Question

What is the slope of the normal to the curve $y=f\left( x \right)$ at the point $x=a$ ?\[\]

Answer 1

Hint: Take two points on the curve $A\left( h,f\left( a \right) \right)$ and $B\left( a+h,f\left( a+h \right) \right)$ . Find the equation of the line with in two point form $\left( y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right) \right)$. Take the limit $h\to 0$ for the slope , you will get the slope of the tangent. Find the slope of the normal using the fact that product slopes of normal and tangent is $-1.$ Find the equation of normal as $y=mx+c$ where $m$ is the slope and $c$ can be obtained by putting $\left( a,f\left( a \right) \right)$.\[\]

Complete step-by-step answer:
Let us have a curve defined by the function $f\left( x \right)$ . We take two points $x=a$ and $x=a+h$ on $x-$axis for a very small positive real number $h$. The value of the function at the point $x=a$ is $f\left( a \right)$ and the value of the function at the point $x=a+h$ is $f\left( a+h \right)$. We name the points $A\left( h,f\left( a \right) \right)$ and $B\left( a+h,f\left( a+h \right) \right)$ and join them. The line AB is now a chord or secant to the curve of $f\left( x \right)$. Using two point formula , the equation of the line AB is given by
\[f\left( a+h \right)-f\left( a \right)=\dfrac{f\left( a+h \right)-f\left( h \right)}{\left( a+h \right)-h}\left( \left( a+h \right)-h \right)\]

We observe the slope in the above equation of line AB as $\dfrac{f\left( a+h \right)-f\left( h \right)}{\left( a+h \right)-h}$. The line AB which cuts the curve at two points will only touch the curve at $x=a$ when $h\to 0$. The derivative is a defined ratio of infinitesimal change in function value and infinitesimal change in variable. That infinitesimal change here is $h$approaches to 0. We define derivative for any point $x=a$ in the domain set is defined as
\[ \dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$\]

Therefore when $h\to 0$,$\dfrac{f\left( a+h \right)-f\left( h \right)}{\left( a+h \right)-h}$ will be the slope of the tangent will be the derivative at that point which is ${{\left. \dfrac{dy}{dx} \right|}_{x=a}}={{f}^{'}}\left( a \right)$
We know that the product of slopes of two lines perpendicular to each other is $-1.$ We also know that the tangent line and normal line are always perpendicular to each other . Let the slope of the normal be $m$. So
\[\begin{align}
  & m{{f}^{'}}\left( a \right)=-1 \\
 & \Rightarrow m=\dfrac{-1}{{{f}^{'}}\left( a \right)} \\
\end{align}\]
The equation of a line in slope-intercept form is given as $y=mx+c$ where $c$is the intercept. We find $c$ putting the point \[\left( a,f\left( a \right) \right)\]

\[\begin{align}
  & y=mx+c \\
 & \Rightarrow f\left( a \right)=\dfrac{-1}{{{f}^{'}}\left( a \right)}a+c \\
 & \Rightarrow c=f\left( a \right){{f}^{'}}\left( a \right)+a \\
\end{align}\]
So the equation of normal for normal to the curve $y=f\left( x \right)$ at the point $x=a$ is $y=\left( \dfrac{-1}{{{f}^{'}}\left( a \right)} \right)x+f\left( a \right){{f}^{'}}\left( a \right)+a$\[\]

Note: We need to be careful of the fact that the limit $h\to 0$ is only right hand limit and we need to check whether the function is differentiable by checking continuity and the existence of left hand derivative $\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$and right hand derivative $\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$.