A find the unit vectors that are parallel to the tangent...

(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) $.

Transcript

00:01 If we need to find the unit vectors parallel to the tangent line to this curve, we first need to find the slope of this tangent line.

00:07 To do so, we take the derivative of our function, so d by d x of 2 sine x.

00:13 That is going to be equal to 2 cosine x.

00:21 Now, we need to find the slope as the point of x equals pi over 6, and we remember from our unit circle that cosine of pi over 6 is squared of 3 over 2, and multiplying that by 2 gives square root of 3.

00:35 So we know that the slope of our unit vector has to be squared of 3.

00:40 Now it's very easy to find a vector, which will have this slope.

00:44 For example, we can just take 1 comma, square root of 3.

00:47 And the reason this works is because square root of 3 is going to be our rise, 1 is going to be our 1, slope is rise over run, and square root of 3 over 1 is obviously square root of 3.

00:58 Now, the issue with this is that this isn't a unit vector.

01:01 It doesn't have a magnitude of 1.

01:04 In fact, we can calculate the magnitude of this just using the pythagorean theorem.

01:11 So we have square root of 1 squared is just 1 plus square root of 3 squared, which is just 3.

01:19 And this is going to be equal to 2.

01:21 So our vector is twice as long.

01:24 It has twice the magnitude of a unit vector.

01:27 So in order to get a unit vector from this, we can just divide both dimensions by 2.

01:31 And we will get 1�, comma, squared of 3 over 2.

01:40 Now, since we're going to need to graph this anyways, we might as well graph it right now.

01:45 And we can see on this graph that i've pasted, we have x equals one half over here, and then the square of three over two is going to be slightly less than one.

01:56 And this point above i've labeled in blue, this is going to, this is pi over six, one.

02:02 That's going to be useful later.

02:04 I've taken the literacy of labeling it now.

02:08 Okay.

02:09 So this is one vector.

02:11 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.

02:18 And now the more precise way of saying this is we multiply both dimensions by negative 1.

02:24 And you can take a moment to convince yourself that this won't change the magnitude or the slope.

02:29 Pretty useful exercise.

02:31 So we're going to have x equals negative 1⁄2 and y equals negative square root of 3 over 2.

02:43 For the endpoint of this vector.

02:45 So x equals negative 1 1 .5, y equals negative root 3 overuse, slightly less than 1 around 0 .86, i believe.

02:52 This is going to be our other vector.

02:56 Cool.

02:57 Now the next thing that we need to do is we need to find the vectors which are parallel to the perpendicular to the tangents of our curve.

03:07 Now recall that the perpendicular to a tangent, if the tangent has a slope, m, the perpendicular is going to have the negative inverse of the slope as its slope.

03:18 So negative 1 over m.

03:21 Since we know the slope of our tangent line, we already found it.

03:25 It's just square of 3.

03:26 We know that the slope of our perpendicular line is going to be negative 1 over square of 3.

03:33 So the vectors that we're looking for now have a slope of negative 1 over square of 3.

03:40 Once again, we can really easily find a vector which will have this slope, just have 1 comma negative 1 over square of 3...

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