2.4 Tangent and normal vectors

Consider a point $P=(x_0,y_0)$ on a curve $C\subset {ℝ}^2$. The tangent line to $C$ at $P$ is the straight line that “just touches” the curve at $P$. A tangent vector at $P$ is a vector in the direction of the tangent line.

One way to think about a tangent vector is via velocity vectors. Suppose a particle has position $(x(t),y(t))$ at time $t$. What does it mean to say that the particle has velocity $(x^{\prime }(t),y^{\prime }(t))$ at time $t$? To measure the velocity of an object one would measure the distance vector $\delta 𝐯$ covered in a period of time $\delta t$: then the average velocity over the period $\delta t$ is $\frac{\delta 𝐯}{\delta t}$. The idea of an ‘instantaneous’ velocity is to take smaller and smaller periods of time, i.e. to let $\delta t$ tend to zero: this is exactly the same thing as differentiating the position vector $(x(t),y(t))$.

Now suppose the particle moves on a curve $C$. The instantaneous velocity of the particle must point in the same direction as the curve. But what does this mean? The curve isn’t a straight line. However, it does look roughly like a straight line if we look in close enough: the best “linear approximation” to the curve at a point $P$ is the tangent line to $C$ at $P$. The same principle can be seen more easily with a surface: the earth is (approximately) a sphere, but at our scales of distance it looks like a flat plane. (More on surfaces later.)

Suppose $C$ is the graph of a function $f(x)$. Then we know how to determine the tangent line to $C$ at a point $(\alpha ,f(\alpha ))$. It has gradient $f^{\prime }(\alpha )$, so is of the form $y=f^{\prime }(\alpha )x+c$ for some $c\in ℝ$. But it also contains the point $(\alpha ,f(\alpha ))$, so the equation of the tangent line is

If $C$ is the image of a parametrized curve $\gamma :I\to {ℝ}^2$ then we can almost always determine the tangent line to a point $\gamma (t_0)$ by differentiating $\gamma$. Indeed, since ${\gamma }^{\prime }(t_0)$ is a tangent vector to the curve at $\gamma (t_0)$, then as long as ${\gamma }^{\prime }(t_0)\ne (0,0)$, the equation of the tangent line to $C$ at $\gamma (t_0)$ is $(x,y)=\gamma (t_0)+\lambda {\gamma }^{\prime }(t_0)$.

A normal vector to the curve $C$ at the point $P$ is a vector which is orthogonal to the tangent line. If $C\subset {ℝ}^2$ then the normal line at $P$ is the line which passes through $P$ in the direction of a normal vector.

If $(u,v)\ne (0,0)$ is a tangent vector, then $(v,-u)$ is a normal vector since $(u,v).(v,-u)=uv-uv=0$. Thus we can often easily determine the normal line.

Normal vectors also give us another way to write the equation of a line $L$ in ${ℝ}^2$. If $𝐧$ is a (non-zero) normal vector to the line and $(x_0,y_0)\in L$, then $(x,y)-(x_0,y_0)$ is a vector in the direction of the line, hence $(x-x_0,y-y_0)\cdot 𝐧=0$, that is: