Answer to Question #149789 in Calculus for Haider

Question

Prove or disprove that the following vector function is discontinuos
r(t)=(cost)i+(sint)j+[t] k

Answer 1

This is from "Multivariable Calculus, Concepts and Contexts" by Stewart.

He says "The parametric equations for this curve are: $x=cos(t), y=sin(t), z=t$

We know from trigonometry that $cos^2t+sin^2t=1$ , and you know from the parametric equation of the curve that .$x=cost,y=sint$. So for any point of the curve, it is true that $x^2+y^2=1$

Just notice that you dropped the equality $z=t$, meaning that you get more $(x,y,z)$

values that satisfy the equalities: you obtain a superset of the curve points. Indeed, the parametric equation corresponds to a linear entity (1D), while $x^2+y^2=1$ defines a surface (2D), which contains the curve.

Therefore the vector function is continuos