Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

The curve with the vector equation \({\rm{r(t) = }}{{\rm{t}}^{\rm{3}}}{\rm{i + 2}}{{\rm{t}}^{\rm{3}}}{\rm{j + 3}}{{\rm{t}}^{\rm{3}}}{\rm{k}}\) is a line.

The given statement is true.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Reparametrize the curve,

The parameter always appears with its third power. If redistributing the curve in this way, they can get a better result

\({\rm{r(t) = }}{{\rm{t}}^{\rm{3}}}{\rm{i + 2}}{{\rm{t}}^{\rm{3}}}{\rm{j + 3}}{{\rm{t}}^{\rm{3}}}{\rm{k}}\)

Step 2: Put \({{\rm{t}}^{\rm{3}}}{\rm{ = s}}\) in the given equation,

Let, \({{\rm{t}}^{\rm{3}}}{\rm{ = s}}\)

\({\rm{r(s) = si + 2sj + 3sk}}\)

This can also write as

\({\rm{r(s) = s}}\langle {\rm{1,2,3}}\rangle \)

The curve is a line whose direction is specified by the vector, as we can see from this form \(\langle {\rm{1,2,3}}\rangle \)

Therefore, the statement is true.

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Essential Calculus: Early Transcendentals

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Short Answer

The curve with the vector equation \({\rm{r(t) = }}{{\rm{t}}^{\rm{3}}}{\rm{i + 2}}{{\rm{t}}^{\rm{3}}}{\rm{j + 3}}{{\rm{t}}^{\rm{3}}}{\rm{k}}\) is a line.

Step by Step Solution

Step 1: Reparametrize the curve,

Step 2: Put \({{\rm{t}}^{\rm{3}}}{\rm{ = s}}\) in the given equation,

Most popular questions for Math Textbooks

Match the vector field F with the plots labelled I–IV. Give reasons for your choices.

\(F(x,y) = \left\langle {x, - y} \right\rangle \)

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Select your language

Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

The curve with the vector equation \({\rm{r(t) = }}{{\rm{t}}^{\rm{3}}}{\rm{i + 2}}{{\rm{t}}^{\rm{3}}}{\rm{j + 3}}{{\rm{t}}^{\rm{3}}}{\rm{k}}\) is a line.

Step by Step Solution

Step 1: Reparametrize the curve,

Step 2: Put \({{\rm{t}}^{\rm{3}}}{\rm{ = s}}\) in the given equation,

Most popular questions for Math Textbooks

Match the vector field F with the plots labelled I–IV. Give reasons for your choices.

\(F(x,y) = \left\langle {x, - y} \right\rangle \)

Want to see more solutions like these?

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