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Q. 11

Expert-verified
Found in: Page 859

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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

The graph of r is contained in a sphere ${x}^{2}+{y}^{2}+{z}^{2}=3{p}^{2}$ where $r\left(t\right)=\left(x\left(t\right),y\left(t\right),z\left(t\right)\right).$

See the step by step solution

Step 1. Given Information.

The given vector-valued function is $r\left(t\right)=\left(x\left(t\right),y\left(t\right),z\left(t\right)\right),t\in \left[a,b\right],$ where a < b are real numbers and the functions x(t), y(t), and z(t) are continuous.

Step 2. Explanation.

As it is given that x(t), y(t), and z(t) are continuous functions, then by the extreme value theorem x(t), y(t), and z(t) have minimum and maximum values.

Let ${P}_{1},{P}_{2}\mathrm{and}{P}_{3}$ be the maximum values of x(t), y(t), and z(t). Now, let P be the highest value of these three values. Then the graph of r(t) on [a.b] lies within the sphere with center (0,0,0) and radius $\sqrt{3}p.$

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