Continuous Vector Functions

Like with continuity of single variable real-valued functions, the continuity of vector-valued functions is as follows:

Definition: Let ${\vec {r}}(t)$ be a vector-valued function. We say that ${\vec {r}}(t)=(x(t),y(t),z(t))$ is Continuous at $a$ if $\lim _{t\to a}{\vec {r}}(t)={\vec {r}}(a)$, that is $\lim _{t\to a}x(t)=x(a)$, $\lim _{t\to a}y(t)=y(a)$ and $\lim _{t\to a}z(t)=z(a)$. We say that ${\vec {r}}(t)$ is Continuous on The Interval $I$ if ${\vec {r}}(t)$ is continuous for all $t\in I$.

We should note that from the definition of ${\vec {r}}(t)$ being continuous at $a$, we must have that the component functions $x(t)$, $y(t)$, and $z(t)$ all be continuous at $a$.

For example, suppose we wanted to show that ${\vec {r}}(t)=(t,-2t^{2},\cos t)$ is continuous at $\pi$. We notice that $x(t)=t$, $y(t)=-2t^{2}$, and $z(t)=\cos t$ are all continuous real-valued functions, and so they're also all continuous at $\pi$. Therefore, we have that:

${\begin{aligned}\quad \quad \lim _{t\to \pi }{\vec {r}}(t)=\left(\lim _{t\to \pi }t,\lim _{t\to \pi }-2t^{2},\lim _{t\to \pi }\cos t\right)=(\pi ,-2\pi ^{2},-1)={\vec {r}}(\pi )\end{aligned}}$

Licensing

Content obtained and/or adapted from:

• Continuity of Vector-Valued Functions, mathonline.wikidot.com under a CC BY-SA license