Position vector functions and derivatives
Vector valued function derivative example Concrete example of the derivative of a vector valued function to better understand what it means
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- What I want to do in this videos is to make to
- parametrizations of essentially the same curve, but we're going
- to go along the curve a different rates.
- And hopefully we'll be able to use that to understand, or get
- a better intuition, behind what exactly it means to take a
- derivative of a position vector valued function.
- So let's say my first parametrization, I have
- x of t is equal to t.
- And let's say that y of t is equal to t squared.
- And this is true for t is greater than or equal to 0,
- and less than or equal to 2.
- And if I want to write this as a position vector valued
- function, let me write this.
- x1, call that y1, and let me write my position vector valued
- function; I could say r1-- I'm numbering them because I'm
- going to do a different version of this exact same curve with a
- slightly different parametrization --so r1 one of
- t, we could say is x1 of t times i-- the unit vector i
- --so we'll just say t times i plus-- this is just x of t
- right here, or x1 of t; I'm numbering them because I'll
- later have an x2 t --plus t squared times j.
- And if I wanted to graph this, I'm going to be very careful
- graphing it because I really want to understand what the
- derivative means here.
- Try to draw it roughly to scale.
- So let's say that this is one, two, three, four.
- Then let me draw my x-axis.
- That's good enough.
- And my x-axis, I want it to be roughly to scale, one and two.
- And so at t equals 0, both my x and y coordinates are at 0-- or
- this is just going to be the 0 vector, so this is where we are
- a t equals 0 --at t equals 1 this is going to be one times
- i-- we're going to be just like that --plus 1 times j.
- 1 squared is j, so we're going to be right there.
- And then at t is equal to 2, we're going to be at 2i.
- So 2i-- you could imagine 2 times i would be this vector
- right there --2 times i plus 4-- 2 squared is 4 --4 times
- j, so plus 4 times j.
- If you add these two vectors heads to tails, you're going
- to get a vector that's end point is right there.
- The vector is going to look something like this.
- So this is what, just to make it clear what we're
- doing, that's r1 of 2.
- This is r1 of 0.
- This is r1 of 1.
- But the bottom line is the path looks like this:
- it's a parabola.
- So the path will look like that.
- Now that's in my first parametrization of it.
- Actually, let me draw a little bit more carefully.
- I want to get rid of this arrows, just because I want it
- to be a nice clean drawing.
- So it's going to be a parabola.
- Let me get rid of that other point, too, just because I
- didn't draw it exactly where it needs to be; it needs
- to be right there.
- And my parabola, or part of my parabola is going to
- look something like that.
- All right.
- Good enough.
- So this is the first parametrization.
- Now I'm going to do this exact same curve, but I'm going to
- do it slightly differently.
- So let's say I'll do it in different colors.
- So x2 of t, let's it equals 2t.
- And y2 of t, let's say it's equal to 2t squared.
- Or we could alternatively write that, that's the same thing as
- 4t squared, just phrasing both of these guys
- to the second power.
- And then let's say instead of going from t equals 0 to 2,
- we're going to go from t goes from 0 to 1.
- But we're going to see, we're going to cover
- the exact same path.
- And our second position vector valued function, r2 of t, is
- going to be equal to 2t times i plus-- I could say 2t squared
- --4t squared times j.
- And if I were to graph this guy right here, it would look
- like-- let me draw my axes again; it's going to look the
- same, but it's I think useful to draw it because I'm going to
- draw the derivatives and all that on it later.
- One, two, three, four.
- One, two.
- And then let's see what happens when t is equal to 0-- or r of
- 0; all these are going to be 0, we're just going to have the
- zero vector; x and y are both equal to 0 --when t is equal
- to 1/2 what are we going to get here?
- 1/2 times 2 is 1.
- And then we're going to get the point 1/2 squared
- is 1/4 times 4 is 1.
- So when t is equal 1/2 we're going to be at the point 1, 1.
- And when the t is equal to 1 we're going to
- be at the point 2, 4.
- So notice the curve is exactly, the path we go
- is exactly the same.
- But before we even do the derivatives, these two
- paths are identical.
- I want to think about something.
- Let's pretend that our parameter, t, really is time.
- And that tends to be the most common, that's
- why they call it t.
- It doesn't have to be time, but let's say it is time.
- So what's happening here?
- In the first parametrization when we go from 0 to 2
- seconds we cover this path.
- You can imagine after 1 second the dot moves here,
- then it moves there.
- You can imagine a dot moving along this curve, and it
- takes two seconds to do so.
- In this situation we have a dot moving along the same curve,
- but it's able to cover the same curve in only one second; and
- half a second it gets here.
- It took this guy one second to get here.
- In a one second, this guy's all the way over here; this guy
- takes two seconds to go over here.
- So in this second parametrization even though the
- path is the same, the curves are the same, the
- dot is faster.
- I want you to keep that in mind when we think about
- the derivatives of both of these position vector
- valued functions.
- So just remember the dot is moving faster for every second
- it's getting further along the curve than here; that's why it
- only took them one second.
- Now let's look at the derivatives of both
- of these guys.
- So the derivative here, so if I were to write r prime, r1 prime
- of t-- let me do that in a different color, actually,
- already used the orange; so let me do it in the blue
- --r1 prime us t.
- So the is the derivative now.
- It's going to be, remember, it's just the derivative of
- each of these times the unit vectors.
- So the derivative of t with respect to t, that's just 1.
- So it's 1 times i.
- I'll just write 1i plus-- I didn't have to write the one
- there --plus the derivative of t squared with respect
- to t is 2t plus 2t j.
- And let me take the derivative over here.
- r2 prime of t.
- The derivative of 2t with respect to t is 2, so
- 2i, plus the derivative of 4t squared is 8t.
- 2 times 4, it is rt.
- Just like that.
- Now the question is, what do their respective derivative
- vectors look like at different points?
- So let's look at, I don't know, let's see how fast they're
- moving when time is equal to 1.
- So let's take it at a specific point.
- This is just the general formula, but let's figure
- out what the derivative is at a specific point.
- So let's take r1 when time is equal to 1.
- And I want to take this specific point on the curve,
- not the specific point in time.
- So this point on the curve here is when time is equal
- to 1, you could say second.
- This point over here, which is the exact corresponding
- point, is when time is equal 1/2 second.
- So r1 of 1 is equal to-- we're taking the derivative
- there --is equal to 1i.
- It's not dependent on t at all.
- So it's 1i plus 2 times 1j, so plus 2j.
- So at this point the derivative of our position vector function
- is going to be 1i plus 2j.
- So we can draw it like this. so if we do 1i is like this: 1i.
- And then 2j.
- Just 2j is like that.
- So our derivative right there, I'll do it in the same
- color that I wrote it in.
- It's in this green color; it's going to look like this.
- And notice it looks like, at least its direction is-- let me
- do it a little bit straighter --its direction looks tangent
- to the curve; it's going in the direction that my
- particle is moving.
- Remember my particle is moving from here to there, so it's
- going in the direction.
- And I'm going to think about, in a second, what this length
- of this to derivative vector is.
- This right here, just to be clear is, r1 prime.
- It's a vector, so it's telling us the instantaneous change in
- our position vector with respect to t, or time, when
- time is equal to 1 second.
- That's this thing right here.
- Now let's take the exact same position here on our curve.
- But that's going to occur at a different time for this guy.
- We already said it only takes him, he's here at time
- is equal to 1/2 second.
- So let's take-- --I'll do it in the same color
- --so here we have r2.
- We're going to evaluate it at 1/2 half because this is at
- time is equal 1/2 second.
- And this is going to be equal to 2i-- this isn't dependent
- at all on time --so 2i plus 8 times the time.
- So time right here is 1/2.
- So 8 times 1/2 is 4.
- So plus 4j.
- So what does this look like?
- The instantaneous derivative here.
- Oh, and this is the derivative; have to be very clear.
- So 2i-- let me draw some more --so 2i maybe gets
- us about that far.
- Plus 4j will get us up to right around there.
- Plus 4j is that factor.
- So when you add those two heads to tails, you get this thing:
- you get something that-- let me like --you get something
- that looks like that.
- I didn't draw it as neatly as I would like to.
- But let's notice something: both of these vectors are going
- in the exact same direction.
- They're both tangential to the path, to our curve.
- But this vector is going, its length, its magnitude, is
- much larger than this vector's magnitude.
- And that makes sense because I hinted at it when we first
- talked about these vector valued position functions and
- their derivatives; is that the length, you can kind of
- view it as the speed.
- The length is equal to the speed if you imagine t being
- time and these parametrizations are representing a dot
- moving along these curves.
- So in this case, the particle only takes a second to go
- there, so at this point in its path, it's moving much faster
- than this particle is.
- So if you think about it, this vector right here, if you
- imagine this is a position factor, this is velocity.
- Velocity is speed plus the direction.
- Speed is just you know, how fast are you going?
- Velocity is how fast you're going in what direction?
- I'm going this fast-- and you could calculate it using the
- Pythagorean Theorem, but I just want to give you the intuition
- right here --I'm going that fast in this direction.
- Here I'm going this fast; I'm going even faster.
- That's my magnitude, but I'm still going in
- the same direction.
- So hopefully you have a gut feeling now of what the
- derivative of these position vectors really are.
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