# Find the unit vectors that are parallel to the tangent

Find the unit vectors that are parallel to the tangent line to the curve $y=2\mathrm{sin}x$ at the point $\left(\frac{\pi }{6},1\right)$
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Stuart Rountree

Step 1
Finding the slope of the tangent line. Recall that the slope of the tangent line for a point in $f\left(x\right)$ is given by ${f}^{\prime }\left(x\right)$. Hence, at point $\left(\frac{\pi }{6},1\right)$ where $x=\frac{\pi }{6}$, the slope of the tangent line is:
$f\left(x\right)=2\mathrm{sin}x⇒{f}^{\prime }\left(x\right)=2\mathrm{cos}x⇒{f}^{\prime }\left(\frac{\pi }{6}\right)=2\mathrm{cos}\left(\frac{\pi }{6}\right)=2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}$
The slope of the tangent line at this point is $m=\sqrt{3}$
Step 2
Finding a unit vector parallel to the tangent line. Since slope is the ratio of rise per run or $\frac{y}{x}$, then we can set an arbitrary vector with slope 4 by letting $x=1$ and $y=\sqrt{3}$ or $v=\left[1,\sqrt{3}\right]$. This vector has "slope" $\sqrt{3}$ and is thus parallel to the tangent line. However, this is not yet a unit vector because its magnitude is obviously not equal to 11. Hence, we need to convert this into unit vector using the formula given in the book:
${u}_{v}=\frac{v}{|v|}$
$=\frac{\begin{array}{cc}1& \sqrt{3}\end{array}}{\sqrt{{1}^{2}+{\left(\sqrt{3}\right)}^{2}}}$
$=\frac{\begin{array}{cc}1& \sqrt{3}\end{array}}{\sqrt{4}}$
$\left[\frac{1}{2},\frac{\sqrt{3}}{2}\right]$
Step 3
Finding another unit vector. However, this is not the only answer! This points toward the first quadrant since both components are positive. Another possible answer is the vector antiparallel to this (parallel but pointing in the opposite direction, the third quadrant in this case). Hence, we negate both components to get another answer:
$-{u}_{v}=\left[-\frac{1}{2},-\frac{\sqrt{3}}{2}\right]$
$\left[\frac{1}{2},\frac{\sqrt{3}}{2}\right]$ and $\left[-\frac{1}{2},-\frac{\sqrt{3}}{2}\right]$

###### Not exactly what you’re looking for?
lalilulelo2k3eq
Step 1
To answer this question, we need to know the slope of the tangent at $\left(\frac{\pi }{6},1\right)$
Recall that: ${f}^{\prime }\left(a\right)$ is the slope of the tangent of $y=f\left(x\right)$ at $x=a$
We know that the derivative of $y=2\mathrm{sin}x$ is ${y}^{\prime }=2\mathrm{cos}x$
Therefore, the slope of the tangent at $\left(\frac{\pi }{6},1\right)$ is
$2\mathrm{cos}\left(\frac{\pi }{6}\right)=2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}$
Let us consider the vector $v={v}_{x}i+{v}_{y}j$
If this vector is parallel to a line with slope $\sqrt{3}$ , we can write
$\frac{{v}_{y}}{{v}_{x}}=±\sqrt{3}$
${v}_{y}=±\sqrt{3}\cdot {v}_{x}⇒\left(Equation1\right)$
Since this is a unit vector, we can also write the following equation
${v}_{\left\{x\right\}}^{2}+{v}_{\left\{y\right\}}^{2}=1$
Use ({Equation 1)(Equation 1) to replace the expression for ${v}_{y}$ in the above equation
${v}_{\left\{x\right\}}^{2}+3{v}_{\left\{x\right\}}^{2}=1$
$4{v}_{\left\{x\right\}}^{2}=1$
Divide both sides by 4
${v}_{\left\{x\right\}}^{2}=\frac{1}{4}$
Take the square root of both sides
${v}_{x}=±\frac{1}{2}$
Substitute this in (Equation 1)), to get
${v}_{y}=±\frac{\sqrt{3}}{2}$
The two unit vectors parallel to the tangent at
$\left(\frac{\pi }{6},1\right)\left(\frac{\pi }{6},1\right)are±<\frac{1}{2},\frac{\sqrt{3}}{2}>$
###### Not exactly what you’re looking for?
nick1337

Step

$\left\{y=2\mathrm{sin}\left(x\right)\right\}$

$\left\{\frac{dy}{dx}=slope=2\mathrm{cos}\left(x\right){\right\}}_{x=\frac{\pi }{6}}$

To find parrallel unit vectors to the line tangent ot point $\left(\frac{\pi }{6},1\right)$ all that is really needed is the slope. At the point in question, slope is $\sqrt{3}$

Step 2

$\left\{y=\sqrt{3}\cdot x\right\}⇒⇒\left(1,\sqrt{3}\right)$

Equation of a parrellel line. Point satisfying equation.

Step 3

$|\stackrel{―}{v}|=\sqrt{\left(\mathrm{\Delta }x{\right)}^{2}+\left(\mathrm{\Delta }y{\right)}^{2}}=2$

magnitude of the vector along our parrallel line, with end at the point $\left(1,\sqrt{3}\right)$

Step 4

$\frac{1}{2}\cdot <1,\sqrt{3}>$

normalizing components of the vector at our chosen point, to find the unit vector. The positve and negative of this vector are parrallel.

$\parallel \stackrel{―}{u}\parallel =\frac{1}{2}i+\frac{\sqrt{3}}{2}j,or-\frac{1}{2}i-\frac{\sqrt{3}}{2}$