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Essential Calculus: Early Transcendentals
Found in: Page 550
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

(a) To find the parallel unit vectors to the tangent line of \(y = 2\sin x\).

(b) To find the perpendicular unit vectors to the tangent line of \(y = 2\sin x\).

(c) To sketch curve of \(y = 2\sin x\) along with vectors \( \pm \frac{1}{2}({\bf{i}} + \sqrt 3 {\bf{j}})\) and \( \pm \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\).

  1. The unit vectors which are parallel to the tangents of \(y = 2\sin x\) at point \(\left( {\frac{\pi }{6},1} \right)\) are \( \pm \frac{1}{2}({\bf{i}} + \sqrt 3 {\bf{j}})\).
  2. The unit vectors which are perpendicular to the tangents of \(y = 2\sin x\) are \( \pm \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\).
  3. The curve \(y = 2\sin x\) along with vectors \( \pm \frac{1}{2}({\bf{i}} + \sqrt 3 {\bf{j}})\) and \( \pm \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\) is sketched.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Concept of unit vector

Unit vectors are the base units of a vector, and they can be used to represent any other vector. Because unit vectors have a magnitude of 1, they solely convey information about a vector's direction. This is why unit vectors are often referred to as 'Direction Vectors.'

Step 2: Calculate parallel unit vector.

(a)

Given:

Equation of curve is \(y = 2\sin x\) and point is \(\left( {\frac{\pi }{6},1} \right)\).

Formula used:

Write the expression for slope of function \(f(x)(m)\).

\(\begin{aligned}{l}m &= {f^\prime }(x)\\m &= \frac{d}{{dx}}(f(x)).......(1)\end{aligned}\)

Here,

\({f^\prime }(x)\) is first derivative of function \(f(x).\)

Write the expression for magnitude of vector \({\bf{a}} = {a_1}{\bf{i}} + {a_2}{\bf{j}}(|{\bf{a}}|)\).

\(|{\bf{a}}| = \sqrt {a_1^2 + a_2^2} .......(2)\)

Write the expression for unit vectors of vector \({\bf{a}}({\bf{u}})\).

\({\bf{u}} = \pm \frac{{\bf{a}}}{{|{\bf{a}}|}}.......{\rm{(3) }}\)

Find the slope of equation \(y = 2\sin x\) by using equation (1).

\(m = \frac{d}{{dx}}(y)\)

Substitute \(2\sin x\) for \(y\),

\(\begin{aligned}{l}m &= \frac{d}{{dx}}(2\sin x)\\ &= 2\cos x\quad \end{aligned}\)

Substitute \(\frac{\pi }{6}\) for \(x\).

\(\begin{aligned}{l}m &= 2\cos \left( {\frac{\pi }{6}} \right)\\ &= 2\left( {\frac{{\sqrt 3 }}{2}} \right)\\ &= \sqrt 3 \end{aligned}\)

The tangent is a line, which touches a curve once a time. Here, we can consider any tangent line vector with the slope of \(\sqrt 3 \). For sample case, consider the vector of line with slope of \(\sqrt 3 \) as \({\bf{i}} + \sqrt 3 {\bf{j}}\).

Find the magnitude of vector \({\bf{i}} + \sqrt 3 {\bf{j}}\) by using equation (2).

\(\begin{aligned}{l}|{\bf{a}}| &= \sqrt {{1^2} + {{(\sqrt 3 )}^2}} \\ &= \sqrt {1 + 3} \\ &= \sqrt 4 &= 2\end{aligned}\)

Substitute \({\bf{i}} + \sqrt 3 {\bf{j}}\) for \({\bf{a}}\) and 2 for \(|{\bf{a}}|\) in equation (3),

\(\begin{aligned}{l}{\bf{u}} &= \pm \frac{{{\bf{i}} + \sqrt 3 {\bf{j}}}}{2}\\ &= \pm \frac{1}{2}({\bf{i}} + \sqrt 3 {\bf{j}})\end{aligned}\)

Thus, the unit vectors which are parallel to the tangents of \(y = 2\sin x\) at point \(\left( {\frac{\pi }{6},1} \right)\) are \( \pm \frac{1}{2}({\bf{i}} + \sqrt 3 {\bf{j}})\).

Step 3: Calculate perpendicular unit vector

(b)

Given:

Equation of curve is \(y = 2\sin x\) and point is \(\left( {\frac{\pi }{6},1} \right)\).

Formula:

Write the expression for slope of line that is perpendicular to tangent line \(\left( {{m_1}} \right)\).

\({m_1} = - \frac{1}{m}........{\rm{(4) }}\)

Substitute \(\sqrt 3 \) for \(m\) in equation (4),

\({m_1} = - \frac{1}{{\sqrt 3 }}\)

The perpendicular line to tangent of curve must possess the slope of \( - \frac{1}{{\sqrt 3 }}\). Consider the vector of line with slope of \( - \frac{1}{{\sqrt 3 }}\) as \(\sqrt 3 {\bf{i}} - {\bf{j}}\).

Find the magnitude of vector \(\sqrt 3 {\bf{i}} - {\bf{j}}\) by using equation (2).

\(\begin{aligned}{l}|{\bf{a}}| &= \sqrt {{{(\sqrt 3 )}^2} + {{( - 1)}^2}} \\ &= \sqrt {3 + 1} \\ &= \sqrt 4 &= 2\end{aligned}\)

Substitute \(\sqrt 3 {\bf{i}} - {\bf{j}}\) for \({\bf{a}}\) and \(2\) for \(|{\bf{a}}|\) in equation (3),

\(\begin{aligned}{l}{\bf{u}} &= \pm \frac{{\sqrt 3 {\bf{i}} - {\bf{j}}}}{2}\\ &= \pm \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\end{aligned}\)

Thus, the unit vectors which are perpendicular to the tangents of \(y = 2\sin x\) are \( \pm \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\)

Step 4: Sketch graph of curve.

Given:

Equation of curve is \(y = 2\sin x\) and point is \(\left( {\frac{\pi }{6},1} \right)\).

Draw the curve \(y = 2\sin x\) in \(x{\rm{ }}y\)plane. Then plot a point \(\left( {\frac{\pi }{6},1} \right)\) on curve.

Draw the unit vectors that is parallel to tangent of curve \( \pm \frac{1}{2}({\bf{i}} + \sqrt 3 {\bf{j}})\) using points \(\left\langle {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right\rangle \) and \(\left\langle { - \frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right\rangle \). Draw the unit vectors that is parallel to tangent of curve \( \pm \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\) using points \(\left\langle { - \frac{{\sqrt 3 }}{2},\frac{1}{2}} \right\rangle \) and \(\left\langle {\frac{{\sqrt 3 }}{2}, - \frac{1}{2}} \right\rangle \).

The sketch of curve \(y = 2\sin x\) along with vectors \( \pm \frac{1}{2}({\bf{i}} + \sqrt 3 {\bf{j}})\) and \( \pm \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\) is drawn as shown in Figure

Thus, the curve \(y = 2\sin x\) along with vectors \( \pm \frac{1}{2}({\bf{i}} + \sqrt 3 {\bf{j}})\) and \( \pm \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\) is sketched.

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