Where do parametric equation for normal line of parametric equation come from?

Question

In my calculus II course, I am currently studying parametric curves. And they are awesome! At some point, the book states the following:

if we set parameters x = f(t), and y = g(t),
the parametric equations for the tangent line at time $t_{0}$ are

x = f ($t_{0}$) + f ' ($t_{0}$)(t - $t_{0}$)
y = g ($t_{0}$) + g ' ($t_{0}$)(t - $t_{0}$)

I understand that these are just the standard line equations, in which f($t_{0}$) is the x coordinate at that time, and g ($t_{0}$) the y coordinate. The derivatives are the slope of the functions at that point.

The next part is what has been puzzling me for a time now

The NORMAL lines at those points are then

x = f ($t_{0}$) + g ' ($t_{0}$)(t - $t_{0}$)
y = g ($t_{0}$) - f ' ($t_{0}$)(t - $t_{0}$)

I would have expected the slope to change from g ' ($t_{0}$) to $\frac{-1}{g ' (t_{0})}$

Am I missing something obvious? Thank you very much for your time!

QUESTION ANSWERED: Thank you very much to all of you, I understand now.

Answer 1

In vector terms, the tangent line has equation

$\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{bmatrix} f(t_0) \\ g(t_0) \end{bmatrix} + (t-t_0) \begin{bmatrix} f'(t_0) \\ g'(t_0) \end{bmatrix} = \overrightarrow{p_0} + (t-t_0)\overrightarrow v$

where $\overrightarrow{p_0}$ is the point of tangency and $\overrightarrow v$ is the direction vector of the tangent line.

Note that the vector $\overrightarrow{v_{\perp}} =\begin{bmatrix} g'(t_0) \\ -f'(t_0) \end{bmatrix}$ is perpendicular to the vector $\overrightarrow v = \begin{bmatrix} f'(t_0) \\ g'(t_0) \end{bmatrix}; $ that is to say, their dot product is equal to $0$.

Hence the equation to the line normal to $\begin{bmatrix} x(t) \\ y(t) \end{bmatrix}$ at the point $\overrightarrow{p_0} = \begin{bmatrix} f'(t_0) \\ g'(t_0) \end{bmatrix}$ is

$$\begin{bmatrix} x_\perp(t) \\ y_\perp(t) \end{bmatrix} = \begin{bmatrix} f(t_0) \\ g(t_0) \end{bmatrix} + (t-t_0) \begin{bmatrix} g'(t_0) \\ -f'(t_0) \end{bmatrix} = \overrightarrow{p_0} + (t-t_0)\overrightarrow{v_\perp}$$

Answer 2

Note that the slope is given by $$\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$$ This should fit your intuition for slope--it's change in $y$ divided by change in $x$. Therefore, in the first formula, the slope is $$\frac{g'(t_0)}{f'(t_0)}$$ and in the second one, the slope is $$-\frac{f'(t_0)}{g'(t_0)}$$ which is consistent with your usual formula for the slope of the perpendicular line.

Answer 3

$\frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac{dx}{dt}}$

With our if we apply this to our tangent lines.

$\frac {dy}{dx} = \frac {g'(t)}{f'(t)}$

and If we apply it to the equations of the normal lines, we get.

$\frac {dy}{dx} = -\frac {f'(t)}{g'(t)}$

which is indeed the negative reciprocal of the our equations for the tangent lines.

However, if you wanted to, you could say:

$x(t) = y(t_0) + \frac 1{f'(t_0)} (t-t_0)\\ y(t) = y(t_0) - \frac 1{g'(t_0)} (t-t_0)$