WEBVTT mathematics/trigonometry/murray
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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.
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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.
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Think of vectors as being arrows.
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Every vector has both magnitude and direction.
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Those are the two things that are important about a vector.
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The magnitude is the length of the vector.
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Magnitude, that's the length of the arrow.
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The direction is if you put it in the coordinate system, there's an angle θ that determines the direction of the vector.
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Every vector has both magnitude and direction.
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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.
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Most vectors, except for the zero vector, have magnitude and direction.
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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.
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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.
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That's the idea of vectors.
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They're used primarily to represent lots of physical concepts about things like velocity.
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Why is velocity a vector?
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We think of velocity as being synonymous with speed but that's really not quite the idea.
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Velocity tells you which direction you're going and how fast you're going.
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There's two ideas imbedded in velocity, one is how fast you're going, and one is the direction you're going.
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That's something that's very natural to measure with a vector.
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Another typical example of a physical concept that is measured with a vector is force.
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If you push on something, you're pushing it in a particular direction and with a particular amount of force.
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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.
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These are two examples of physical concepts.
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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.
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There's some equations to be associated with vectors.
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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 ...
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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.
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The direction is slightly more tricky.
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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.
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Let me show you why it's more subtle than that.
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Remember that arctangent always gives you an angle between -π/2 and π/2.
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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 ...
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Let me label my quadrants here.
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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.
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The way you fix that is you add 180 to the arctan(y/x).
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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.
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You use this formula when in the quadrants 2 or 3, which is the same as saying, when x is less than 0.
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This standard formula, θ=arctan(y/x).
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I'm running out of space here.
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Let me make a little space over here.
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Use when in quadrants 1 or 4, when x is bigger than 0.
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That's a little tricky.
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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.
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That's probably the trickiest formula.
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If you know r and θ, then it's easy to find x and y, this just comes from SOH CAH TOA.
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There's the x-coordinate and the y-coordinate.
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Remember sin(θ) is equal to the opposite, this is from SOH CAH TOA, over the hypotenuse.
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Here, the opposite is y, the side opposite to y, the hypotenuse is r, you can solve that out to y=rsin(θ).
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Similarly, cos(θ) is equal adjacent over hypotenuse, which is equal to x/r, you can solve that out to x=rcos(θ).
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Those just come from old-fashioned right triangle trigonometry and SOH CAH TOA, x=rcos(θ), y=rsin(θ).
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Those are always true no matter what is positive or negative.
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You can be very safe with the x and y formulas.
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The only tricky one is when you know x and y and you're solving for θ, it's a little tricky.
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You have to check the sin(x), and you either have to use the arctan(y/x) or arctan(y/x)+180.
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Let's practice that with some actual vectors and some examples.
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The first problem is find the magnitude and direction of a vector whose horizontal and vertical components are -3 and 4.
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Let me draw that vector.
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X component is -3, y component is 4.
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There's that vector (-3,4).
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We want to find the magnitude and direction, that's r and θ, for that vector.
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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.
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That's 9+16, 25.
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The magnitude of that vector is 5 units.
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The direction of that vector, θ=arctan(y/x), 4/-3.
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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.
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This vector is in the second quadrant, we have to do plus 180 here.
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That's because the x-coordinate is negative.
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That's not a common value, let me work that out on my calculator.
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I'm going to set degree mode here because I'm talking in terms of degrees.
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When I said the 180, it's kind of determined that I have to be using degrees here.
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Arctan(4/-3)=-53.1+180.
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Of course, -53.1 degrees would be down there, that's -53.1 degrees.
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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.
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When we add the 180 to that, we get 126.9 degrees.
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If you want to give that as an angle from the x-axis, there's your answer, 126.9 degrees.
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In a lot of these applications, we're going to give North, South, East, West compass directions for our directions.
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If you think about that, that's 90 degrees plus 36.9 degrees, that means that angle right there is 36.9 degrees.
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If you want to give your answers in terms of a compass direction, this will be 36.9 degrees west of north.
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It kind of depends on the kind of answer you're looking for.
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If you're looking for an answer as an angle around from the x-axis, there's your answer right there, 126.9.
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If you want a compass direction oriented from the North Pole, then I would call this 36.9 degrees west of north.
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Let's recap there.
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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.
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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.
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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.
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Let's look at another example here.
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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.
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Certainly, a problem like this, you want to draw a picture.
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You definitely want to draw a picture.
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Let me draw a 30-degree ramp, there's the ground, there's my 30-degree ramp.
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I've got a box on this ramp.
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Apparently, I'm pulling on this box horizontally with a force of 30 Newtons.
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There's my horizontal force and I know that that's 30 Newtons.
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What I want to do is find the components of the force parallel to the ramp and perpendicular to the ramp.
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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.
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One of which goes straight in to the ramp.
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If you see, I'm finding two vectors that are parallel and perpendicular to the ramp that add up to the force there.
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Then I want to figure out what those vectors are.
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I'm going to translate this angle 30 degrees.
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This is probably a little easier if I get rid of some of the elements of the picture.
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Let me just write this as a right triangle like that, with a 30-degree angle there.
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That's the hypotenuse of the triangle, it's 30 Newtons there.
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And I'm trying to find the lengths of the other sides.
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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.
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SOH CAH TOA, that says that sin(θ) is equal to the opposite over the hypotenuse.
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The opposite angle to the angle θ here or the opposite side is that.
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That's the adjacent angle to angle, if we call that angle θ.
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Of course, this is the hypotenuse.
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Sin(30) is equal to the opposite over the hypotenuse, is 30, the opposite is equal to 30sin(30).
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That's one of my common values, I remember the sin(30).
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I don't have to go to the calculator for this one.
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The sin(30), that's π/6, that's equal to the sin(π/6)=1/2, 30×1/2 is 15.
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To find the adjacent side, I'm going to use the cosine part of SOH CAH TOA.
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Cosine is equal to adjacent over hypotenuse.
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Cos(30) is equal to ...
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I don't know the adjacent yet but I know that the hypotenuse is 30.
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If I solve that for the adjacent side, I get 30cos(30).
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Again, 30 is one of my common values, it's π/6.
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I know that the cos(30) is root 3/2, so I get 15 root 3.
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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.
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Let's recap what made that problem work.
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Again, we started with the word problem, first thing you want to do is draw a picture.
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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.
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I was asked to find the components both parallel and perpendicular to the ramp.
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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.
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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.
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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.
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We have another word problem here.
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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.
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Meanwhile, the wind is blowing from north 60 degrees east at 40 kilometers per hour.
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We're told the Great Falls is 180 kilometers south of the Canadian Border.
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We want to know how long until the plane enters Canadian air space.
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This is quite a complicated vector problem.
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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.
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We're told that the plane is heading 60 degrees west of north.
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Let me just write in my compass directions here, North, West, South, East.
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We're told that the plane is heading 60 degrees west of north.
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There it is, that's a 60-degree angle.
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It's heading at an air speed of 160 kilometers per hour, that means that the magnitude of that vector is 160.
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Meanwhile, the wind is blowing from the direction north 60 degrees east at 40 kilometers per hour.
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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.
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We want to figure those out separately.
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I want to find the components of each one of those vectors.
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Let's find the components of the plane.
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If that's 60 degrees west of north, then that's really 60+90, that's 150 degrees.
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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).
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Let's figure out those first.
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Cos(150), that's one of the common values, that's 5π/6.
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I remember that it's cosine is negative square root of 3 over 2.
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It's negative because the x-coordinate is negative, over there in the second quadrant.
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Sin(150), that's sine of 5π/6, that's 1/2, that's a common value that I remember.
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This is -80 square root of 3, and 80 for the plane's motion.
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Those are the x and y components of the plane's motion.
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Meantime, the wind, let me fork out the wind in blue.
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The wind is blowing from north 60 degrees east.
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The wind is coming from 60 degrees east of nort, it's coming from that direction.
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What I'm going to do is move that vector down so that it starts at the origin.
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There's the wind in blue.
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We know that it's blowing at 40 kilometers per hour.
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We want to figure out the components of the wind.
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Let me figure out the wind.
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Now, that is a 30-degree angle there, because that's 60.
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The whole angle is 210, because it's 30 degrees past 180.
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The angle is 210.
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Using rcos(θ), rsin(θ), we have 40cos(210), and 40sin(210).
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We got to work out what those are.
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The cos(210), again that's a common value, that's 7π/6.
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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.
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Sin(210)=-1/2, it's negative because again third quadrant and the y-coordinate is negative there too.
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This is -20 root 3, and -20.
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That's the components of the wind's vector.
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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.
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We have the plane in black.
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We have the wind blowing it in blue.
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In red, I'll show the net travel is the sum of the two vectors.
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I want to add up the x and y components of the plane and the wind there.
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I'm going to do all these in red.
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The net is equal to the plane plus the wind.
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The net velocity of the plane, that's equal to ...
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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.
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That came from adding up those two vectors.
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The net travel is -100 square root of 3 in the x direction, and 60 in the y direction.
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Let's go back and read some more of the problem.
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It says the Great Falls is 180 kilometers south of the Canadian border, and the border runs due east-west.
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Let me draw a very rough map of the United States and Canada here.
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Here's the US, and here's Canada.
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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.
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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.
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Let me draw my net vector a little bit shorter here to make a little more to scale.
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The y-component there is 60 kilometers per hour.
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The question is how long will it take to go 180 kilometers north if its y-component is increasing at 60 kilometers per hour.
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At 60 kilometers per hour, that's an easy division problem, it will take 3 hours to go 180 kilometers north.
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Let's go back and look at that problem again and recap it.
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Lots of words to start with, we're greeted by this problem that's very long and very wordy.
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First thing to do is draw a picture and try to isolate the different quantities involved.
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First we drew a picture of the plane's motion.
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The plane's motion is this one in black here, this black vector we know it's 160, it has a 160 magnitude.
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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.
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That's where the 150 comes in.
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To find its components we used x=rcos(θ), y=rsin(θ).
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We plug those in and we get the components of the plane's motion.
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Then we want to try to find the components of the wind's vector.
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The wind is blowing from north 60 degrees east at 40 kilometers per hour.
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That's where we get this vector is blowing from the northeast, I showed it blowing to the southwest.
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From the northeast is the same as to the southwest.
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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.
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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.
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I add those two together, and in red, I get a vector representing the net velocity of the plane.
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That's my vector representing the net velocity of the plane.
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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.
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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.
00:29:49.000 --> 00:29:51.000
I think I said miles before but our unit is kilometers.
00:29:51.000 --> 00:46:42.000
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 ...