Trigonometry Vectors

We are learning about vectors and we are trying some more examples.0000

We are trying to find the components of a vector of magnitude 8 that points 30 degrees West of South.0003

First thing to do is to draw a picture here, let me put my North, East, South, West (Never Eat Shredded Wheat).0009

I want a vector that points 30 degrees West of South and that would be down there.0023

That is 30 degrees but I like to think of these things in terms of a reference angle from a positive x axis.0029

That would be a 60 degree angle there, the reference from the positive x axis would be a 180 + 60 that would 240 degrees.0037

Now I want to find the components of that vector, I know its magnitude is 8 so r=8.0050

Remember we got the components, these useful formulas x=arcos(theta), y=arcsin(theta).0058

We are going to use those, x=r(8)cos(theta) the reference angle is 240 degrees.0069

That is 8(cos) is the x coordinate of that angle and the x coordinate is -1/2.0080

Negative because its on the left hand side where the x value is negative so that is -4.0089

The y component is (8)sin(240degrees), again common value, I memorized the answers for multiples of 30 degrees so I do not need to check the calculator for this.0097

This is (8)(-square root of 3/2) that is -4 square root of 3.0112

The components there are xy=-4 square root of 3.0122

That one was one of the simpler ones, we are given a vector, magnitude, direction.0136

The magnitude of course is the r, the direction essentially is the theta but we want to draw a picture and make sure what theta we are talking about.0140

It says that this direction is given in terms of West of South0150

That is why I drew my vector down here, 30 degrees to the West of South and then I worked that out to be 240 degrees from the x axis.0155

Then I dropped those into arcos(theta) and arcsin(theta).0165

Simplify those down using the common values and I get the components of the vector.0170

Finally we have another one, it is a ship this time, sailing at a current that flows at 5km/hr NE.0000

The ship’s captain steers the course 30 degrees East of South at 40km/hr relative to water.0009

I want to find out the true speed and true course of the ship.0017

This is again one or more we are going to have two velocities.0021

There is the ship’s velocity and the current’s velocity, I want to figure out each one separately.0025

Find the components and add them up and then convert that back into a vector.0030

We are given the current first, I’m going to draw a graph of that and find its components.0037

The current flows 5km NE so this is the current, I will do this in black.0046

NE so that makes that this is N, this is E, that is a 45 degree angle.0057

The components we can get using arcos(theta) and arcsin(theta).0064

The current is arcos(theta), (phi)cos(45), (phi)sin(45) which is a common value that I recognized.0072

Remember that 45 is pi/4 and its cos is square root/2, so it its sin.0092

Now, I’m going to add up the vectors and the numbers are not going to be very nice.0103

I’m going to have to throw that in to my calculator.0107

Let me find the decimal approximation for (phi)(square root of 2)/2 it is about 3.54.0111

That is the components of the vector for the component.0133

Now, I will do the ship in blue, the ship is going 30 degrees East of South.0137

There is South down there, the ship is going 30 degrees East of South so that is a 30 degree angle.0144

It is steering a course at 40 km/hr relative to the water so the magnitude is 40 km.0154

I want to find the components of that vector.0165

I want to get a reference angle from the positive x axis, I know that it is 270 degrees down to the -y axis and then 30 more degrees, 270 + 30 would be 300 degrees.0169

To find the components of the ship’s vector we are going to use arcos(theta) and arcsin(theta).0182

The ship is (40)cos(300 degrees) and (40)sin(300 degrees).0190

300 degrees common value, I recognize that one that is 5pi/3 and its cos is +1/2 because its x coordinate is positive, its sin is (–square root of 3)/2 because its y coordinate is negative.0204

That is a common value, I do not need to check it on a calculator.0228

That simplifies down to 20 and -20 square root of 3.0232

I am going to be adding that together with the component for the current.0237

I am going to go ahead and throw that in to my calculator.0243

Of course 20 is still 20, 20 square root of 3 is 34.64, those are the components of the vectors for the ship.0247

Now, what I want to do is add these two up to find the net components of the ship’s travel.0266

Let me move the current vector down, I am going to draw a copy of the current vector, same magnitude and same direction.0276

I will draw the net in red here, I add these two vectors together and I will get the net.0285

Let me remind you that the blue one is the ship and the red one is the net.0294

For the net, I will just going to add those two together.0302

I take those two and I add them together, I get 23.54 for the (x).0308

For the (y) I get -34.64, so I get -31.10, that is the x and y components of the net travel of the ship.0316

We had its travel relative to the water but the water is also moving so the net travel, you add up those two velocities and you will get the component of the ship’s motion.0343

We are asked to find the true speed, speed is the magnitude of the velocity.0353

Remember velocity is a vector and speed is a number, and the true direction.0361

The speed is the (r), the magnitude of the velocity so that is the square root of (x² + y²).0369

Where this is my (x), the (x) I am talking about is 23.54 and the y is -31.10.0378

I am going to work those out.0386

It works out to be just about 39.0 and it looks like our units here are km/hr.0400

That is the magnitude of that vector which is the true speed of the ship, that is half of our answer there.0413

Remember, we find the direction by finding (theta) is arctan(y/x) and then there is this question of whether you have to add 180.0425

I said you have to add 180 depending on which quadrant you are in.0439

You have to add 180 if you are in the second or third quadrant.0444

But we are in the fourth quadrant here, our (x) coordinate here so we are in the fourth quadrant, we do not have to add 180 here.0452

That is a small mercy there this is because we are in quadrant for or alternatively you can think of that as because the (x) is greater than 0, the (x) is positive there so you do not have to add 180.0464

We can just figure out the arctan(y/x), tan inverse, make sure you are in degree mode here.0484

The arctan(-31.1/23.54) I will work that out.0492

It works out to -52.9 degrees, of course I’m rounding there.0514

That really tells you the angle from the positive (x) axis, I do not really like giving a direction in terms of a negative angle.0525

If this is a real ship I would not want to say it is sailing in the negative angle.0532

Let me draw that out, that tells us that, that angle right there is 52.9 degrees.0537

A useful way to describe that would be as some angle East of South and so I can find the other complimentary angle by doing 90-52.9.0553

I worked that out and what it tells me is that its 37.1 degrees as reference in terms of East of South.0581

The direction that this ship is actually moving is 37.1 degrees East of South.0607

Just to remind you there, we calculated this angle right here that is where the 37.1 came from.0621

That was a pretty tricky problem, we had two different things going on here.0630

We had a ship sailing and a current, so we had a velocity vector for the current and then we had another velocity vector for the way the ship is steering relative to the water.0634

What we have to do there was to take the current’s vector 5km/hr NE and use arcos (theta) and arcsin(theta) to find the components of the current.0646

We took the ship’s velocity vector relative to the water and we use arcos(theta) arcsin(theta) here, the (r) is 40 because that is the magnitude, the speed of the ship relative to the water.0663

The 300 we got from the fact that it was 30 degrees East of South.0677

We graph that out and we found that it was 300 degrees from the positive (x) axis, that is where we get 300.0682

Plot that through arcos(theta) and arscin(theta) and we get a vector for the ship.0690

We add those two set of components together and we get the net components of the ship’s motion in terms of (x) and (y).0697

But we want to find the speed and direction, so we know the (x) and (y) components to find the speed that is the magnitude of the velocity.0708

We use square root (x² + y²), that came out to be 39.10, and it is km/hr, because we solved it all our distances were in terms of km.0716

To find the direction, we used arctan(y/x) and there is always this issue of do you have to add 180? We are in quadrant 4 so we do not have to add 180.0730

We add 180 when if we were in quadrant 2 or 3, we do not have to add 180.0740

Another way to check that is to check whether the (x) is positive or negative, but our (x) is positive so ok no adding of 180.0747

We find the arctan (y/x) and it comes out to be -52.9 degrees.0754

That means that your 52.9 degrees below the positive (x) axis.0761

But if you want to give that in terms of the major compass directions North, South, East, West, we looked down there and discovered that is 37.1 degrees East of the South direction.0766

That is how we finally gave our answer as 37.1 degrees East of South.0780

That is the end of the vector lecture on trigonometry as part of the trigonometry series on www.educator.com.0786

Thanks for watching.0792

Hi, these are the trigonometry lectures on educator.com, and today we're going to talk about vectors which is a big topic in most trigonometry classes, and they also come up in a lot of physics applications as well.0000

You really want to think of vectors as being a way of measuring physical concepts that have both magnitude and direction, and kind of the graphical way you think about vectors is as arrows.0012

Think of vectors as being arrows.0029

Every vector has both magnitude and direction.0033

Those are the two things that are important about a vector.0035

The magnitude is the length of the vector.0039

Magnitude, that's the length of the arrow.0044

The direction is if you put it in the coordinate system, there's an angle θ that determines the direction of the vector.0051

Every vector has both magnitude and direction.0061

Of course, if the magnitude is 0, then you have an arrow that just reduces down to a dot, that doesn't really have a direction.0063

Most vectors, except for the zero vector, have magnitude and direction.0071

You can draw these vectors anywhere you want, you can move them around as long as you don't change the length or the magnitude, and as long as you don't change the direction.0075

You can move them around but you aren't allowed to move their direction, and you aren't allowed to stretch them or shrink them.0087

That's the idea of vectors.0095

They're used primarily to represent lots of physical concepts about things like velocity.0097

Why is velocity a vector?0106

We think of velocity as being synonymous with speed but that's really not quite the idea.0108

Velocity tells you which direction you're going and how fast you're going.0115

There's two ideas imbedded in velocity, one is how fast you're going, and one is the direction you're going.0122

That's something that's very natural to measure with a vector.0130

Another typical example of a physical concept that is measured with a vector is force.0135

If you push on something, you're pushing it in a particular direction and with a particular amount of force.0141

There's a magnitude that tells how hard you're pushing it, and there's a direction which tells you in which direction you're pushing it.0148

These are two examples of physical concepts.0157

There are lots and lots that when you try to describe them, you really need to talk about both magnitude and direction, so we keep track of them using vectors.0161

There's some equations to be associated with vectors.0171

If you think of vectors having components x and y, think of it as the terminal point of a vector where the arrow is going to as x and y in the initial point as the origin, then the magnitude ...0178

We often use r for the magnitude, you can figure that out using the Pythagorean theorem, it's just the square root of x²+y².0197

The direction is slightly more tricky.0206

To find the direction of a vector, you have to remember that tanθ=y/x, so it's tempting to say that θ=arctan(y/x), that's sometimes true but more subtle than that.0209

Let me show you why it's more subtle than that.0227

Remember that arctangent always gives you an angle between -π/2 and π/2.0232

Arctangent always gives you an angle in the fourth quadrant or the first quadrant, -π/2 to π/2, which means that if you take a vector in the second quadrant or the third quadrant ...0240

Let me label my quadrants here.0262

If you take a vector in the second quadrant or the third quadrant, and you take y/x, then you take arctan of that, it will give you an answer, an angle in the fourth or first quadrant and that will clearly be wrong.0265

The way you fix that is you add 180 to the arctan(y/x).0282

You want to do that whenever the vector is in the second or third quadrant, that in turn turns out to be whenever the x-coordinate is negative.0288

You use this formula when in the quadrants 2 or 3, which is the same as saying, when x is less than 0.0306

This standard formula, θ=arctan(y/x).0331

I'm running out of space here.0338

Let me make a little space over here.0342

Use when in quadrants 1 or 4, when x is bigger than 0.0349

That's a little tricky.0367

The formula for θ, it's usually arctan(y/x), but if you're in the second or third quadrant, or in other words, if your x-coordinate is less than 0, then it's arctan(y/x)+180.0369

That's probably the trickiest formula.0381

If you know r and θ, then it's easy to find x and y, this just comes from SOH CAH TOA.0385

There's the x-coordinate and the y-coordinate.0402

Remember sin(θ) is equal to the opposite, this is from SOH CAH TOA, over the hypotenuse.0404

Here, the opposite is y, the side opposite to y, the hypotenuse is r, you can solve that out to y=rsin(θ).0413

Similarly, cos(θ) is equal adjacent over hypotenuse, which is equal to x/r, you can solve that out to x=rcos(θ).0431

Those just come from old-fashioned right triangle trigonometry and SOH CAH TOA, x=rcos(θ), y=rsin(θ).0448

Those are always true no matter what is positive or negative.0457

You can be very safe with the x and y formulas.0462

The only tricky one is when you know x and y and you're solving for θ, it's a little tricky.0465

You have to check the sin(x), and you either have to use the arctan(y/x) or arctan(y/x)+180.0470

Let's practice that with some actual vectors and some examples.0478

The first problem is find the magnitude and direction of a vector whose horizontal and vertical components are -3 and 4.0483

Let me draw that vector.0490

X component is -3, y component is 4.0495

There's that vector (-3,4).0505

We want to find the magnitude and direction, that's r and θ, for that vector.0508

I'll use the formulas that we learned on the previous slide, r is equal to the square root of x²+y², which is the square root of -3²+4².0518

That's 9+16, 25.0533

The magnitude of that vector is 5 units.0540

The direction of that vector, θ=arctan(y/x), 4/-3.0547

Remember that the formula for θ is tricky because it can be arctan(y/x) or, depending on where the vector is, it can be that plus 180.0563

This vector is in the second quadrant, we have to do plus 180 here.0574

That's because the x-coordinate is negative.0579

That's not a common value, let me work that out on my calculator.0583

I'm going to set degree mode here because I'm talking in terms of degrees.0591

When I said the 180, it's kind of determined that I have to be using degrees here.0597

Arctan(4/-3)=-53.1+180.0603

Of course, -53.1 degrees would be down there, that's -53.1 degrees.0613

That's certainly wrong because that's 180 degrees away from the vector that we're looking for, that's why we have to add the 180.0623

When we add the 180 to that, we get 126.9 degrees.0629

If you want to give that as an angle from the x-axis, there's your answer, 126.9 degrees.0644

In a lot of these applications, we're going to give North, South, East, West compass directions for our directions.0650

If you think about that, that's 90 degrees plus 36.9 degrees, that means that angle right there is 36.9 degrees.0655

If you want to give your answers in terms of a compass direction, this will be 36.9 degrees west of north.0673

It kind of depends on the kind of answer you're looking for.0686

If you're looking for an answer as an angle around from the x-axis, there's your answer right there, 126.9.0691

If you want a compass direction oriented from the North Pole, then I would call this 36.9 degrees west of north.0698

Let's recap there.0706

We were given the two components of a vector, the x and y, we find the r just by this Pythagorean formula, we throw those in and we get the magnitude of the vector.0708

We find the direction of the vector by using arctan(y/x), but because the vector is in the second quadrant, because its x is negative, we have to add 180.0718

We worked that through, we get an angle, and then if we want to describe it in terms of compass directions, we write it as 36.9 degrees west of north.0732

Let's look at another example here.0746

We're told that a horizontal force of 30 Newtons is applied to a box on a 30-degree ramp, and we want to find the components of the force parallel to the ramp and perpendicular to the ramp.0749

Certainly, a problem like this, you want to draw a picture.0760

You definitely want to draw a picture.0763

Let me draw a 30-degree ramp, there's the ground, there's my 30-degree ramp.0768

I've got a box on this ramp.0779

Apparently, I'm pulling on this box horizontally with a force of 30 Newtons.0785

There's my horizontal force and I know that that's 30 Newtons.0795

What I want to do is find the components of the force parallel to the ramp and perpendicular to the ramp.0802

I want to break this vector up into two pieces, one of which is parallel to the ramp, one of which is going straight up the ramp, and one of which is perpendicular to the ramp.0808

One of which goes straight in to the ramp.0822

If you see, I'm finding two vectors that are parallel and perpendicular to the ramp that add up to the force there.0824

Then I want to figure out what those vectors are.0834

I'm going to translate this angle 30 degrees.0840

This is probably a little easier if I get rid of some of the elements of the picture.0844

Let me just write this as a right triangle like that, with a 30-degree angle there.0849

That's the hypotenuse of the triangle, it's 30 Newtons there.0862

And I'm trying to find the lengths of the other sides.0866

This is a right triangle, it's okay to use SOH CAH TOA, I don't have to use law of sines or law of cosines because I have a right triangle.0868

SOH CAH TOA, that says that sin(θ) is equal to the opposite over the hypotenuse.0881

The opposite angle to the angle θ here or the opposite side is that.0889

That's the adjacent angle to angle, if we call that angle θ.0897

Of course, this is the hypotenuse.0905

Sin(30) is equal to the opposite over the hypotenuse, is 30, the opposite is equal to 30sin(30).0908

That's one of my common values, I remember the sin(30).0924

I don't have to go to the calculator for this one.0927

The sin(30), that's π/6, that's equal to the sin(π/6)=1/2, 30×1/2 is 15.0929

To find the adjacent side, I'm going to use the cosine part of SOH CAH TOA.0951

Cosine is equal to adjacent over hypotenuse.0955

Cos(30) is equal to ...0964

I don't know the adjacent yet but I know that the hypotenuse is 30.0969

If I solve that for the adjacent side, I get 30cos(30).0972

Again, 30 is one of my common values, it's π/6.0980

I know that the cos(30) is root 3/2, so I get 15 root 3.0982

If I fill those in, this is 15 root 3, and this is 15, the answer to the original question is that the parallel force, force parallel to the ramp is 15 root 3 Newtons, and the force perpendicular to the ramp is exactly 15 Newtons.0994

Let's recap what made that problem work.1047

Again, we started with the word problem, first thing you want to do is draw a picture.1050

I drew my picture of my box on a 30-degree ramp, I drew in my horizontal force as a vector, vector 30 units long, because it's a force of 30 Newtons.1055

I was asked to find the components both parallel and perpendicular to the ramp.1065

What I really had to do is break this arrow up into two arrows, one of which was parallel to the ramp, one of which is perpendicular to the ramp.1071

I broke it up into those two arrows, those two vectors then I abstracted into another triangle, then I could use SOH CAH TOA to find the lengths of those two sides of the triangles.1081

That's what I did here, I found the lengths of those two sides of the triangles, I plugged them back in to my original picture, then I interpreted them in the context of the original problem.1096

We have another word problem here.1112

It says a small plane leaves Great Falls heading north, 60 degrees west, that means 60 degrees west of north, an air speed of 160 kilometers per hour.1116

Meanwhile, the wind is blowing from north 60 degrees east at 40 kilometers per hour.1127

We're told the Great Falls is 180 kilometers south of the Canadian Border.1134

We want to know how long until the plane enters Canadian air space.1140

This is quite a complicated vector problem.1142

As usual with a problem, when you're overwhelmed by lots of words, you want to draw a picture, so let me try to draw a picture here.1146

We're told that the plane is heading 60 degrees west of north.1158

Let me just write in my compass directions here, North, West, South, East.1164

We're told that the plane is heading 60 degrees west of north.1174

There it is, that's a 60-degree angle.1182

It's heading at an air speed of 160 kilometers per hour, that means that the magnitude of that vector is 160.1186

Meanwhile, the wind is blowing from the direction north 60 degrees east at 40 kilometers per hour.1196

I think what we're going to be doing is we're going to be adding the contribution of the plane's engines and the contribution of the wind.1203

We want to figure those out separately.1213

I want to find the components of each one of those vectors.1216

Let's find the components of the plane.1221

If that's 60 degrees west of north, then that's really 60+90, that's 150 degrees.1225

The vector of the plane's motion is, remember x=arccos(θ) and y=arcsin(θ), we know our θ is 150 degrees, so x=160cos(150), and the y component is 160sin(150).1239

Let's figure out those first.1280

Cos(150), that's one of the common values, that's 5π/6.1283

I remember that it's cosine is negative square root of 3 over 2.1289

It's negative because the x-coordinate is negative, over there in the second quadrant.1297

Sin(150), that's sine of 5π/6, that's 1/2, that's a common value that I remember.1300

This is -80 square root of 3, and 80 for the plane's motion.1312

Those are the x and y components of the plane's motion.1320

Meantime, the wind, let me fork out the wind in blue.1325

The wind is blowing from north 60 degrees east.1329

The wind is coming from 60 degrees east of nort, it's coming from that direction.1338

What I'm going to do is move that vector down so that it starts at the origin.1350

There's the wind in blue.1358

We know that it's blowing at 40 kilometers per hour.1361

We want to figure out the components of the wind.1365

Let me figure out the wind.1369

Now, that is a 30-degree angle there, because that's 60.1372

The whole angle is 210, because it's 30 degrees past 180.1381

The angle is 210.1388

Using rcos(θ), rsin(θ), we have 40cos(210), and 40sin(210).1390

We got to work out what those are.1406

The cos(210), again that's a common value, that's 7π/6.1408

I can work that out, that's negative root 3/2, negative because we're over there in the third quadrant, the x-coordinate is negative.1414

Sin(210)=-1/2, it's negative because again third quadrant and the y-coordinate is negative there too.1428

This is -20 root 3, and -20.1437

That's the components of the wind's vector.1447

We found the plane's vector and the wind's vector, the net travel of the plane will be the sum of those two vectors.1452

We have the plane in black.1468

We have the wind blowing it in blue.1474

In red, I'll show the net travel is the sum of the two vectors.1480

I want to add up the x and y components of the plane and the wind there.1486

I'm going to do all these in red.1492

The net is equal to the plane plus the wind.1495

The net velocity of the plane, that's equal to ...1501

If I add up -80 square root of 3 and -20 square root of 3, I get -100 square root of 3, then 80-20=60.1515

That came from adding up those two vectors.1527

The net travel is -100 square root of 3 in the x direction, and 60 in the y direction.1533

Let's go back and read some more of the problem.1547

It says the Great Falls is 180 kilometers south of the Canadian border, and the border runs due east-west.1550

Let me draw a very rough map of the United States and Canada here.1558

Here's the US, and here's Canada.1564

We're told that this plane, now we know it's travelling according to that vector, we're really only interested when it crosses the Canadian border.1568

That is a matter of how long it takes, its y-component, to get us across the Canadian border, and we're told that the vertical distance is 180 kilometers.1581

Let me draw my net vector a little bit shorter here to make a little more to scale.1604

The y-component there is 60 kilometers per hour.1611

The question is how long will it take to go 180 kilometers north if its y-component is increasing at 60 kilometers per hour.1616

At 60 kilometers per hour, that's an easy division problem, it will take 3 hours to go 180 kilometers north.1628

Let's go back and look at that problem again and recap it.1657

Lots of words to start with, we're greeted by this problem that's very long and very wordy.1660

First thing to do is draw a picture and try to isolate the different quantities involved.1667

First we drew a picture of the plane's motion.1671

The plane's motion is this one in black here, this black vector we know it's 160, it has a 160 magnitude.1674

We're told that the angle is 60 degrees east of north, but in terms of a reference angle from the x-axis, that's really 150 degrees because it's 90+60.1683

That's where the 150 comes in.1701

To find its components we used x=rcos(θ), y=rsin(θ).1702

We plug those in and we get the components of the plane's motion.1708

Then we want to try to find the components of the wind's vector.1712

The wind is blowing from north 60 degrees east at 40 kilometers per hour.1719

That's where we get this vector is blowing from the northeast, I showed it blowing to the southwest.1726

From the northeast is the same as to the southwest.1730

It has magnitude 40, I put that in to rcos(θ) and rsin(θ), and I work out the components of the wind's motion.1733

The net motion of the plane is the way its flying, it's air speed, or it's velocity relative to the wind, plus the wind's velocity.1744

I add those two together, and in red, I get a vector representing the net velocity of the plane.1757

That's my vector representing the net velocity of the plane.1765

Finally, the question when you blow all the words out of it is asking how long does it take the plane to go 180 miles north.1769

Since the border runs east-west, it really doesn't matter how far the plane moves east or west during that time, we only care how long it takes to go 180 kilometers north.1779

I think I said miles before but our unit is kilometers.1789

We want to go 180 kilometers north, we don't care about the east-west motion which is why I never really did anything with the -100 square root of 3, instead we looked at their vertical motion, the 60 ...1791