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

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

Explanation

Solution

Take two points on the curve A(h,f(a))A\left( h,f\left( a \right) \right) and B(a+h,f(a+h))B\left( a+h,f\left( a+h \right) \right) . Find the equation of the line with in two point form (y−y1=y2−y1x2−x1(x−x1))\left( y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right) \right). Take the limit h→0h\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.-1. Find the equation of normal as y=mx+cy=mx+c where mm is the slope and cc can be obtained by putting (a,f(a))\left( a,f\left( a \right) \right).$$$$

Complete step-by-step answer:
Let us have a curve defined by the function f(x)f\left( x \right) . We take two points x=ax=a and x=a+hx=a+h on x−x-axis for a very small positive real number hh. The value of the function at the point x=ax=a is f(a)f\left( a \right) and the value of the function at the point x=a+hx=a+h is f(a+h)f\left( a+h \right). We name the points A(h,f(a))A\left( h,f\left( a \right) \right) and B(a+h,f(a+h))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(x)f\left( x \right). Using two point formula , the equation of the line AB is given by
f(a+h)−f(a)=f(a+h)−f(h)(a+h)−h((a+h)−h)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 f(a+h)−f(h)(a+h)−h\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=ax=a when h→0h\to 0. The derivative is a defined ratio of infinitesimal change in function value and infinitesimal change in variable. That infinitesimal change here is hhapproaches to 0. We define derivative for any point x=ax=a in the domain set is defined as
dydx=lim⁡h→0,f(a+h)−f(a)h \dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }},\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$

Therefore when h→0h\to 0,f(a+h)−f(h)(a+h)−h\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 dydx∣x=a=f′(a){{\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.-1. We also know that the tangent line and normal line are always perpendicular to each other . Let the slope of the normal be mm. So

& m{{f}^{'}}\left( a \right)=-1 \\\ & \Rightarrow m=\dfrac{-1}{{{f}^{'}}\left( a \right)} \\\ \end{aligned}$$ The equation of a line in slope-intercept form is given as $y=mx+c$ where $c$is the intercept. We find $c$ putting the point $$\left( a,f\left( a \right) \right)$$ $$\begin{aligned} & 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{aligned}$$ So the equation of normal for normal to the curve $y=f\left( x \right)$ at the point $x=a$ is $y=\left( \dfrac{-1}{{{f}^{'}}\left( a \right)} \right)x+f\left( a \right){{f}^{'}}\left( a \right)+a$$$$$ **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 $\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$.