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

Still in other if you see A written bold for A vector.1692

A that is not bold may indicate the magnitude of a vector if you do not add that extra symbology to explain it is a vector could be magnitude.1696

Take a look at your book that you are using and try and find out what it is using1706

and maintain some consistency with that throughout the course.1710

We will use a couple of these just you get used to all the different forms of notation.1713

We can add vectors we can subtract vectors we can also multiply vectors.1718

But there are two types of vector notation.1724

The Dot product also known as a scalar product takes two vectors you multiply them and what you get is an output as a scalar.1727

The cross products or vector product gives you a vector as the output of the multiplication.1734

We will talk about these different types starting with the Dot product and the Scalar product.1740

What it does is it tells you the component of a given a vector in the direction of the second vector really multiplied by the magnitude of that second vector.1745

You would write it as A • B and the result is the component of vector A that is in the direction of vector B multiplied by the size of vector B.1754

How can we define that a little bit better?1765

A • B = to the magnitude of A magnitude × magnitude of B × cos of the angle between those two vectors.1768

Another definition that you are really going to want to know.1777

Let us see how we can calculate this.1781

In unit vector notation let us assume we have some vector A where A = x component of A × i ̂ the unit vector in x direction + the y component of A × the unit vector1784

in the y direction j ̂ + the Z component of A × the unit vector in Z direction known as k ̂.1799

We are going to also define vector B as the x component of B I ̂ + y component of B j ̂ + the Z component of B k ̂.1810

Therefore A • B we want to do these two together that is just going to be Ax Bx multiply the two x components together + ay by.1826

Multiply the y components together + az bz.1847

And no unit vectors because it is scalar.1853

The dot product gives you a scalar output.1856

In vector component notation if we written A as Ax Ay Az and written B as Bx By Bz then A • B would still be AxBx + AyBy + AzBz.1860

Depending on the type of notation you prefer still get the dot product of the exact same formula.1896

Let us do it.1904

Find the dot product of the following vectors A and B where A is 123 in the x, y, and z directions B is 321.1906

Couple ways we could do this.1915

We will start off doing it the easy way.1918

A • B is going to be the x components multiplied 3 + the y components multiplied 4 + z components multiplied 3 or 10.1925

We also mentioned that you can find that by AB cos θ.1940

Let us do that while you are here but first we need to find the magnitude of A.1945

The magnitude of that vector A we can use our Pythagorean theorem that is going to be the square root of the component squared.1949

1² + 2² + 3² which would be √14.1956

The magnitude of B is going to be √(3² + 2² + 1²) also √14.1965

We could also look at A • B as AB cos θ their magnitude × cos of the angle between them1978

which implies that we know A • B is 10 must be equals √14 × √14 is this going to be 14 cos θ.1993

Or 10 = 14 cos θ therefore θ = the inverse cos of 10/14 which is 44.4° and we just found the angle between A and B.2006

Couple different ways you can do this.2025

Let us take a look at a couple of special dot products.2030

The dot product of perpendicular vectors is always 0 because there is no component of 1 that lie on the other.2033

If their dot product is 0 they are perpendicular.2040

The dot product of parallel vectors is just the product of their magnitudes.2043

One way we could look at this is if we are talking about A • B is AB cos θ.2049

If the angle between θ is 0 cos of 0 is 1 you just get the product of their magnitudes.2061

If however they are perpendicular write that specifically if θ = 0 they are parallel.2069

AB cos θ however if they are perpendicular and θ = 90° cos 90 is going to be 0 therefore you would get 0 for your dot product.2076

A couple of dot product properties.2091

First off the commutative property A • B = B • A that works it is commutative.2094

You have A + B vectors .C = A • C + B • C associative property.2108

If you are taking a derivative, the derivative of A • B = to the derivative of A • B + the derivative of B • A.2125

Let us do a couple more examples here.2146

If A -2, 3 and B is 4, by find a value of By such A and B are perpendicular vectors.2150

The way to start this is recognizing that if they are perpendicular then their dot product A • B must be = 0.2158

A • B =0 let us take a look what happens when we do our dot product.2170

The x components multiplied we get -8 + 3By = 0 or 3By = 8.2177

By = 8/3 if they are going to be perpendicular.2191

That is the dot product.2201

The cross product of two vectors gives you a vector perpendicular to both because magnitude is equal2204

to the area of the parallelogram defined by the two initial vectors.2209

Sounds complicated but let us say that we are talking about a couple of vectors where A cross x symbol with B gives you some vector C.2213

The area of a parallelogram defined by those vectors let us see if we can scope that out a little bit.2230

I will draw something kind of like that and the area of that parallelogram defined by AB is the magnitude of your vector C.2236

And C going to be perpendicular to both A and B where its direction is given by the right hand rule.2252

We have to do as is if you have A cross B take the fingers of your right hand, point them in the direction of vector A2258

bend them in the direction of vector B and your thumb points in the direction that is positive for C.2266

A cross with B gives you C a right-hand rule for cross products.2274

And that is going to come up in this course multiple times as well as in the ENM course.2279

Now interesting to note the cross product of parallel vectors is 0 which it has to be because they cannot define parallelogram.2284

Defining the cross product.2295

A × B the magnitude of A cross B is AB sin θ.2297

The direction given by right-hand rule where is before A • B is value was AB cos θ.2302

The cross product only the magnitude of the vector is AB sin θ still have to worry about direction2309

because the cross product it outputs a vector not a scalar.2315

If we look at the cross product with unit vector notation A cross B = Ay Bz –Az By on the I ̂ direction the x component.2321

The y component Az Bx – Ax Bz in the y direction the j ̂ component.2335

The z component is Ax By - Ay Bx so you could memorize that formula that is one way to do it because calculating cross product is considerably more complex than dot products.2344

Or the way I tend to do it is with some linear algebra looking at the determinant.2357

What we are going to do is we are going to take the determinant of these vectors such that if we have A cross B where is x is Ax Ay Az the components of B are Bx By Bz.2363

I would start by drawing A 3 × 3 matrix I ̂ at the top, j, ̂ and k ̂ is my first row.2375

My second row Ax Ay Az and my third row Bx By Bz.2387

And that is we are going to take the determinant of.2405

When you do that, the way I do this to help me understand and to help do this a little bit more easily is2409

I also repeat these over to the right and left.2414

Ax Ay Az what comes next in the pattern would be Ax Ay and down here we have Bx By.2417

I also need that over here to the left so over here I will have Az before we get Ax.2427

We will have Ay there, we have a By, and the Bz here.2433

To take this determinant when I do it is I startup here and for the I ̂ direction I am going to go down into the right and those are going to be positives.2442

I am going to start with Ay Bz × I ̂ - Az By and all of those are multiplied by I ̂ + for the j component I started to j I go down to the right.2452

I have Az Bx – Ax Bz × j ̂ + Ax By – Ay Bx J ̂.2481

That is another way you can come up with the formula.2519

And typically this is a lot easier for me to remember how to do than memorizing that entire previous formula.2521

An example, find a cross product of the following vectors if we are given A and B we want to find C which is A cross B.2530

A couple ways we can do this.2539

First let us start with looking at the magnitude of A and pretty easy to see2541

that the magnitude of A is just going to be 2 or you could go on the Pythagorean theorem √0² + 2² + 0² still give you 2.2546

B is written in a slightly different notation but that is just the equivalent to 2, 0, 0 and the magnitude of B must be 2.2555

If we want to know the magnitude of C, magnitude of C is going to be magnitude of A.2571

Magnitude of B × the sin of the angle between them is going to be 2 × 2 × the sin of the angle between them.2579

If this is in the y direction this is in the x direction they are perpendicular that is 90° sin 90 is 1 so that is just going to be 4.2589

We know that we are going to have a magnitude of 4 on our answer.2600

Using the right-hand rule we got to be perpendicular to both i and j, to both x and y that means it is in the z direction.2605

If I we are to graph this out quickly.2613

Let us put our x here y, z, if we are doing this we start off with the y.2618

We are bending our fingers in the direction of x that tells me that down is going to be the direction of my positive for my z by right-hand rule.2619

I could just by thinking through this one state that z going to be 0, 0, and that is going to be -4 because of our right-hand rule.2637

A little bit shaky on doing that so instead let us do the determinant.2648

Let us find out analytically how we can do that.2652

We start off with i ̂, j ̂, k ̂.2656

Our first vector is going to be 0, 2, 0 and then for our B vector we have 2, 0, 0.2662

We are going to take our determinant and repeat our pattern.2673

0, 2, 2, 0 we would have a 0, 2 over here and the 0, 0.2675

Our C vector is going to start at i ̂ that is going to be 0-0 that is easy.2684

J ̂ 0 -0 that is easy.2695

K ̂ 0-4 k ̂ C is just -4 k ̂ or 0, 0, -4.2699

Couple of ways we can go about solving them.2712

Let us take a look at a couple of cross product properties.2715

A cross with B = -B × A.2720

A × B + C= A × B + A × C.2732

If we have a constant some C × A cross B = C × A cross B or = A cross C × B.2749

If we take the derivative of a cross product the derivative of A cross B is equal to the derivative of A cross B + A cross B.2772

I think that is good on vector math for the time being.2796

Let us talk for a few minutes about units.2799

The fundamental units in physics there are 7 of them.2802

Our length which is measured in meters, mass in kilograms, time is in seconds, temperature is in kelvins.2806

The amount of the substance is measured in moles, you might remember that from chemistry.2815

Electrical current is in the amperes and luminosity is in candela.2819

And here we are talking about mechanics we are mostly going to be dealing with meters, kilograms, and seconds.2826

All over other units are derived units they are combinations of these fundamental units.2834

Given units we can oftentimes use not to help us check our answers you will see if we have done things right.2842

If displacement is in meters, time is in seconds, velocity would be derived unit meters per second2848

Acceleration the meter per second, per second or meter per second squared, force is measured in Newton’s which is really a kg m/s².2854

The gravitational constant capital G is in N m/s² / kg².2863

You can go and see if all the units match up, if they do not you have probably made a mistake.2868

For example this first one does this dimension correct or are there errors?2873

A meter per second is equal to a meter per second times a second + m/s².2877

No, it does not work.2882

If you came up with that formula you probably messed up somewhere.2884

Distance displacement in meters is equal to a meter per second of velocity times a second squared.2888

No, that is not going to work out because that will cancel.2896

Meter equals meter per second is not going to work out so that can not be right.2900

This one over here though the force which is a kg m/s² is n equivalent to gravitational constant2906

which is a N m²/ kg² × mass which is a kilogram × mass which is a kilogram divided by a distance squared.2914

Let us see kg², kg², m², m².2925

Newton which is a kg m/s²= N.2930

Yes, this one works so that formula would be valid.2934

A great way to check your answers as you are going through and doing these problems called dimensional analysis.2937

The part you all have been waiting for calculus.2944

AP physics C is not really about calculus.2948

We will use calculus as a tool throughout the course.2952

We are going to cover just a few basic calculus applications here and2956

you might have seen some of them before you might have not.2959

You can find a much more thorough and detailed accounting of calculus2963

on the www.educator.com courses AP calculus AB and BC right here.2966

Believe it or not you have probably done a lot of calculus already.2973

Ever taken tangent line to find the slope, that is differential calculus or ever look at the area under a graph that is integral calculus.2976

You might not have known it you have probably done some calculus in your life already.2984

Let us take a look at differential calculus first.2990

Differentiation is finding the slope of a line tangent to a curve.2994

The derivative is the slope of the line tangent to a function at any given point, the result of differentiation.2998

If we have a curve here we can really do is take a point of the curve and we are going to try our best to find the slope of that line.3005

And given that function what we are doing when we take a derivative is finding the slope at a given point3015

or finding the function that tells you the slope of the original function that is differentiation.3021

If we say that we have some function a value y which is a function of x which is A some constant A × X ⁺n3028

then the first derivative of y is equal to the derivative of y with respect to x as for notation.3037

Or the first derivative of x, that function x all mean the same thing is equal to n Ax ⁺n -1.3046

This basic formula for a polynomial differentiation that you will become very familiar with throughout the course.3054

The derivative with respect to E ⁺x is just E ⁺x.3061

The derivative with respect to x, the natural log of x is 1/x.3066

The derivative of the sin is the cos, the derivative of the cos is the opposite of the sin.3071

All the things that you will become familiar with throughout the course.3077

Let us do a couple of derivatives and if this is troubling to you, you probably been learning it as you go along in the course.3080

Let us start with y = 4 x³.3089

If we wanted to know the first derivative of y or y would respect x we can write as y prime or dy dx.3091

The derivative with respect to x of our y function which is 4 x³ are all just different forms of notation for the same thing.3100

It is going to be 12 x² using that formula for polynomial from the previous slide.3107

Here we got the same basic idea y prime is going to be equal to -2.4 × 0.75 = -1.8x we will subtract 1 from that - 3.4.3116

The derivative of e ⁺2x, y prime would be 2e ⁺2x.3131

Derivative of the sin of 7x², a y prime is going to be equal to 14x through the sin is the cos of 7x².3139

and cos to 2x⁶, y prime is going to be equal to -12x sin 2x⁶.3152

If these are given you some trouble you are probably going to check out some of the calculus lessons before we get too deep into the math here.3169

We will start with very little calculus but it is going to grow as we go to the course you are going to need it as tool before we are done.3176

Integral calculus, integration is finding the area under the curve or adding up lots of little things to get a whole.3185

The integral is the area under a curve at any given point, the result of an integration.3192

Integration and derivation are inverse functions.3197

The integral is the anti derivative or the derivative is the anti integral.3201

If we have a curve like this and we want an integral between a couple points we are actually doing3206

is finding the area between those points going the opposite direction.3212

Suppose the derivative of a function is given can you determine the original function?3218

That is what integration is.3223

If you know the derivative of something is 2x you really need to think what is the original function if derivative was 2x?3225

And that would be y = x² the derivative x² is 2x.3233

You could write this in integral form y =∫ of 2x with respect to x.3238

Which when we do that is going to be x² + this constant of integration where that constant come from.3247

If we had a constant over here if this was x² + 3 the derivative of 3 is 0 so we still have a derivative 2x.3254

We do not know if there was a 3 or not there.3262

This constant says you know there could be some constant there we do not know what it is yet.3264

It could be -5 it could be 0 it could be 37,000.3269

Just being there as a piece we do not know about and we have some different tricks to make that go away3273

in order to find out what those constants are later on.3278

Let us look at a couple of common integrations.3284

The integral of x ⁺ndx follows this formula 1/n + 1 x ⁺n + 1 + some constant and n cannot be -1 or else you would have an undefined function.3287

The integral of the E ⁺x dx would just be e ⁺x.3301

The integral of the dx/x is the natural log of x.3305

The integral of the cos of x is the sin of x and the integral of the sin of x is the opposite of the cos of x.3308

Let us do a couple quick integrations for practice.3316

The integral of 3x² dx is just going to be x³ as to think about the derivative of x³ is going to be 3x².3320

We have to remember our constant of integration + C.3328

The integral of 2 cos 6x dx now what I would do there is recognize that3332

when I do this we will have to do integral of 2 cos 6x dx.3338

We are going to have a 6 in here in order to integrate I need to put 16 over there so they are maintained my same value so that is going to be = sin 6x/3 + C.3347

They are more involved integration.3363

The integral e ⁺4x dx is going to be integral of e ⁺4x dx now what this in the form e ⁺xdx.3366

We are going to need a 4 here which means I am going to have to put of 1⁄4 out there to maintain my original value.3376

That is going to be e ⁺4x / 4 + constant of integration.3383

And finally integral here with some limits.3390

When we have these limits that is really telling us what values we are integrate and allows us to get rid of that constant of integration.3393

I would write that as integral of 2xdx well that is x² evaluated from x = 0 to x = 4.3400

What that means is what you are going to do is you are going to take your x²3409

and you are going to plug your top value that 4 in the first 1 - your same thing but with this value plug in for x.3413

That is going to be equal to 4² – 0² which is 16.3427

If these are troubling the www.educator.com lessons on calculus are outstanding.3435

Let us look at this one more way.3442

If we said this was the area under the curve how could that look?3444

Let us draw a quick graph and assume that we have some y function where y = 2x.3448

There is y there is x.3461

We are going to look from the limits where x = 0 to x = 4 that means that over here at 4 the y value is going to be 8.3464

We said that integration give you the area under the curve.3477

Let us figure out what the area is under this curve.3480

That is a triangle so the area of that triangle formula that for the area triangle is ½ base × height is going to be ½ × our base 4 × height 8 which is 16.3485

Analytical version and graphical version gives you the same value.3500

What you are really doing is finding the area under that graph.3506

Let us take a look at our last example for this lesson.3511

The velocity of a particle is a function of time is given by the equation v of t = 3t².3517

The particle starts at position 0 at time 0.3523

Find the slope of the velocity time graph as a function of time?3526

That will give you the particles acceleration function.3531

Acceleration is the slope of that vt graph, Velocity vs. time which we could write is V prime of t.3535

The first derivative of V which is going to be the derivative with respect to T and derivative with respect to time of whatever that function is.3544

Vt² which happens to be 60 so the acceleration of the particle is equal to 6 times whatever the time happens to be.3553

When you know the velocity function you can find the acceleration, calculus.3562

Find the area under the velocity time graph as a function of time to give your particles position function.3568

Position we will call that r is going to be the integral of our velocity function with respect to time will be the integral of 3t² dt which should be t³ + our consonant of integration.3574

But here is a little trick to making a consonant of integration go way.3592

We know by one of our boundary conditions that the position of our particle at time 0 equals 0 because that is given up here in the problem.3596

If r=0 at time 0 that means if we plug 0 in here for time 0 r must equal z.3607

But r is 0 at that time so z must equal 0.3616

Therefore our total function is r = t³ z is 0 in this problem.3620

We are able to come up with the particles position function based on velocity.3627

We took the derivative in one direction to find the acceleration.3633

We took the integral going the other way in order to find it is change in position.3636

Alright that helpfully gets you a feel for the type of math we will be using in this course.3642

Thank you for watching www.educator.com.3646

Come back real soon and make it a great day everyone.3648