AP Physics C: Mechanics Math Review

Hello, everyone, and welcome back to www.educator.com.0000

I am Dan Fullerton and in this lesson we are going to talk about some of the math skills that we are going to need to be successful in this course.0003

To begin with, our objectives are going to be explaining how vectors and scalars are used to describe physical quantities.0009

Calculating the dot and cross products of vectors, a little bit of vector multiplication.0017

Utilizing dimensional analysis to evaluate the units of a quantity.0022

Calculating the derivative, a basic functions, calculating the integral of basic functions and0027

explaining the meaning of the derivative and integral in terms of graphical analysis.0033

I know at this point a lot of folk start to get worried about the math involved in physics.0037

This is a calculus based math, a calculus base physics course.0042

However that does not mean it is a math course.0047

The math is used as the language of physics.0050

To help explain things much more efficiently than you could in words.0053

It is not really about math, it is about the physics applying those math principles and putting them to good use.0057

Let us start by talking about vectors and scalars.0064

Scalars are physical quantities that have a magnitude only.0068

They do not need to be described at the direction.0072

Things like temperature, mass, time, all of those are scalar quantities.0075

We could say when time move forward and backward but we are talking North, South, East, West directions.0082

If it has a direction we call it a vector quantity, things like velocity.0088

You are driving 55mph down the highway west.0093

Force, I pushed Susie forward.0098

Acceleration, I accelerated to the east.0103

They have a direction as well as a magnitude.0107

Vectors are typically represented by arrows where the direction is given by the direction of the arrow and the longer the arrow is the larger the magnitude,0110

if we want to show vectors graphically.0118

Let us take a look at a couple vector representations.0121

Let us assume that we have some vector in blue here.0124

Let us call it A and we have another vector here in red - let us call it B.0127

They both have the same direction but B has a larger magnitude than A, that is pretty straightforward.0134

Now one of the rules of vectors though is you are allowed to move it in space as long as you do not change it's direction or it's size you can slide it anywhere you want.0141

We could take this A vector if we wanted to.0149

Let us say we slide it down here and I am going to redraw our A vector.0152

I think it is roughly that length.0157

We will call that our new A vector and make that one go away.0160

Perfectly reasonable thing to do as a long as you keep the same magnitude and directions, vectors are free to move around in space.0165

We can also add two vectors.0175

We have vector A here in red and we have vector B here in blue.0178

We wanted to add A + B to get the vector C.0186

The way we do that is by aligning these two vectors up tip to tail.0196

They are not connected but what if instead of having it just like this I am going to redraw this over here and then try my best to do about the same length.0201

Here is a vector B but I'm also going to move vector A the same direction, same magnitude, and I'm going to slide over here so that vector A is now aligned tip to tail.0210

Its tip is touching the tail of vector B.0216

Once you have the vectors lined up tip to tail in order to find the addition of those two,0229

what we call the resultant, we draw a line from the starting point of our first vector to the ending point of our last vector.0235

That would be vector C, the resultant of A + B or what happens when you add vectors A and B.0245

It does not matter which order you do this or how many vectors you do with it.0253

As long as you add them all up to tip to tail it will work for 100 vectors as easily as it will for two.0257

Let us demonstrate that for a second by taking our B vector I am going to redraw again0262

and going to draw it down here, roughly the same length and same direction.0271

But now instead of having A point to its end, I'm going to have that point it to the end of A.0276

A was right about there so we will try drawing A here again tip to tail but with a different vector in front.0281

Once again to find the resultant I go from the starting point of my first vector to the endpoint of my last vector.0292

There is C notice regardless of how I did it which order I have roughly the same magnitude and same direction.0303

You can tell they are a little bit off on the drawing because I'm doing it by hand0311

and very carefully with the protractor so we are not here all day.0314

But that also shows the order of addition for vectors does not matter.0318

We could just as easily have written B + A = C vector.0322

There is graphical vector addition.0333

Let us take a look at graphical vector subtraction.0336

We will have A in red again.0339

We will put B over here in blue and now if we want to know what A – B.0342

The trick to doing that is realizing that subtraction is just addition of a negative that is the same as saying A + -B.0352

We are going to have A + -B we are going to call that vector D.0365

We want to add A + -B because we know how to add vectors.0370

We got b here but how do we get the negative?0374

It as easy as you might think.0377

If that direction B, if that is vector B, all we have to do to get –B is switch its direction.0378

There is –B.0387

To add A + -B all I do is I line them up tip to tail again and I am going to take A and slide A over0389

so it is right about there roughly the same length and direction.0397

Line them up tip to tail so if this is A+ -B is the same A – B.0404

Draw a line from the starting point of my first vector to the ending point of my last.0413

I suppose we made D in purple.0420

Let us do that, there is D in purple.0421

There would be vector D, graphical vector subtraction.0425

Sometimes dealing with these graphically can get little bit tedious.0432

If you have a vector at an angle lot of times what you might want to do is break it down into components that are parallel perpendicular to the primary axis you are dealing with.0436

Let us say that we have a vector A here at some angle θ with the horizontal.0445

It can be considerably more efficient to break into a component that is along the x axis.0451

Let us do the y axis first just to make it easier to draw.0459

We call this the y component of vector A and we will have some x component of vector A.0462

Noticed that A axis a vector + y as a vector gives you the total A.0480

You could replace vector A with the equivalent set of vectors Ax and Ay where Ax is in the x direction.0487

Ay is along the y axis so we could break that up into components.0494

If you want to know how big those components are to find their magnitudes, the size of them, their length, Ay in trigonometry that is the side that is opposite our angle.0500

That is going to be equal to the magnitude of A our total vector × sin angle θ.0511

In a similar fashion, to find this component, the adjacent side of our right triangle Ax is going to be equal to A cos θ where A is the magnitude of this vector.0518

How about finding the angle then?0533

Here is a right triangle, we have an adjacent and opposite side and the hypotenuse.0536

If we know 2 or 3 sides we can find the angle.0542

The tan of θ is the opposite side over the adjacent side.0545

If we know those two sides then the angle between them, θ is going to be the inverse tan0553

of the opposite side over the adjacent side not the angle between the angle of the triangle.0559

If you know the opposite and the hypotenuse you can use the sin θ.0567

Sin θ that is opposite over hypotenuse therefore θ will be the inverse sin of the opposite over the hypotenuse.0573

If you know the adjacent side and hypotenuse, cos θ is adjacent over hypotenuse.0586

When you know those two you can find θ is the inverse cos of the adjacent divided by the hypotenuse.0595

If you know any two of the three sides you can go find the angles.0604

It is so wonderful.0609

As we talk about all these vectors and in this course we are going to be dealing with vectors in three dimensions in the x, y, and the z planes.0611

It is helpful to have some standard notation to help you deal with it.0619

Also if you are using a textbook, a lot of times they are different notational styles in different textbooks.0622

Probably we are talking about those for a little bit to have some consistency throughout the course.0627

First thing I am going to do is I'm going to draw a three dimensional axis up here.0632

We will give ourselves y, x and z.0639

There is a three dimensional axis to begin with.0655

If we have some vector let us call it A, it can have components in the x, y, and z directions.0658

We could write that as in this bracket notation is its x value, its y value, and that z value.0665

That is a fairly common way of writing these and one of my favorites.0674

If you take a moment you define what we call some unit vectors, there are some other ways we can deal with this two.0677

That is our x, let us call a vector of the unit length 1, a vector that has a length of 1 in the x direction0684

that special vector we are going to call I ̂.0694

The hat means is the unit vector, its length and its magnitude is always 1.0699

In the x direction we call it I ̂.0703

In the y direction, we are going to do the same basic thing.0706

Make a unit vector of length 1 and we are going to call it j ̂.0710

In the z direction I am sure you have not guess by now.0716

A unit vector of length 1 we are going to call k ̂.0720

We could also write our A vector now whatever happens to be as having components of x coordinate × I ̂.0726

That is the unit of vector 1 in the x direction + y value × j ̂ + z value × k ̂.0740

That will get a little funky until you get used to it.0760

That is pretty common in a bunch of textbooks to see these unit vectors along with the vector in front of them.0763

These two types of notation really mean the exact same thing.0770

Let us see how that can be useful down here.0777

Let us try it again with a fairly simple vector addition problem.0779

I am going to draw my axis again, give ourselves y, x, and the z axis.0784

What we are going to do is we are going to define a couple of vectors.0805

The first one I am going to define is called P.0808

Where P goes from the origin to some point that is 4/x, 3 in the y, 1 in the z.0812

It might be somewhere about there.0824

We will draw our vector from the origin there and that is our point P 4, 3, 1.0828

Let us take a second vector and we will call it q.0836

We are going to go 2 points in the x, 0 in the y, 4 in the z.0842

We call this the vector q which leads us to q 2, 0, 4.0849

If we want to add this two to get r well when you are in three dimensional space0859

it is starting get pretty tough to see what is going on if you we want to line this up tip to tail0864

and come up with a reasonable solution that you can actually make sense of graphically.0869

What we are going to do is we are going to say that our r vector is equal to P + q which implies then our P vector is equal to 4, 3, 1, if we use bracket notation.0874

Our q vector is equal to 2, 0, 4.0899

If we want to add these up to find r all we have to do is in its bracket notation is to add up the individual components.0907

4 + 2= 6, 3 + 0= 3, 1 + 4= 5 so 6, 3, 5 will give us the vector to our resultant.0917

Let us draw that in here.0928

We go 1, 2, 3, 4 about 6 on the x, 3 in the y, 5 on the z.0930

As I draw it is going to come out somewhere around there depending on your perspective we will call that r.0939

What if you just did that graphically which you can get by lining these up?0949

It is going to be a really tough to see what is going on.0953

We get in the three dimensions especially analytically looking at these vectors sure makes a lot more sense and it makes life a whole lot easier.0956

A little bit more with vector components.0967

A soccer player kicks the ball with an initial velocity of 10 m/s if an angle of 30° above the horizontal.0970

Find the magnitude of the horizontal component in vertical component of the balls velocity.0977

I like to start with graphs wherever possible to help me visualize the problem.0983

And this look like it is in two dimensions.0989

We got a vertical and horizontal component to our problem.0991

We call this our x and y and it has an initial velocity of 10 m/s in angle of 30° above the horizontal.0996

Let us see if we can eyeball roughly 30°.1006

The magnitude of our vector is 10 m/s at some angle 30°.1010

We want the horizontal component and vertical component.1016

Let us draw these in first.1020

Our vertical component will be that piece and our horizontal component will be that piece.1022

The x component of V, our initial velocity V = 10 m/s.1033

Our x component is going to be V cos 30° because we are looking for the adjacent side of that right triangle.1042

It is going to be about 8.66 m/s.1051

The vertical component Vy is going to be V sin 30°.1055

We got the opposite side there which is going to be about 5 m/s.1063

And of course if we ever want to put these back together, if we have the components in one of the whole we can just use the Pythagorean theorem.1069

Let us take a look at the second example.1076

An airplane flies with the velocity of 750 kph 30° South East.1080

What is the magnitude of the planes eastward velocity?1086

Lets draw a diagram again and we would call that north and south so this must be east and west.1090

It is traveling the velocity 750 kph 30° South of East in this basic direction.1109

Our V = 750 kph and an angle of 30° South of East.1123

The magnitude of the planes eastward velocity, it looks like we are after an x component here.1133

Let us draw that in, we are after just this piece, the x component.1139

Vx equals that is the adjacent sides and that going to be V cos θ which is going to be 750 kph.1148

The magnitude of our entire vector × the cos 30° and that can give you about 650 kph.1157

How about the problem with the magnitude of vector?1170

A dog walks a lady 8m due north and then 6m due east.1175

You probably all seen that before.1179

It is a really big dog and little even lady the dogs doing the controlling.1181

Determine the magnitude of the dog’s total displacement.1185

The way I do that is looks like we have a couple of vectors that we can add up.1190

Dogs walk lady 8m due north so I draw vector magnitude 8m due north and then 6m due east.1194

Determine the magnitude of the dog’s total displacement.1211

Displacement being the straight line distance from where you start to where you finish.1214

The way I will do that then so we are going to from the starting point of our first to the ending point of the last.1219

That red vector represents the total displacement.1228

How do I find the magnitude of that?1233

It is a right triangle I can use that Pythagorean theorem.1236

A² + B²=C².1239

Our hypotenuse is going to be =√(a² + b² ) which is the √(8² + 6²).1246

64 + 36 is going to be 100 m² which implies then that C is the square root of that which is going to be 10 m.1257

If we want to find the angle, let us define an x axis.1271

What if we wanted to find that angle θ?1275

We could if we wanted to θ equals inverse tan of the opposite side of a right triangle divided by1278

the adjacent side that could be the inverse tan of.1286

The opposite side is this piece here that is not shown but we can see that is 8m ÷ the adjacent side that is going to be the length of 6m which comes out to be about 53.1°.1290

We are not asked for that but if we were we would know how to go calculate it.1305

Let us look at the vector addition problem.1310

A frog hops 4m angle of 30° North of East.1312

He then hops 6m angle of 60° North of West.1316

What was the frog total displacement from his starting position?1320

Alright we are getting a little bit more challenging here.1324

Let us draw what we have north, east, and west.1328

Frog hops 4m angle of 30° North of East1346

He goes 4m North of East.1351

Let us called about 4m assuming that is 30° and I'm just estimating these.1359

He then hops 6m an angle of 60° North of West.1366

To figure out what that is let us draw x and y here.1372

At 60° North of West that is going to be roughly this direction and it goes 6m that way and we will say that is something like that.1375

There is our second piece of 60°.1400

What is the frog’s total displacement from his starting position?1403

Thought displacement goes from the starting point of our first to the ending point of our last.1407

We could be fairly accurate if we are doing this with the protractor we will call that C, A, and B but it is a lot easier to do analytically.1414

Let us take a look at how we add A and B up to get C in vector notation.1426

A vector has an x component that is going to be 4m cos 30° and the y component that is going to be 4 m sin 30°.1431

4m and our angle is 30° we can break that to x and y components.1448

Our B vector the 6m and angle of 60° North of West is just going to be, since we can tell it is going left to1453

begin the x piece is going to be -6 m cos 60° in the y component 6m sin 60°.1462

If we then want to find out our total C, our C vector is just going to be the sum of those.1477

That is going to be our x components 4m cos 30° + -6m cos 60° for the x and for the y we have 4m sin 30° +6m sin 60°.1487

And if we put that all together I find that our C vector is about 0.46m, 7.2m.1515

Our estimation here of ½ m to the right, 7.2 to the left, it is roughly in the ballpark for just a eyeballing that one.1524

If we want to know the magnitude of our answer the magnitude of C like that is going to be.1533

Well we have these two components x and y it is going to be the √ x component 4.6m² + 7 ⁺2m = 7.21m for the magnitude.1542

If want the angle from the origin we can go back to our trig.1562

Θ is the inverse tan of the opposite over the adjacent which is going to be our 7.2m ÷ 4.6m or about 86.3° North of East.1569

You can see how valuable, how useful these vectors can be and breaking them up into components in order to manipulate them.1587

Find the angle θ depicted by the blue vector below given the x and y components.1596

Let just hit this again because it is often times a trouble spot as we are getting started.1602

We know the opposite side and the adjacent sides so θ is going to be the inverse tangent of the opposite side over the adjacent.1607

Put that in your calculator and making sure it is a degree mode for a question like this where we want an answer in degrees and I come up with 60°.1615

Let us go hit vector notation a little bit more.1626

Unit vector notation we said can be written as A vector = x × I ̂ the unit vector in x direction + its y value × j ̂ + z value × k ̂.1629

The vector component notation we would write that S or bracket notation x, y, z.1646

You will see vectors written in many different ways in many different textbooks.1654

Some of the standard ones I used most often are the capital letter with a line over it or lower case letter with the line over it is the vector.1658

Sometimes you will see just a very old letter in something like a textbook, if it is bold that usually means it is a vector.1670

If you want the magnitude of the vector you have the vector symbol inside absolute value signs that would be a magnitude.1679

In other books you will see double lines surrounded to indicate magnitude.1687