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Expert-verified Found in: Page 550 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # (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. See the step by step solution

## 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. ### Want to see more solutions like these? 