Speed of a parametric function?

Question

I know speed = |velocity|

Why is speed of parametric defined as $$speed = \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$$

How is this derived? What is the principle here? Is this Pythag? Thinking of this in terms of vectors?

Answer 1

That's the resultant of $(2D)$ speeds in $i,j$ directions and basically its Pythagoras theorem for small parts of velocity in given directions

Answer 2

When you have a parametric equation in $2D$ it is usually defined as $\alpha(t)=(\alpha_{1}(t);\alpha_{2}(t))$.

The velocity vector at time $t_{0}$ would then be the derivative of $\alpha$ at $t_{0}$, which is, $$ \alpha'(t_{0})= \lim_{t\to t_{0}}\dfrac{\alpha(t)-\alpha(t_{0})}{t-t_{0}} $$ Assuming that $\alpha_{1}(t)$ and $\alpha_{2}(t)$ are both derivable, you get $$\alpha'(t_{0})= \lim_{t\to t_{0}} \bigg( \dfrac{\alpha_{1}(t)-\alpha_{1}(t_{0})}{t-t_{0}};\dfrac{\alpha_{2}(t)-\alpha_{2}(t_{0})}{t-t_{0}}\bigg)$$ Using the fact that $\lim_{x\to x_{0}}\big(a(x);b(x)\big)=\big(\lim_{x\to x_{0}}a(x);\lim_{x\to x_{0}}b(x)\big)$, you have $$\alpha'(t_{0})= \big( \alpha_{1}'(t_{0});\alpha_{2}'(t_{0})\big)$$

What you call speed at time $t_{0}$ is in fact the norm of the velocity vector, $\alpha'(t_{0})$. And so, $$\operatorname{speed}(t_{0})= ||\alpha'(t_{0})||=\sqrt{(\alpha_{1}'(t_{0}))^{2}+(\alpha_{2}'(t_{0}))^{2}}$$