Normal straight line to a curve at a point

It is the straight line that, when crossing the curved line, it is perpendicular to the curve.

The following figure shows the normal straight line to the curve $\displaystyle y=\frac{1}{x-1}+1$:

Two functions $f(x),g(x)$ will be normal in a point if, in the crossing point $a$, it is satisfied that:$$f'(a)\cdot g'(a)=-1$$

The following table shows several values of slopes of perpendicular straight lines:

$f'(a)$	$g'(a)$
$1$	$-1$
$2$	$\displaystyle -\frac{1}{2}$
$-3$	$\displaystyle \frac{1}{3}$
$\displaystyle \frac{3}{8}$	$\displaystyle -\frac{8}{3}$

The general expression of the normal straight line to $f(x)$ at a point $a$ is:$$\displaystyle y-f(a)=-\frac{1}{f'(a)}\cdot (x-a)$$

Solve the figure showed previously, that is, find the normal straight line to $f(x)=\displaystyle \frac{1}{x-1}+1$ at the point $a=2$:

a) The slope of the curve at the crossing point is:$$\begin{array}{rcl} \displaystyle f'(x)& =& -\frac{1}{(x-1)^2} \\ f'(2)& = &-1\end{array}$$And the slope of the straight line is: $$\displaystyle m=-\frac{1}{f'(2)}=1$$

b) Using the above mentioned, the straight line will go through $$(a,f(a))=(2,2)$$

Finally, the equation of the normal straight line is:$$\begin{array}{rcl}y-2 & = & 1\cdot (x-2) \\ y & = & x \end{array}$$Consistent with what we can observe in the figure.

Find the tangent straight line to the function $y=\sqrt{x}$ at the point $x=0$, as well as its normal straight line.

a) We start by looking at the derivative of the function at $x=0$.

However we can see that it does not exist. We then have to compute the limit as $x$ goes to zero from the right: $$\displaystyle \begin{array}{l} y'(x)=\frac{1}{2\sqrt{x}} \\ \lim_{x \to 0} y'(x)=\lim_{x \to 0} \frac{1}{2\sqrt{x}}=\infty\end{array}$$

b) Since the formula $y=a\cdot x+b$ is not useful when we have an infinite slope, we have to realize that it is the axis that is normal to curve. That is if we take the straight line $x=0$ we obtain a straight line with infinite slope.

c) Finally, we should realize that the line perpendicular to the straight line defined by the $y$ axis is the $x$ axis. That is, by the line defined by $y=0$.