WEBVTT physics/ap-physics-c-mechanics/fullerton
00:00:00.000 --> 00:00:03.300
Hello, everyone, and welcome back to www.educator.com.
00:00:03.300 --> 00:00:09.600
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.
00:00:09.600 --> 00:00:17.100
To begin with, our objectives are going to be explaining how vectors and scalars are used to describe physical quantities.
00:00:17.100 --> 00:00:22.200
Calculating the dot and cross products of vectors, a little bit of vector multiplication.
00:00:22.200 --> 00:00:27.100
Utilizing dimensional analysis to evaluate the units of a quantity.
00:00:27.100 --> 00:00:33.000
Calculating the derivative, a basic functions, calculating the integral of basic functions and
00:00:33.000 --> 00:00:37.500
explaining the meaning of the derivative and integral in terms of graphical analysis.
00:00:37.500 --> 00:00:42.500
I know at this point a lot of folk start to get worried about the math involved in physics.
00:00:42.500 --> 00:00:47.000
This is a calculus based math, a calculus base physics course.
00:00:47.000 --> 00:00:50.400
However that does not mean it is a math course.
00:00:50.400 --> 00:00:53.000
The math is used as the language of physics.
00:00:53.000 --> 00:00:57.000
To help explain things much more efficiently than you could in words.
00:00:57.000 --> 00:01:03.900
It is not really about math, it is about the physics applying those math principles and putting them to good use.
00:01:03.900 --> 00:01:08.100
Let us start by talking about vectors and scalars.
00:01:08.100 --> 00:01:12.200
Scalars are physical quantities that have a magnitude only.
00:01:12.200 --> 00:01:15.300
They do not need to be described at the direction.
00:01:15.300 --> 00:01:21.900
Things like temperature, mass, time, all of those are scalar quantities.
00:01:21.900 --> 00:01:28.100
We could say when time move forward and backward but we are talking North, South, East, West directions.
00:01:28.100 --> 00:01:33.000
If it has a direction we call it a vector quantity, things like velocity.
00:01:33.000 --> 00:01:38.300
You are driving 55mph down the highway west.
00:01:38.300 --> 00:01:43.400
Force, I pushed Susie forward.
00:01:43.400 --> 00:01:47.500
Acceleration, I accelerated to the east.
00:01:47.500 --> 00:01:50.400
They have a direction as well as a magnitude.
00:01:50.400 --> 00:01:58.400
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,
00:01:58.400 --> 00:02:00.900
if we want to show vectors graphically.
00:02:00.900 --> 00:02:04.000
Let us take a look at a couple vector representations.
00:02:04.000 --> 00:02:07.600
Let us assume that we have some vector in blue here.
00:02:07.600 --> 00:02:14.400
Let us call it A and we have another vector here in red - let us call it B.
00:02:14.400 --> 00:02:20.800
They both have the same direction but B has a larger magnitude than A, that is pretty straightforward.
00:02:20.800 --> 00:02:29.200
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.
00:02:29.200 --> 00:02:32.400
We could take this A vector if we wanted to.
00:02:32.400 --> 00:02:37.300
Let us say we slide it down here and I am going to redraw our A vector.
00:02:37.300 --> 00:02:40.500
I think it is roughly that length.
00:02:40.500 --> 00:02:45.200
We will call that our new A vector and make that one go away.
00:02:45.200 --> 00:02:55.100
Perfectly reasonable thing to do as a long as you keep the same magnitude and directions, vectors are free to move around in space.
00:02:55.100 --> 00:02:57.800
We can also add two vectors.
00:02:57.800 --> 00:03:06.300
We have vector A here in red and we have vector B here in blue.
00:03:06.300 --> 00:03:16.100
We wanted to add A + B to get the vector C.
00:03:16.100 --> 00:03:20.700
The way we do that is by aligning these two vectors up tip to tail.
00:03:20.700 --> 00:03:30.600
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.
00:03:30.600 --> 00:03:36.500
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.
00:03:36.500 --> 00:03:49.200
Its tip is touching the tail of vector B.
00:03:49.200 --> 00:03:55.000
Once you have the vectors lined up tip to tail in order to find the addition of those two,
00:03:55.000 --> 00:04:04.800
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.
00:04:04.800 --> 00:04:12.700
That would be vector C, the resultant of A + B or what happens when you add vectors A and B.
00:04:12.700 --> 00:04:17.000
It does not matter which order you do this or how many vectors you do with it.
00:04:17.000 --> 00:04:22.200
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.
00:04:22.200 --> 00:04:31.100
Let us demonstrate that for a second by taking our B vector I am going to redraw again
00:04:31.100 --> 00:04:36.600
and going to draw it down here, roughly the same length and same direction.
00:04:36.600 --> 00:04:41.600
But now instead of having A point to its end, I'm going to have that point it to the end of A.
00:04:41.600 --> 00:04:51.700
A was right about there so we will try drawing A here again tip to tail but with a different vector in front.
00:04:51.700 --> 00:05:02.900
Once again to find the resultant I go from the starting point of my first vector to the endpoint of my last vector.
00:05:02.900 --> 00:05:11.100
There is C notice regardless of how I did it which order I have roughly the same magnitude and same direction.
00:05:11.100 --> 00:05:14.300
You can tell they are a little bit off on the drawing because I'm doing it by hand
00:05:14.300 --> 00:05:18.100
and very carefully with the protractor so we are not here all day.
00:05:18.100 --> 00:05:22.500
But that also shows the order of addition for vectors does not matter.
00:05:22.500 --> 00:05:32.900
We could just as easily have written B + A = C vector.
00:05:32.900 --> 00:05:36.000
There is graphical vector addition.
00:05:36.000 --> 00:05:39.500
Let us take a look at graphical vector subtraction.
00:05:39.500 --> 00:05:42.000
We will have A in red again.
00:05:42.000 --> 00:05:52.400
We will put B over here in blue and now if we want to know what A – B.
00:05:52.400 --> 00:06:04.700
The trick to doing that is realizing that subtraction is just addition of a negative that is the same as saying A + -B.
00:06:04.700 --> 00:06:10.300
We are going to have A + -B we are going to call that vector D.
00:06:10.300 --> 00:06:14.200
We want to add A + -B because we know how to add vectors.
00:06:14.200 --> 00:06:16.900
We got b here but how do we get the negative?
00:06:16.900 --> 00:06:18.500
It as easy as you might think.
00:06:18.500 --> 00:06:26.700
If that direction B, if that is vector B, all we have to do to get –B is switch its direction.
00:06:26.700 --> 00:06:29.100
There is –B.
00:06:29.100 --> 00:06:37.200
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
00:06:37.200 --> 00:06:43.800
so it is right about there roughly the same length and direction.
00:06:43.800 --> 00:06:53.200
Line them up tip to tail so if this is A+ -B is the same A – B.
00:06:53.200 --> 00:06:59.700
Draw a line from the starting point of my first vector to the ending point of my last.
00:06:59.700 --> 00:07:01.300
I suppose we made D in purple.
00:07:01.300 --> 00:07:05.200
Let us do that, there is D in purple.
00:07:05.200 --> 00:07:12.300
There would be vector D, graphical vector subtraction.
00:07:12.300 --> 00:07:16.200
Sometimes dealing with these graphically can get little bit tedious.
00:07:16.200 --> 00:07:25.400
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.
00:07:25.400 --> 00:07:31.000
Let us say that we have a vector A here at some angle θ with the horizontal.
00:07:31.000 --> 00:07:38.900
It can be considerably more efficient to break into a component that is along the x axis.
00:07:38.900 --> 00:07:42.300
Let us do the y axis first just to make it easier to draw.
00:07:42.300 --> 00:08:00.500
We call this the y component of vector A and we will have some x component of vector A.
00:08:00.500 --> 00:08:07.000
Noticed that A axis a vector + y as a vector gives you the total A.
00:08:07.000 --> 00:08:14.100
You could replace vector A with the equivalent set of vectors Ax and Ay where Ax is in the x direction.
00:08:14.100 --> 00:08:19.700
Ay is along the y axis so we could break that up into components.
00:08:19.700 --> 00:08:31.000
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.
00:08:31.000 --> 00:08:38.500
That is going to be equal to the magnitude of A our total vector × sin angle θ.
00:08:38.500 --> 00:08:53.200
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.
00:08:53.200 --> 00:08:56.500
How about finding the angle then?
00:08:56.500 --> 00:09:01.800
Here is a right triangle, we have an adjacent and opposite side and the hypotenuse.
00:09:01.800 --> 00:09:04.900
If we know 2 or 3 sides we can find the angle.
00:09:04.900 --> 00:09:13.000
The tan of θ is the opposite side over the adjacent side.
00:09:13.000 --> 00:09:19.300
If we know those two sides then the angle between them, θ is going to be the inverse tan
00:09:19.300 --> 00:09:27.500
of the opposite side over the adjacent side not the angle between the angle of the triangle.
00:09:27.500 --> 00:09:33.200
If you know the opposite and the hypotenuse you can use the sin θ.
00:09:33.200 --> 00:09:45.800
Sin θ that is opposite over hypotenuse therefore θ will be the inverse sin of the opposite over the hypotenuse.
00:09:45.800 --> 00:09:55.000
If you know the adjacent side and hypotenuse, cos θ is adjacent over hypotenuse.
00:09:55.000 --> 00:10:04.300
When you know those two you can find θ is the inverse cos of the adjacent divided by the hypotenuse.
00:10:04.300 --> 00:10:08.900
If you know any two of the three sides you can go find the angles.
00:10:08.900 --> 00:10:11.200
It is so wonderful.
00:10:11.200 --> 00:10:19.400
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.
00:10:19.400 --> 00:10:22.400
It is helpful to have some standard notation to help you deal with it.
00:10:22.400 --> 00:10:27.500
Also if you are using a textbook, a lot of times they are different notational styles in different textbooks.
00:10:27.500 --> 00:10:32.600
Probably we are talking about those for a little bit to have some consistency throughout the course.
00:10:32.600 --> 00:10:39.100
First thing I am going to do is I'm going to draw a three dimensional axis up here.
00:10:39.100 --> 00:10:55.100
We will give ourselves y, x and z.
00:10:55.100 --> 00:10:57.700
There is a three dimensional axis to begin with.
00:10:57.700 --> 00:11:04.700
If we have some vector let us call it A, it can have components in the x, y, and z directions.
00:11:04.700 --> 00:11:13.700
We could write that as in this bracket notation is its x value, its y value, and that z value.
00:11:13.700 --> 00:11:17.200
That is a fairly common way of writing these and one of my favorites.
00:11:17.200 --> 00:11:24.200
If you take a moment you define what we call some unit vectors, there are some other ways we can deal with this two.
00:11:24.200 --> 00:11:34.200
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
00:11:34.200 --> 00:11:38.900
that special vector we are going to call I ̂.
00:11:38.900 --> 00:11:43.300
The hat means is the unit vector, its length and its magnitude is always 1.
00:11:43.300 --> 00:11:46.000
In the x direction we call it I ̂.
00:11:46.000 --> 00:11:50.100
In the y direction, we are going to do the same basic thing.
00:11:50.100 --> 00:11:56.100
Make a unit vector of length 1 and we are going to call it j ̂.
00:11:56.100 --> 00:11:59.700
In the z direction I am sure you have not guess by now.
00:11:59.700 --> 00:12:06.400
A unit vector of length 1 we are going to call k ̂.
00:12:06.400 --> 00:12:20.500
We could also write our A vector now whatever happens to be as having components of x coordinate × I ̂.
00:12:20.500 --> 00:12:40.100
That is the unit of vector 1 in the x direction + y value × j ̂ + z value × k ̂.
00:12:40.100 --> 00:12:43.100
That will get a little funky until you get used to it.
00:12:43.100 --> 00:12:50.000
That is pretty common in a bunch of textbooks to see these unit vectors along with the vector in front of them.
00:12:50.000 --> 00:12:56.900
These two types of notation really mean the exact same thing.
00:12:56.900 --> 00:12:59.200
Let us see how that can be useful down here.
00:12:59.200 --> 00:13:04.000
Let us try it again with a fairly simple vector addition problem.
00:13:04.000 --> 00:13:25.300
I am going to draw my axis again, give ourselves y, x, and the z axis.
00:13:25.300 --> 00:13:28.500
What we are going to do is we are going to define a couple of vectors.
00:13:28.500 --> 00:13:32.000
The first one I am going to define is called P.
00:13:32.000 --> 00:13:43.900
Where P goes from the origin to some point that is 4/x, 3 in the y, 1 in the z.
00:13:43.900 --> 00:13:47.900
It might be somewhere about there.
00:13:47.900 --> 00:13:55.700
We will draw our vector from the origin there and that is our point P 4, 3, 1.
00:13:55.700 --> 00:14:02.200
Let us take a second vector and we will call it q.
00:14:02.200 --> 00:14:08.800
We are going to go 2 points in the x, 0 in the y, 4 in the z.
00:14:08.800 --> 00:14:19.500
We call this the vector q which leads us to q 2, 0, 4.
00:14:19.500 --> 00:14:24.500
If we want to add this two to get r well when you are in three dimensional space
00:14:24.500 --> 00:14:28.800
it is starting get pretty tough to see what is going on if you we want to line this up tip to tail
00:14:28.800 --> 00:14:34.200
and come up with a reasonable solution that you can actually make sense of graphically.
00:14:34.200 --> 00:14:58.800
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.
00:14:58.800 --> 00:15:06.700
Our q vector is equal to 2, 0, 4.
00:15:06.700 --> 00:15:16.900
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.
00:15:16.900 --> 00:15:28.000
4 + 2= 6, 3 + 0= 3, 1 + 4= 5 so 6, 3, 5 will give us the vector to our resultant.
00:15:28.000 --> 00:15:30.300
Let us draw that in here.
00:15:30.300 --> 00:15:38.700
We go 1, 2, 3, 4 about 6 on the x, 3 in the y, 5 on the z.
00:15:38.700 --> 00:15:48.900
As I draw it is going to come out somewhere around there depending on your perspective we will call that r.
00:15:48.900 --> 00:15:52.800
What if you just did that graphically which you can get by lining these up?
00:15:52.800 --> 00:15:55.800
It is going to be a really tough to see what is going on.
00:15:55.800 --> 00:16:07.500
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.
00:16:07.500 --> 00:16:10.300
A little bit more with vector components.
00:16:10.300 --> 00:16:17.500
A soccer player kicks the ball with an initial velocity of 10 m/s if an angle of 30° above the horizontal.
00:16:17.500 --> 00:16:22.900
Find the magnitude of the horizontal component in vertical component of the balls velocity.
00:16:22.900 --> 00:16:28.900
I like to start with graphs wherever possible to help me visualize the problem.
00:16:28.900 --> 00:16:31.500
And this look like it is in two dimensions.
00:16:31.500 --> 00:16:35.700
We got a vertical and horizontal component to our problem.
00:16:35.700 --> 00:16:45.700
We call this our x and y and it has an initial velocity of 10 m/s in angle of 30° above the horizontal.
00:16:45.700 --> 00:16:50.100
Let us see if we can eyeball roughly 30°.
00:16:50.100 --> 00:16:56.200
The magnitude of our vector is 10 m/s at some angle 30°.
00:16:56.200 --> 00:16:59.800
We want the horizontal component and vertical component.
00:16:59.800 --> 00:17:02.200
Let us draw these in first.
00:17:02.200 --> 00:17:13.300
Our vertical component will be that piece and our horizontal component will be that piece.
00:17:13.300 --> 00:17:21.700
The x component of V, our initial velocity V = 10 m/s.
00:17:21.700 --> 00:17:30.900
Our x component is going to be V cos 30° because we are looking for the adjacent side of that right triangle.
00:17:30.900 --> 00:17:35.600
It is going to be about 8.66 m/s.
00:17:35.600 --> 00:17:43.500
The vertical component Vy is going to be V sin 30°.
00:17:43.500 --> 00:17:48.700
We got the opposite side there which is going to be about 5 m/s.
00:17:48.700 --> 00:17:56.000
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.
00:17:56.000 --> 00:18:00.100
Let us take a look at the second example.
00:18:00.100 --> 00:18:05.900
An airplane flies with the velocity of 750 kph 30° South East.
00:18:05.900 --> 00:18:10.000
What is the magnitude of the planes eastward velocity?
00:18:10.000 --> 00:18:29.500
Lets draw a diagram again and we would call that north and south so this must be east and west.
00:18:29.500 --> 00:18:43.300
It is traveling the velocity 750 kph 30° South of East in this basic direction.
00:18:43.300 --> 00:18:53.400
Our V = 750 kph and an angle of 30° South of East.
00:18:53.400 --> 00:18:58.800
The magnitude of the planes eastward velocity, it looks like we are after an x component here.
00:18:58.800 --> 00:19:08.100
Let us draw that in, we are after just this piece, the x component.
00:19:08.100 --> 00:19:17.500
Vx equals that is the adjacent sides and that going to be V cos θ which is going to be 750 kph.
00:19:17.500 --> 00:19:30.400
The magnitude of our entire vector × the cos 30° and that can give you about 650 kph.
00:19:30.400 --> 00:19:35.300
How about the problem with the magnitude of vector?
00:19:35.300 --> 00:19:39.400
A dog walks a lady 8m due north and then 6m due east.
00:19:39.400 --> 00:19:41.400
You probably all seen that before.
00:19:41.400 --> 00:19:45.200
It is a really big dog and little even lady the dogs doing the controlling.
00:19:45.200 --> 00:19:49.900
Determine the magnitude of the dog’s total displacement.
00:19:49.900 --> 00:19:54.100
The way I do that is looks like we have a couple of vectors that we can add up.
00:19:54.100 --> 00:20:11.500
Dogs walk lady 8m due north so I draw vector magnitude 8m due north and then 6m due east.
00:20:11.500 --> 00:20:14.400
Determine the magnitude of the dog’s total displacement.
00:20:14.400 --> 00:20:19.200
Displacement being the straight line distance from where you start to where you finish.
00:20:19.200 --> 00:20:28.200
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.
00:20:28.200 --> 00:20:33.400
That red vector represents the total displacement.
00:20:33.400 --> 00:20:36.000
How do I find the magnitude of that?
00:20:36.000 --> 00:20:39.400
It is a right triangle I can use that Pythagorean theorem.
00:20:39.400 --> 00:20:46.100
A² + B²=C².
00:20:46.100 --> 00:20:57.600
Our hypotenuse is going to be =√(a² + b² ) which is the √(8² + 6²).
00:20:57.600 --> 00:21:10.700
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.
00:21:10.700 --> 00:21:15.600
If we want to find the angle, let us define an x axis.
00:21:15.600 --> 00:21:18.200
What if we wanted to find that angle θ?
00:21:18.200 --> 00:21:25.800
We could if we wanted to θ equals inverse tan of the opposite side of a right triangle divided by
00:21:25.800 --> 00:21:30.600
the adjacent side that could be the inverse tan of.
00:21:30.600 --> 00:21:44.900
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°.
00:21:44.900 --> 00:21:49.700
We are not asked for that but if we were we would know how to go calculate it.
00:21:49.700 --> 00:21:52.400
Let us look at the vector addition problem.
00:21:52.400 --> 00:21:56.600
A frog hops 4m angle of 30° North of East.
00:21:56.600 --> 00:22:00.300
He then hops 6m angle of 60° North of West.
00:22:00.300 --> 00:22:04.200
What was the frog total displacement from his starting position?
00:22:04.200 --> 00:22:07.700
Alright we are getting a little bit more challenging here.
00:22:07.700 --> 00:22:26.100
Let us draw what we have north, east, and west.
00:22:26.100 --> 00:22:31.500
Frog hops 4m angle of 30° North of East
00:22:31.500 --> 00:22:39.200
He goes 4m North of East.
00:22:39.200 --> 00:22:45.800
Let us called about 4m assuming that is 30° and I'm just estimating these.
00:22:45.800 --> 00:22:52.400
He then hops 6m an angle of 60° North of West.
00:22:52.400 --> 00:22:55.400
To figure out what that is let us draw x and y here.
00:22:55.400 --> 00:23:19.700
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.
00:23:19.700 --> 00:23:23.000
There is our second piece of 60°.
00:23:23.000 --> 00:23:27.000
What is the frog’s total displacement from his starting position?
00:23:27.000 --> 00:23:33.700
Thought displacement goes from the starting point of our first to the ending point of our last.
00:23:33.700 --> 00:23:45.700
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.
00:23:45.700 --> 00:23:51.200
Let us take a look at how we add A and B up to get C in vector notation.
00:23:51.200 --> 00:24:08.000
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°.
00:24:08.000 --> 00:24:13.400
4m and our angle is 30° we can break that to x and y components.
00:24:13.400 --> 00:24:22.400
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
00:24:22.400 --> 00:24:37.100
begin the x piece is going to be -6 m cos 60° in the y component 6m sin 60°.
00:24:37.100 --> 00:24:46.800
If we then want to find out our total C, our C vector is just going to be the sum of those.
00:24:46.800 --> 00:25:15.400
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°.
00:25:15.400 --> 00:25:24.500
And if we put that all together I find that our C vector is about 0.46m, 7.2m.
00:25:24.500 --> 00:25:33.500
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.
00:25:33.500 --> 00:25:42.300
If we want to know the magnitude of our answer the magnitude of C like that is going to be.
00:25:42.300 --> 00:26:01.800
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.
00:26:01.800 --> 00:26:08.700
If want the angle from the origin we can go back to our trig.
00:26:08.700 --> 00:26:27.100
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.
00:26:27.100 --> 00:26:36.000
You can see how valuable, how useful these vectors can be and breaking them up into components in order to manipulate them.
00:26:36.000 --> 00:26:42.100
Find the angle θ depicted by the blue vector below given the x and y components.
00:26:42.100 --> 00:26:46.800
Let just hit this again because it is often times a trouble spot as we are getting started.
00:26:46.800 --> 00:26:55.200
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.
00:26:55.200 --> 00:27:06.100
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°.
00:27:06.100 --> 00:27:09.300
Let us go hit vector notation a little bit more.
00:27:09.300 --> 00:27:26.400
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 ̂.
00:27:26.400 --> 00:27:34.400
The vector component notation we would write that S or bracket notation x, y, z.
00:27:34.400 --> 00:27:38.200
You will see vectors written in many different ways in many different textbooks.
00:27:38.200 --> 00:27:50.400
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.
00:27:50.400 --> 00:27:59.600
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.
00:27:59.600 --> 00:28:06.700
If you want the magnitude of the vector you have the vector symbol inside absolute value signs that would be a magnitude.
00:28:06.700 --> 00:28:12.500
In other books you will see double lines surrounded to indicate magnitude.
00:28:12.500 --> 00:28:16.500
Still in other if you see A written bold for A vector.
00:28:16.500 --> 00:28:26.500
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.
00:28:26.500 --> 00:28:30.200
Take a look at your book that you are using and try and find out what it is using
00:28:30.200 --> 00:28:33.200
and maintain some consistency with that throughout the course.
00:28:33.200 --> 00:28:38.300
We will use a couple of these just you get used to all the different forms of notation.
00:28:38.300 --> 00:28:43.800
We can add vectors we can subtract vectors we can also multiply vectors.
00:28:43.800 --> 00:28:46.900
But there are two types of vector notation.
00:28:46.900 --> 00:28:54.300
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.
00:28:54.300 --> 00:28:59.900
The cross products or vector product gives you a vector as the output of the multiplication.
00:28:59.900 --> 00:29:05.300
We will talk about these different types starting with the Dot product and the Scalar product.
00:29:05.300 --> 00:29:14.300
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.
00:29:14.300 --> 00:29:25.400
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.
00:29:25.400 --> 00:29:27.800
How can we define that a little bit better?
00:29:27.800 --> 00:29:36.900
A • B = to the magnitude of A magnitude × magnitude of B × cos of the angle between those two vectors.
00:29:36.900 --> 00:29:41.400
Another definition that you are really going to want to know.
00:29:41.400 --> 00:29:44.200
Let us see how we can calculate this.
00:29:44.200 --> 00:29:59.300
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
00:29:59.300 --> 00:30:10.300
in the y direction j ̂ + the Z component of A × the unit vector in Z direction known as k ̂.
00:30:10.300 --> 00:30:26.500
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 ̂.
00:30:26.500 --> 00:30:47.300
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.
00:30:47.300 --> 00:30:53.500
Multiply the y components together + az bz.
00:30:53.500 --> 00:30:55.800
And no unit vectors because it is scalar.
00:30:55.800 --> 00:30:59.800
The dot product gives you a scalar output.
00:30:59.800 --> 00:31:36.300
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.
00:31:36.300 --> 00:31:44.100
Depending on the type of notation you prefer still get the dot product of the exact same formula.
00:31:44.100 --> 00:31:46.300
Let us do it.
00:31:46.300 --> 00:31:55.500
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.
00:31:55.500 --> 00:31:57.800
Couple ways we could do this.
00:31:57.800 --> 00:32:05.500
We will start off doing it the easy way.
00:32:05.500 --> 00:32:20.600
A • B is going to be the x components multiplied 3 + the y components multiplied 4 + z components multiplied 3 or 10.
00:32:20.600 --> 00:32:25.400
We also mentioned that you can find that by AB cos θ.
00:32:25.400 --> 00:32:29.300
Let us do that while you are here but first we need to find the magnitude of A.
00:32:29.300 --> 00:32:36.100
The magnitude of that vector A we can use our Pythagorean theorem that is going to be the square root of the component squared.
00:32:36.100 --> 00:32:44.800
1² + 2² + 3² which would be √14.
00:32:44.800 --> 00:32:58.600
The magnitude of B is going to be √(3² + 2² + 1²) also √14.
00:32:58.600 --> 00:33:13.400
We could also look at A • B as AB cos θ their magnitude × cos of the angle between them
00:33:13.400 --> 00:33:26.200
which implies that we know A • B is 10 must be equals √14 × √14 is this going to be 14 cos θ.
00:33:26.200 --> 00:33:45.500
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.
00:33:45.500 --> 00:33:49.700
Couple different ways you can do this.
00:33:49.700 --> 00:33:53.300
Let us take a look at a couple of special dot products.
00:33:53.300 --> 00:34:00.200
The dot product of perpendicular vectors is always 0 because there is no component of 1 that lie on the other.
00:34:00.200 --> 00:34:03.300
If their dot product is 0 they are perpendicular.
00:34:03.300 --> 00:34:08.900
The dot product of parallel vectors is just the product of their magnitudes.
00:34:08.900 --> 00:34:21.400
One way we could look at this is if we are talking about A • B is AB cos θ.
00:34:21.400 --> 00:34:29.200
If the angle between θ is 0 cos of 0 is 1 you just get the product of their magnitudes.
00:34:29.200 --> 00:34:36.000
If however they are perpendicular write that specifically if θ = 0 they are parallel.
00:34:36.000 --> 00:34:51.400
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.
00:34:51.400 --> 00:34:53.900
A couple of dot product properties.
00:34:53.900 --> 00:35:07.700
First off the commutative property A • B = B • A that works it is commutative.
00:35:07.700 --> 00:35:24.900
You have A + B vectors .C = A • C + B • C associative property.
00:35:24.900 --> 00:35:46.600
If you are taking a derivative, the derivative of A • B = to the derivative of A • B + the derivative of B • A.
00:35:46.600 --> 00:35:49.700
Let us do a couple more examples here.
00:35:49.700 --> 00:35:58.600
If A -2, 3 and B is 4, by find a value of By such A and B are perpendicular vectors.
00:35:58.600 --> 00:36:10.300
The way to start this is recognizing that if they are perpendicular then their dot product A • B must be = 0.
00:36:10.300 --> 00:36:16.800
A • B =0 let us take a look what happens when we do our dot product.
00:36:16.800 --> 00:36:30.900
The x components multiplied we get -8 + 3By = 0 or 3By = 8.
00:36:30.900 --> 00:36:41.100
By = 8/3 if they are going to be perpendicular.
00:36:41.100 --> 00:36:43.700
That is the dot product.
00:36:43.700 --> 00:36:48.800
The cross product of two vectors gives you a vector perpendicular to both because magnitude is equal
00:36:48.800 --> 00:36:53.000
to the area of the parallelogram defined by the two initial vectors.
00:36:53.000 --> 00:37:10.000
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.
00:37:10.000 --> 00:37:16.200
The area of a parallelogram defined by those vectors let us see if we can scope that out a little bit.
00:37:16.200 --> 00:37:32.200
I will draw something kind of like that and the area of that parallelogram defined by AB is the magnitude of your vector C.
00:37:32.200 --> 00:37:38.500
And C going to be perpendicular to both A and B where its direction is given by the right hand rule.
00:37:38.500 --> 00:37:46.400
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
00:37:46.400 --> 00:37:54.300
bend them in the direction of vector B and your thumb points in the direction that is positive for C.
00:37:54.300 --> 00:37:59.300
A cross with B gives you C a right-hand rule for cross products.
00:37:59.300 --> 00:38:04.600
And that is going to come up in this course multiple times as well as in the ENM course.
00:38:04.600 --> 00:38:14.700
Now interesting to note the cross product of parallel vectors is 0 which it has to be because they cannot define parallelogram.
00:38:14.700 --> 00:38:17.400
Defining the cross product.
00:38:17.400 --> 00:38:22.200
A × B the magnitude of A cross B is AB sin θ.
00:38:22.200 --> 00:38:29.100
The direction given by right-hand rule where is before A • B is value was AB cos θ.
00:38:29.100 --> 00:38:35.400
The cross product only the magnitude of the vector is AB sin θ still have to worry about direction
00:38:35.400 --> 00:38:40.700
because the cross product it outputs a vector not a scalar.
00:38:40.700 --> 00:38:55.600
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.
00:38:55.600 --> 00:39:04.500
The y component Az Bx – Ax Bz in the y direction the j ̂ component.
00:39:04.500 --> 00:39:17.400
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.
00:39:17.400 --> 00:39:22.700
Or the way I tend to do it is with some linear algebra looking at the determinant.
00:39:22.700 --> 00:39:34.800
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.
00:39:34.800 --> 00:39:47.600
I would start by drawing A 3 × 3 matrix I ̂ at the top, j, ̂ and k ̂ is my first row.
00:39:47.600 --> 00:40:05.400
My second row Ax Ay Az and my third row Bx By Bz.
00:40:05.400 --> 00:40:08.800
And that is we are going to take the determinant of.
00:40:08.800 --> 00:40:14.200
When you do that, the way I do this to help me understand and to help do this a little bit more easily is
00:40:14.200 --> 00:40:17.300
I also repeat these over to the right and left.
00:40:17.300 --> 00:40:27.600
Ax Ay Az what comes next in the pattern would be Ax Ay and down here we have Bx By.
00:40:27.600 --> 00:40:33.500
I also need that over here to the left so over here I will have Az before we get Ax.
00:40:33.500 --> 00:40:41.700
We will have Ay there, we have a By, and the Bz here.
00:40:41.700 --> 00:40:52.600
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.
00:40:52.600 --> 00:41:20.700
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.
00:41:20.700 --> 00:41:58.900
I have Az Bx – Ax Bz × j ̂ + Ax By – Ay Bx J ̂.
00:41:58.900 --> 00:42:01.300
That is another way you can come up with the formula.
00:42:01.300 --> 00:42:10.200
And typically this is a lot easier for me to remember how to do than memorizing that entire previous formula.
00:42:10.200 --> 00:42:19.400
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.
00:42:19.400 --> 00:42:21.200
A couple ways we can do this.
00:42:21.200 --> 00:42:25.900
First let us start with looking at the magnitude of A and pretty easy to see
00:42:25.900 --> 00:42:35.500
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.
00:42:35.500 --> 00:42:51.500
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.
00:42:51.500 --> 00:42:59.500
If we want to know the magnitude of C, magnitude of C is going to be magnitude of A.
00:42:59.500 --> 00:43:08.900
Magnitude of B × the sin of the angle between them is going to be 2 × 2 × the sin of the angle between them.
00:43:08.900 --> 00:43:20.600
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.
00:43:20.600 --> 00:43:24.700
We know that we are going to have a magnitude of 4 on our answer.
00:43:24.700 --> 00:43:32.900
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.
00:43:32.900 --> 00:43:38.500
If I we are to graph this out quickly.
00:43:38.500 --> 00:43:39.400
Let us put our x here y, z, if we are doing this we start off with the y.
00:43:39.400 --> 00:43:57.200
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.
00:43:57.200 --> 00:44:08.600
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.
00:44:08.600 --> 00:44:12.300
A little bit shaky on doing that so instead let us do the determinant.
00:44:12.300 --> 00:44:15.900
Let us find out analytically how we can do that.
00:44:15.900 --> 00:44:22.200
We start off with i ̂, j ̂, k ̂.
00:44:22.200 --> 00:44:32.800
Our first vector is going to be 0, 2, 0 and then for our B vector we have 2, 0, 0.
00:44:32.800 --> 00:44:34.700
We are going to take our determinant and repeat our pattern.
00:44:34.700 --> 00:44:44.200
0, 2, 2, 0 we would have a 0, 2 over here and the 0, 0.
00:44:44.200 --> 00:44:54.800
Our C vector is going to start at i ̂ that is going to be 0-0 that is easy.
00:44:54.800 --> 00:44:59.000
J ̂ 0 -0 that is easy.
00:44:59.000 --> 00:45:12.500
K ̂ 0-4 k ̂ C is just -4 k ̂ or 0, 0, -4.
00:45:12.500 --> 00:45:15.500
Couple of ways we can go about solving them.
00:45:15.500 --> 00:45:20.400
Let us take a look at a couple of cross product properties.
00:45:20.400 --> 00:45:31.700
A cross with B = -B × A.
00:45:31.700 --> 00:45:49.300
A × B + C= A × B + A × C.
00:45:49.300 --> 00:46:12.100
If we have a constant some C × A cross B = C × A cross B or = A cross C × B.
00:46:12.100 --> 00:46:36.200
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.
00:46:36.200 --> 00:46:39.000
I think that is good on vector math for the time being.
00:46:39.000 --> 00:46:42.100
Let us talk for a few minutes about units.
00:46:42.100 --> 00:46:46.600
The fundamental units in physics there are 7 of them.
00:46:46.600 --> 00:46:54.900
Our length which is measured in meters, mass in kilograms, time is in seconds, temperature is in kelvins.
00:46:54.900 --> 00:46:59.400
The amount of the substance is measured in moles, you might remember that from chemistry.
00:46:59.400 --> 00:47:05.800
Electrical current is in the amperes and luminosity is in candela.
00:47:05.800 --> 00:47:13.900
And here we are talking about mechanics we are mostly going to be dealing with meters, kilograms, and seconds.
00:47:13.900 --> 00:47:22.300
All over other units are derived units they are combinations of these fundamental units.
00:47:22.300 --> 00:47:28.400
Given units we can oftentimes use not to help us check our answers you will see if we have done things right.
00:47:28.400 --> 00:47:34.200
If displacement is in meters, time is in seconds, velocity would be derived unit meters per second
00:47:34.200 --> 00:47:43.200
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.
00:47:43.200 --> 00:47:48.300
The gravitational constant capital G is in N m/s² / kg².
00:47:48.300 --> 00:47:52.800
You can go and see if all the units match up, if they do not you have probably made a mistake.
00:47:52.800 --> 00:47:57.100
For example this first one does this dimension correct or are there errors?
00:47:57.100 --> 00:48:02.300
A meter per second is equal to a meter per second times a second + m/s².
00:48:02.300 --> 00:48:03.800
No, it does not work.
00:48:03.800 --> 00:48:08.600
If you came up with that formula you probably messed up somewhere.
00:48:08.600 --> 00:48:16.400
Distance displacement in meters is equal to a meter per second of velocity times a second squared.
00:48:16.400 --> 00:48:20.200
No, that is not going to work out because that will cancel.
00:48:20.200 --> 00:48:26.000
Meter equals meter per second is not going to work out so that can not be right.
00:48:26.000 --> 00:48:33.900
This one over here though the force which is a kg m/s² is n equivalent to gravitational constant
00:48:33.900 --> 00:48:45.200
which is a N m²/ kg² × mass which is a kilogram × mass which is a kilogram divided by a distance squared.
00:48:45.200 --> 00:48:49.700
Let us see kg², kg², m², m².
00:48:49.700 --> 00:48:53.800
Newton which is a kg m/s²= N.
00:48:53.800 --> 00:48:57.600
Yes, this one works so that formula would be valid.
00:48:57.600 --> 00:49:04.400
A great way to check your answers as you are going through and doing these problems called dimensional analysis.
00:49:04.400 --> 00:49:08.600
The part you all have been waiting for calculus.
00:49:08.600 --> 00:49:12.500
AP physics C is not really about calculus.
00:49:12.500 --> 00:49:15.900
We will use calculus as a tool throughout the course.
00:49:15.900 --> 00:49:19.100
We are going to cover just a few basic calculus applications here and
00:49:19.100 --> 00:49:23.000
you might have seen some of them before you might have not.
00:49:23.000 --> 00:49:26.600
You can find a much more thorough and detailed accounting of calculus
00:49:26.600 --> 00:49:32.700
on the www.educator.com courses AP calculus AB and BC right here.
00:49:32.700 --> 00:49:36.000
Believe it or not you have probably done a lot of calculus already.
00:49:36.000 --> 00:49:44.100
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.
00:49:44.100 --> 00:49:50.400
You might not have known it you have probably done some calculus in your life already.
00:49:50.400 --> 00:49:53.800
Let us take a look at differential calculus first.
00:49:53.800 --> 00:49:58.400
Differentiation is finding the slope of a line tangent to a curve.
00:49:58.400 --> 00:50:05.000
The derivative is the slope of the line tangent to a function at any given point, the result of differentiation.
00:50:05.000 --> 00:50:15.300
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.
00:50:15.300 --> 00:50:21.400
And given that function what we are doing when we take a derivative is finding the slope at a given point
00:50:21.400 --> 00:50:28.600
or finding the function that tells you the slope of the original function that is differentiation.
00:50:28.600 --> 00:50:37.600
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
00:50:37.600 --> 00:50:45.900
then the first derivative of y is equal to the derivative of y with respect to x as for notation.
00:50:45.900 --> 00:50:53.900
Or the first derivative of x, that function x all mean the same thing is equal to n Ax ⁺n -1.
00:50:53.900 --> 00:51:01.200
This basic formula for a polynomial differentiation that you will become very familiar with throughout the course.
00:51:01.200 --> 00:51:05.800
The derivative with respect to E ⁺x is just E ⁺x.
00:51:05.800 --> 00:51:11.500
The derivative with respect to x, the natural log of x is 1/x.
00:51:11.500 --> 00:51:17.000
The derivative of the sin is the cos, the derivative of the cos is the opposite of the sin.
00:51:17.000 --> 00:51:20.400
All the things that you will become familiar with throughout the course.
00:51:20.400 --> 00:51:29.400
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.
00:51:29.400 --> 00:51:31.300
Let us start with y = 4 x³.
00:51:31.300 --> 00:51:39.700
If we wanted to know the first derivative of y or y would respect x we can write as y prime or dy dx.
00:51:39.700 --> 00:51:47.600
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.
00:51:47.600 --> 00:51:56.000
It is going to be 12 x² using that formula for polynomial from the previous slide.
00:51:56.000 --> 00:52:11.100
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.
00:52:11.100 --> 00:52:19.600
The derivative of e ⁺2x, y prime would be 2e ⁺2x.
00:52:19.600 --> 00:52:32.400
Derivative of the sin of 7x², a y prime is going to be equal to 14x through the sin is the cos of 7x².
00:52:32.400 --> 00:52:49.100
and cos to 2x⁶, y prime is going to be equal to -12x sin 2x⁶.
00:52:49.100 --> 00:52:56.200
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.
00:52:56.200 --> 00:53:04.700
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.
00:53:04.700 --> 00:53:11.700
Integral calculus, integration is finding the area under the curve or adding up lots of little things to get a whole.
00:53:11.700 --> 00:53:17.400
The integral is the area under a curve at any given point, the result of an integration.
00:53:17.400 --> 00:53:20.900
Integration and derivation are inverse functions.
00:53:20.900 --> 00:53:26.400
The integral is the anti derivative or the derivative is the anti integral.
00:53:26.400 --> 00:53:31.800
If we have a curve like this and we want an integral between a couple points we are actually doing
00:53:31.800 --> 00:53:38.100
is finding the area between those points going the opposite direction.
00:53:38.100 --> 00:53:43.100
Suppose the derivative of a function is given can you determine the original function?
00:53:43.100 --> 00:53:45.100
That is what integration is.
00:53:45.100 --> 00:53:53.200
If you know the derivative of something is 2x you really need to think what is the original function if derivative was 2x?
00:53:53.200 --> 00:53:58.400
And that would be y = x² the derivative x² is 2x.
00:53:58.400 --> 00:54:07.200
You could write this in integral form y =∫ of 2x with respect to x.
00:54:07.200 --> 00:54:14.400
Which when we do that is going to be x² + this constant of integration where that constant come from.
00:54:14.400 --> 00:54:22.100
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.
00:54:22.100 --> 00:54:24.500
We do not know if there was a 3 or not there.
00:54:24.500 --> 00:54:28.800
This constant says you know there could be some constant there we do not know what it is yet.
00:54:28.800 --> 00:54:33.100
It could be -5 it could be 0 it could be 37,000.
00:54:33.100 --> 00:54:37.700
Just being there as a piece we do not know about and we have some different tricks to make that go away
00:54:37.700 --> 00:54:43.800
in order to find out what those constants are later on.
00:54:43.800 --> 00:54:47.000
Let us look at a couple of common integrations.
00:54:47.000 --> 00:55:01.200
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.
00:55:01.200 --> 00:55:04.700
The integral of the E ⁺x dx would just be e ⁺x.
00:55:04.700 --> 00:55:08.600
The integral of the dx/x is the natural log of x.
00:55:08.600 --> 00:55:16.100
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.
00:55:16.100 --> 00:55:19.900
Let us do a couple quick integrations for practice.
00:55:19.900 --> 00:55:27.900
The integral of 3x² dx is just going to be x³ as to think about the derivative of x³ is going to be 3x².
00:55:27.900 --> 00:55:32.600
We have to remember our constant of integration + C.
00:55:32.600 --> 00:55:38.400
The integral of 2 cos 6x dx now what I would do there is recognize that
00:55:38.400 --> 00:55:47.400
when I do this we will have to do integral of 2 cos 6x dx.
00:55:47.400 --> 00:56:03.500
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.
00:56:03.500 --> 00:56:06.200
They are more involved integration.
00:56:06.200 --> 00:56:16.200
The integral e ⁺4x dx is going to be integral of e ⁺4x dx now what this in the form e ⁺xdx.
00:56:16.200 --> 00:56:23.200
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.
00:56:23.200 --> 00:56:30.000
That is going to be e ⁺4x / 4 + constant of integration.
00:56:30.000 --> 00:56:32.900
And finally integral here with some limits.
00:56:32.900 --> 00:56:39.800
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.
00:56:39.800 --> 00:56:48.800
I would write that as integral of 2xdx well that is x² evaluated from x = 0 to x = 4.
00:56:48.800 --> 00:56:52.900
What that means is what you are going to do is you are going to take your x²
00:56:52.900 --> 00:57:07.000
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.
00:57:07.000 --> 00:57:14.700
That is going to be equal to 4² – 0² which is 16.
00:57:14.700 --> 00:57:22.200
If these are troubling the www.educator.com lessons on calculus are outstanding.
00:57:22.200 --> 00:57:23.900
Let us look at this one more way.
00:57:23.900 --> 00:57:27.800
If we said this was the area under the curve how could that look?
00:57:27.800 --> 00:57:41.100
Let us draw a quick graph and assume that we have some y function where y = 2x.
00:57:41.100 --> 00:57:44.000
There is y there is x.
00:57:44.000 --> 00:57:57.100
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.
00:57:57.100 --> 00:57:59.800
We said that integration give you the area under the curve.
00:57:59.800 --> 00:58:04.700
Let us figure out what the area is under this curve.
00:58:04.700 --> 00:58:20.500
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.
00:58:20.500 --> 00:58:26.400
Analytical version and graphical version gives you the same value.
00:58:26.400 --> 00:58:31.300
What you are really doing is finding the area under that graph.
00:58:31.300 --> 00:58:37.100
Let us take a look at our last example for this lesson.
00:58:37.100 --> 00:58:43.000
The velocity of a particle is a function of time is given by the equation v of t = 3t².
00:58:43.000 --> 00:58:46.500
The particle starts at position 0 at time 0.
00:58:46.500 --> 00:58:50.900
Find the slope of the velocity time graph as a function of time?
00:58:50.900 --> 00:58:54.700
That will give you the particles acceleration function.
00:58:54.700 --> 00:59:03.900
Acceleration is the slope of that vt graph, Velocity vs. time which we could write is V prime of t.
00:59:03.900 --> 00:59:13.200
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.
00:59:13.200 --> 00:59:21.700
Vt² which happens to be 60 so the acceleration of the particle is equal to 6 times whatever the time happens to be.
00:59:21.700 --> 00:59:28.500
When you know the velocity function you can find the acceleration, calculus.
00:59:28.500 --> 00:59:34.400
Find the area under the velocity time graph as a function of time to give your particles position function.
00:59:34.400 --> 00:59:52.100
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.
00:59:52.100 --> 00:59:56.200
But here is a little trick to making a consonant of integration go way.
00:59:56.200 --> 01:00:07.300
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.
01:00:07.300 --> 01:00:16.300
If r=0 at time 0 that means if we plug 0 in here for time 0 r must equal z.
01:00:16.300 --> 01:00:19.700
But r is 0 at that time so z must equal 0.
01:00:19.700 --> 01:00:27.400
Therefore our total function is r = t³ z is 0 in this problem.
01:00:27.400 --> 01:00:32.800
We are able to come up with the particles position function based on velocity.
01:00:32.800 --> 01:00:36.500
We took the derivative in one direction to find the acceleration.
01:00:36.500 --> 01:00:42.000
We took the integral going the other way in order to find it is change in position.
01:00:42.000 --> 01:00:46.400
Alright that helpfully gets you a feel for the type of math we will be using in this course.
01:00:46.400 --> 01:00:48.600
Thank you for watching www.educator.com.
01:00:48.600 --> 01:00:51.000
Come back real soon and make it a great day everyone.