Vectorizing a tangent line to a curve

Question

I am watching the video series "Multivariable Calculus" by Prof. Herbert Gross. In Part III, Lecture 2 it is mentioned that if you have some function: $w = f(y)$, the tangent line to that curve at a point $P_0$ can be written in vector form using Cartesian Coordinates and more specifically one vector that represents the tangent line would be the following: $$\vec{v_1} = \vec{j} + \frac{dw}{dy} \vec{k}$$

where $\vec{j}$ is a unit vector in the y-direction and $\vec{k}$ is a unit vector in the w-direction (the derivative is evaluated at $P_0$).

I can't seem to grasp why this would be the case. How is this vector obtained step by step?

Answer 1

The slope of the tangent line at $P_0$ is $\dfrac{\operatorname dw}{\operatorname dy}$, as is the slope of $\vec v_1=(1, \dfrac{\operatorname dw}{\operatorname dy})$.

So $\vec v_1$ has the direction of the tangent line to the curve at $P_0$. And $P_0+\vec v_1\cdot t$ would be a parametrization of that tangent line.

Answer 2

What you have is a curve given by $w = f(y)$, that is, your curve is the graph of $f$ plotted in $yw$-plane. So, the problem is to determine tangent line of a graph of a single variable function at point $P_0 = (y_0,w_0)$. As I'm sure you know, equation of this line is given by $w-w_0 = f'(y_0)(y-y_0)$ and we can parametrize this line as $$(y,w) = (y, w_0 + f'(y_0)(y-y_0)) = (y_0+t, w_0 + tf'(y_0)) = (y_0,w_0)+t(1,f'(y_0)),$$ where I substituted $t = y - y_0$ as parameter. This tells us that the tangent line at $P_0 = (y_0,w_0)$ is given by $$P_0 + t(1,f'(y_0)),\, t\in \mathbb R.$$

We conclude that the wanted tangent vector is $(1,f'(y_0)) = \vec j + f'(y_0)\vec k = \vec j + \frac{dw}{dy}(y_0)\vec k$.