How is the normal vector to the parabola $r(t)=(t,t^2)$ calculated?

Question

I think the parametrisation of a parabola ought to be like so:

• $r(t)=(t,t^2)$

When I compute $r'(t)$ I get $r'(t)=(1,2t)$, and when I compute $r''(t)$ I get $r''(t)=(0,2)$. At each point $0i + 2j$ isn't orthogonal to the tangent everywhere. What is the correct normal vector calculation?

Answer 1

There is a fact that if $r(t)$ is a vector such that $\| r(t) \|$ is constant, then $r'(t)$ is orthogonal to $r(t)$. So, to fix your calculations you can first normalize $r'(t)$ to have constant magnitude, and then calculate the second derivative.

Edit: the start of the first step is to calculate $$\| r'(t) \| = \sqrt{1^2 + (2t)^2 } = \sqrt{1 + 4t^2}, $$ so our unit tangent vector is $$T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}} \right\rangle.$$

Answer 2

Rotating a vector $(a,b)$ by 90 degrees counterclockwise yields $(-b,a)$. The unit tangent vector to $r(t)=(t,t^2)$ is $$T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{(1,2t)}{\sqrt{1+4t^2}}.$$Rotate to obtain $$N(t)=\frac{(-2t,1)}{\sqrt{1+4t^2}}.$$

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Bonus: plane curves can be studied with complex numbers. Then, rotation by 90 degrees counterclockwise amounts to multiplying by $\rm i$. So if $r(t)=t+t^2{\rm i}$, divide $r'(t)$ and ${\rm i}r'(t)$ by $|r'(t)|$ to reobtain $T(t)$ and $N(t)$.