime }( \pi )\) and \(\displaystyle f^{\prime }( -\pi )\).
\[ \begin{equation*} \begin{aligned} f^{\prime }( x) & =\pi \left[\frac{x\cos x-\sin x}{x^{2}}\right] \end{aligned} \Longrightarrow \begin{aligned} f^{\prime }( \pi ) & =-1\\ f^{\prime }( -\pi ) & =1 \end{aligned} \end{equation*} \]
Since the slopes satisfy \(\displaystyle f^{\prime }( \pi ) \cdot f^{\prime}( -\pi ) =-1\), the two tangents are perpendicular to each other. Hence the angle between them is \(\displaystyle \boxed{90^{\circ }}\).