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For more information, please see full course syllabus of AP Physics C: Mechanics
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Lecture Comments (16)

2 answers

Last reply by: Professor Dan Fullerton
Tue Mar 21, 2017 6:12 AM

Post by Woong Ryeol Yoo on March 20 at 07:44:12 PM

Hi Mr. Fullerton.

I was wondering to what extent of knowlege of trig identities need for Ap Physics C mech exam.
Do I have to know like coscos-sinsin=cos and other complex forumulas?
Or is knowing only sin, cos, tan, sin sq + cos sq = 1 enough for the to exam?

1 answer

Last reply by: Professor Dan Fullerton
Wed Aug 17, 2016 1:08 PM

Post by Cathy Zhao on August 16, 2016

I still feel confused about the right hand rule. Can you explain it a little bit more? Thanks

2 answers

Last reply by: Shikha Bansal
Fri May 27, 2016 3:46 PM

Post by Shikha Bansal on May 24, 2016

Hi Mr.Fullerton
I have finished ap physics 1 as well as algebra 2 this year in school, and I was hoping to study physics C on here over the summer as I would love to go in more depth on physics. However, I do not know much precal currrently. Should I learn precal first or is this math review enough to more or less get me ready for the math in this course?

1 answer

Last reply by: Professor Dan Fullerton
Wed Apr 1, 2015 9:50 AM

Post by Luvivia Chang on March 31, 2015

Hello Professor Dan Fullerton
Can we take the derivative of a vector? Because in the part of dot product properties, you write "d(vector A)/dt (dot)times vector B" while in the properties of cross product, you just write "dA/dt (cross)times vector B" .Is there any difference between the two?
Thank you.

3 answers

Last reply by: Professor Dan Fullerton
Tue Dec 23, 2014 6:53 AM

Post by John Powell on December 20, 2014

Should dot product property given as "associative" be distributive?

1 answer

Last reply by: Daniel Fullerton
Sun Dec 7, 2014 3:25 PM

Post by Shaina M on December 6, 2014

On l
Slide with examples of derivatives the last one is wrong. Instead of 12x it's supposed to be 12x^5.

Related Articles:

Math Review

  • Matter is anything that has mass and takes up space. It is the amount of "stuff" making up an object. Mass is measured in kilograms.
  • Energy is the ability or capacity to do work. Work is the process of moving an object. Therefore, energy is the ability or capacity to move an object.
  • Mass-Energy Equivalence states that the mass of an object is really a measure of its energy.
  • The source of all energy on Earth is the conversion of mass into energy.
  • Physics is the study of matter and energy. This course will focus on mechanics, fluids, thermal physics, electricity and magnetism, waves and optics, and selected topics in modern physics.

Math Review

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  1. Intro
    • Objectives
      • Vectors and Scalars
      • Vector Representations
      • Graphical Vector Addition
      • Graphical Vector Subtraction
      • Vector Components
      • Angle of a Vector
      • Vector Notation
      • Example I: Magnitude of the Horizontal & Vertical Component
        • Example II: Magnitude of the Plane's Eastward Velocity
          • Example III: Magnitude of Displacement
            • Example IV: Total Displacement from Starting Position
              • Example V: Find the Angle Theta Depicted by the Diagram
                • Vector Notation, cont.
                • Vector Multiplication
                • Dot Product
                • Defining the Dot Product
                • Calculating the Dot Product
                • Example VI: Calculating a Dot Product
                • Special Dot Products
                • Dot Product Properties
                • Example VII: Perpendicular Vectors
                  • Cross Product
                  • Defining the Cross Product
                  • Calculating the Cross Product Unit Vector Notation
                  • Calculating the Cross Product Matrix Notation
                  • Example VII: Find the Cross Product of the Following Vectors
                    • Cross Product Properties
                    • Units
                    • Example IX: Dimensional Analysis
                      • Calculus
                      • Differential Calculus
                      • Example X: Derivatives
                        • Integral Calculus
                        • Common Integrations
                        • Example XI: Integrals
                          • Example XII: Calculus Applications
                            • Intro 0:00
                            • Objectives 0:10
                            • Vectors and Scalars 1:06
                              • Scalars
                              • Vectors
                            • Vector Representations 2:00
                              • Vector Representations
                            • Graphical Vector Addition 2:54
                              • Graphical Vector Addition
                            • Graphical Vector Subtraction 5:36
                              • Graphical Vector Subtraction
                            • Vector Components 7:12
                              • Vector Components
                            • Angle of a Vector 8:56
                              • tan θ
                              • sin θ
                              • cos θ
                            • Vector Notation 10:10
                              • Vector Notation 1
                              • Vector Notation 2
                            • Example I: Magnitude of the Horizontal & Vertical Component 16:08
                            • Example II: Magnitude of the Plane's Eastward Velocity 17:59
                            • Example III: Magnitude of Displacement 19:33
                            • Example IV: Total Displacement from Starting Position 21:51
                            • Example V: Find the Angle Theta Depicted by the Diagram 26:35
                            • Vector Notation, cont. 27:07
                              • Unit Vector Notation
                              • Vector Component Notation
                            • Vector Multiplication 28:39
                              • Dot Product
                              • Cross Product
                            • Dot Product 29:03
                              • Dot Product
                            • Defining the Dot Product 29:26
                              • Defining the Dot Product
                            • Calculating the Dot Product 29:42
                              • Unit Vector Notation
                              • Vector Component Notation
                            • Example VI: Calculating a Dot Product 31:45
                              • Example VI: Part 1 - Find the Dot Product of the Following Vectors
                              • Example VI: Part 2 - What is the Angle Between A and B?
                            • Special Dot Products 33:52
                              • Dot Product of Perpendicular Vectors
                              • Dot Product of Parallel Vectors
                            • Dot Product Properties 34:51
                              • Commutative
                              • Associative
                              • Derivative of A * B
                            • Example VII: Perpendicular Vectors 35:47
                            • Cross Product 36:42
                              • Cross Product of Two Vectors
                              • Direction Using the Right-hand Rule
                              • Cross Product of Parallel Vectors
                            • Defining the Cross Product 38:13
                              • Defining the Cross Product
                            • Calculating the Cross Product Unit Vector Notation 38:41
                              • Calculating the Cross Product Unit Vector Notation
                            • Calculating the Cross Product Matrix Notation 39:18
                              • Calculating the Cross Product Matrix Notation
                            • Example VII: Find the Cross Product of the Following Vectors 42:09
                            • Cross Product Properties 45:16
                              • Cross Product Properties
                            • Units 46:41
                              • Fundamental Units
                              • Derived units
                            • Example IX: Dimensional Analysis 47:21
                            • Calculus 49:05
                              • Calculus
                            • Differential Calculus 49:49
                              • Differentiation & Derivative
                            • Example X: Derivatives 51:21
                            • Integral Calculus 53:03
                              • Integration
                              • Integral
                              • Integration & Derivation are Inverse Functions
                              • Determine the Original Function
                            • Common Integrations 54:45
                              • Common Integrations
                            • Example XI: Integrals 55:17
                            • Example XII: Calculus Applications 58:32

                            Transcription: Math Review

                            Hello, everyone, and welcome back to

                            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 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 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

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