# Let r(t) = x(t), y(t), z(t) , t \in [a, b], be a vector-valued function, where a < b are real...

## Question:

Let {eq}r(t) = x(t), y(t), z(t) , t \in [a, b] {/eq}, be a vector-valued function, where {eq}a < b {/eq} are real numbers and the functions {eq}x(t), y(t), {/eq} and {eq}z(t) {/eq} are continuous. Explain why the graph of {eq}r {/eq} is contained in some sphere centered at the origin.

## Parametrically Defined Function

Given a function defined in parametric form, whose x, y and z components are all continuous functions we form an upper bound for the graph of the function.

## Answer and Explanation:

Since x, y and z are all continuous functions on the given interval {eq}[a, b] {/eq} therefore they are bounded on {eq}[a, b]. {/eq}

Therefore, {eq}\displaystyle M = \max_{t \in [a,b]} \left\{ |x(t)|,|y(t)|,|z(t)| \right\}. {/eq}

Hence,

{eq}|x(t)|, \; |y(t)|, \; |z(t)| \leq M \; \forall \; t \in [a,b] \qquad (1) {/eq}

Therefore, all points the graph of the function will lie inside a sphere of radius M centered at (0, 0, 0).

#### Learn more about this topic:

Evaluating Parametric Equations: Process & Examples

from Precalculus: High School

Chapter 24 / Lesson 3
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