eozoischgc
2021-12-16
Answered

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

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Stuart Rountree

Answered 2021-12-17
Author has **29** answers

Step 1

Finding the slope of the tangent line. Recall that the slope of the tangent line for a point in

The slope of the tangent line at this point is

Step 2

Finding a unit vector parallel to the tangent line. Since slope is the ratio of rise per run or

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:

Answer

lalilulelo2k3eq

Answered 2021-12-18
Author has **38** answers

Step 1

To answer this question, we need to know the slope of the tangent at$(\frac{\pi}{6},1)$

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$(\frac{\pi}{6},1)$ 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

$\frac{{v}_{y}}{{v}_{x}}=\pm \sqrt{3}$

${v}_{y}=\pm \sqrt{3}\cdot {v}_{x}\Rightarrow \left(Equation1\right)$

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

$v}_{x}=\pm \frac{1}{2$

Substitute this in (Equation 1)), to get

$v}_{y}=\pm \frac{\sqrt{3}}{2$

Answer

The two unit vectors parallel to the tangent at

$(\frac{\pi}{6},1)(\frac{\pi}{6},1)are\pm <\frac{1}{2},\frac{\sqrt{3}}{2}>$

To answer this question, we need to know the slope of the tangent at

Recall that:

We know that the derivative of

Therefore, the slope of the tangent at

Let us consider the vector

If this vector is parallel to a line with slope

Since this is a unit vector, we can also write the following equation

Use ({Equation 1)(Equation 1) to replace the expression for

Divide both sides by 4

Take the square root of both sides

Substitute this in (Equation 1)), to get

Answer

The two unit vectors parallel to the tangent at

nick1337

Answered 2021-12-27
Author has **575** answers

Step

To find parrallel unit vectors to the line tangent ot point

Step 2

Equation of a parrellel line. Point satisfying equation.

Step 3

magnitude of the vector along our parrallel line, with end at the point

Step 4

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

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