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Numerade Educator



Problem 42 Hard Difficulty

(a) Find the unit vectors that are parallel to the tangent line to the curve $ y = 2 \sin x $ at the point $ (\frac{\pi}{6}, 1) $.
(b) Find the unit vectors that are perpendicular to the tangent line.
(c) Sketch the curve $ y = 2 \sin x $ and the vectors in parts (a) and (b), all starting at $ (\frac{\pi}{6}, 1) $.


a) $$
\overline{u}_{ \|}=\frac{1}{2} i+\frac{\sqrt{3}}{2} j, \text { or }-\frac{1}{2} i-\frac{\sqrt{3}}{2} j
b) $$
\left\langle-\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle \text { or }\left\langle\frac{\sqrt{3}}{2},-\frac{1}{2}\right\rangle
c) The blue curve is the graph of $y=2 \sin x .$ The purple vectors are the unit tangent vectors: $\langle 1 / 2, \sqrt{3} / 2\rangle$ and $\langle- 1 / 2,-\sqrt{3} / 2\rangle$ at the point $(\pi / 6,1) .$ (The light purple line is the tangle line at $(\pi / 6,1) . )$ The red vectors are the unit vectors that are perpendicular to the tangent line: $(-\sqrt{3} / 2,1 / 2)$ and $\langle\sqrt{3} / 2,-1 / 2\rangle$
at $(\pi / 6,1) .$ (The light red line is the normal line at $(\pi / 6,1) . )$


You must be signed in to discuss.

Alex G.

September 22, 2020

Can someone explain what the unit vectors is?


Nadia H.

September 22, 2020

Hey Alex, In mathematics, a unit vector in a normed vector space is a vector of length 1. A unit vector is often denoted by a lowercase letter with a circumflex, or "hat", as in.


Howie C.

September 22, 2020

What is a tangent line?


Samantha T.

September 22, 2020

I know this one! In geometry, the tangent line to a plane curve at a given point is the straight line that "just touches" the curve at that point. Leibniz defined it as the line through a pair of infinitely close points on the curve.


Cam R.

September 22, 2020

Anyone else confused by the perpendicular, can someone explain?


Lindsey P.

September 22, 2020

I see how that could be confusing. In elementary geometry, the property of being perpendicular is the relationship between two lines which meet at a right angle. The property extends to other related geometric objects. A line is said to be perpendicular t

Video Transcript

if we need to find the unit vectors parallel to the tangent line to this curve, we first need to find the slope of the tangent line to do so we take the derivative of our function so G by dx of two. Simex, that is going to be equal to to cosign X. Now, if we need to find the slope as the point uh X equals pi over six. And we remember from our unit circle that co sign of pi over six is squared of 3/2. And we'll find that by two gives squared of three. So we know that the slope of our unit vector has to be squared of three. Now it's very easy to find a vector which will have this look for example, we can just take one comma squared of three. And the reason this works because squared of three is going to be our rise, one is going to be our run slope is rise over run and squared of 3/1 is obviously squared of three. Now, the issue with this is that this isn't a unit factory doesn't have a magnitude of one. Ah In fact, we can calculate the magnitude of this just using the pythagorean theorem. So we have square root of one square, it is just one plus Squared of three squared which is just three. This is going to be equal to two. So our vector is twice as long it has twice the magnitude of a unit factor. So in order to get a unit vector from this, we can just divide both dimensions by two and we will get one half comma it's greater 3/2. Cool. Now, since we're going to need to draft this anyways, we might as well record right now. Um and we can see on this graph here that I've pasted. Uh we have X equals one half over here and then 3 to 3/2 is going to be slightly less than one. And this point above, I've labelled in blue, this is going to this is uh pi over 61. That's going to be useful later. I've taken delivery of labeling it now. Yeah. So this is one vector and the other vector which is going to have the same magnitude and slope is just going to be this vector but pointing in the opposite direction. And now the more precise way of saying this is we multiply in both dimensions by -1 and you can take a moment to convince yourself that this won't change the magnitude or the slope. Uh Pretty useful exercise. So we're going to have X equals negative one half. And why it was negative squared of 3/2 for the end point of the specter. Alright, so X equals negative one half, Y equals negative. Right here was used slightly less than one or 2.86. I believe this is going to be your other vector. Cool. Now the next thing that we need to do is we need to find the vectors which are parallel to the perpendicular to the tangent of our curve. Now recall that the perpendicular to extensions. If the tangent has a slope mm particular is going to have the negative inverse of the slope As its slope. So negative one over M. Since we know the slope of our tangent line, we already found it. It's just squared of three. We have a slope of are perpendicular line is going to be negative one over squared of three. So the victims that we're looking for now Have a slope of negative one over squared of three. Once again, we can really easily find a vector which will have this slope. I just have one comma negative one over squared of three. That we once again find the time to do this Square root of one, screwed with screws one plus negative one over root three. That's just squared. That's going to be just one third. Which is to say squirt of 4/3 or two over route three. And we don't have to regulate this right now. Since this isn't our final answer. Okay, so We scale this victory down by two thirds now, just as we previously scaled by two And we take one over sorry, about 2/3 to over rule three. So we still down by two over route three. Just as previously, we still down by two and our first sector is going to be Great. Yeah comma then if you divide negative one over root three by two of the route three, if you cancel things up correctly you're going to get negative one half. And then once again our second vector is going to be Distracted by what multiplied by -1 on both dimensions. Or negative route 3/2 comma one half, Cool. Now we can also draw this, we're going to do it in a separate color. So as we established, Route 3/2 is just slightly less than one and our second dimension is negative one half. So this is going to be roughly here and our second vector is that But in the other direction I'm going to slightly go off my graph here, it's okay. All right now we can we? Re found these vectors and we are told in the problem I believe to uh sketch them starting at this point because of labels um and that is just the same thing as translating the all the all of these factors up to this point. Um Right, so what we do is we schedule that going to look roughly like this and we can he translated this picture just now. And the second back here, this one going to look roughly like this that we translate these factors uh huh And if you are going to be required by the problem to label the endpoints of these vectors, what you can do is we already found the end points of the vectors before we translated them and then translating them was just the same as adding Pi over 6 to the x value and one to the white value since we translated them over two, pi over six come up one. And now one last thing that we need to do is we need to sketch the curve itself. The curve itself is to sin X, which is just sign of X but skilled vertically by two. So we start at 00 and we know that at pi we're going to have Okay pie, we're going to have returns to zero so we can label that point as well. And the next time we will return to zero at to buy which is roughly 6.28. So over here and then the peak is going to be a pi over two so roughly here and normally sign of expects that one. So the two side effects is going to be at two. And then over here we're going to have a trough at the negative to also slightly I figure after this Okay, then after that we just try that. My drawing is rather sloppy but that's not too bad. It's close enough and that should be all the audio problems asking us to do. Okay.