For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

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### 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.

- 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
- 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
- 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

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: 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 and *0027

*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 again *0262

*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 hand *0311

*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 over *0389

*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 tan *0553

*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 direction *0684

*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 space *0859

*it is starting get pretty tough to see what is going on if you we want to line this up tip to tail *0864

*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 by *1278

*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 East*1346

*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 to *1453

*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 using *1706

*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 vector *1784

*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 them*1978

*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 equal *2204

*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 A *2258

*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 direction *2309

*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 is*2409

*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 see *2541

*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 second*2848

*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 constant*2906

*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 and *2956

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

*You can find a much more thorough and detailed accounting of calculus *2963

*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 point*3015

*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 ⁺n*3028

*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 doing *3206

*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 away *3273

*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 that*3332

*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 x² *3409

*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

1 answer

Last reply by: Professor Dan Fullerton

Wed Aug 17, 2016 1:08 PM

Post by Cathy Zhao on August 16 at 11:02:37 PM

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 at 03:05:15 PM

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?

Thanks!

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.