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Vector valued function derivative example

Concrete example of the derivative of a vector valued function to better understand what it means. Created by Sal Khan.

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  • blobby green style avatar for user Michel André Breau
    At the end ( and beyond) shouldn't the velocity vector for r2 be the EXACT SAME as the velocity vector in R1... since 2(1/2)i + 4(1/2)j = 1+2j?
    (6 votes)
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  • winston baby style avatar for user Andrew
    at , Sal says that the path is the same. Yes, the paths appears to look the same, but how can they truly be the same if the graphs use different scalings (in x axis). If the graph on the right were scaled the same way as the graph on the left (each "notch" on the x and y axis represents 1 unit), the graph on the right would be much steeper and thus the "path" would -not- be the same.
    (5 votes)
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    • leaf green style avatar for user Joe Kern
      He is not using a different scalar multiple. Think of the vector r as (x, (y)^2). Now what he did was multiply r by 2 (2*r) which would multiply x and y in the same fashion (2*x, (2*y)^2), However (2*y)^2 = 4*y^2. Remember order of operation is parenthesis then exponents.
      (0 votes)
  • piceratops sapling style avatar for user juliana.harmatiuk
    Hello, I have a question. If I decide to plot this path as a function F(t) perpendicularly to the xy plane, wouldn't the derivative of r(t) behave just like the gradient of the F(t)? For me it looks so much like a gradient of some function, given that it would be directed perpendicularly to it and point in the direction of highest increase in its value. I might be confused though. Any help?
    (2 votes)
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    • aqualine ultimate style avatar for user Kyler Kathan
      A gradient implies you're dealing with a surface or a function that has 2 inputs, but in this case we only really have 1 input that changes everything, which is t. Since there's only the input t, the parametric system forms a path like a line rather than a surface.
      (2 votes)
  • male robot johnny style avatar for user Ralph Hebgen
    I am new to this site, guys - I would like to find exercises associated with the lectures, and I know there are some on the site, but I have no idea how to identify / access them. Can anybody help and show me where to go (for example if I wanted to find an exercise on the exact content of this video, how would I do it)? Sorry for being thick.
    (2 votes)
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    • leaf green style avatar for user Joe Kern
      I assume you already figured this out since it has been 8 months, but if not no worries. The practice problems are only for certain sections of math. While they haven't added many for Multivariable calculus, they do have practice problems for Integration, Differentiation, and pre-k12 math (up to Pre calc.) I hope they add problems for all math and science related subjects in the future, but as of 7/6/2016 they do not have practice questions for everything. As for accessing them, go to the section you desire in the subjects tab at top and click on it in the drop down menu. Then it should have an option on that page to do practice problems. It will say something like continue or start your mission.
      (2 votes)
  • blobby green style avatar for user izzyde360
    the velocity vector for the second graph is wrong, it should be double the length that is drawn.
    (2 votes)
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  • leafers seed style avatar for user Hiletso
    Aren't the two curves just look the same because they have different x scales? I don't understand why Sal said they are essentially the same.
    (2 votes)
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  • piceratops tree style avatar for user Harshul Mehta
    Umm this probably gonna be a stupid one but why does the vector originate from the point (1,1) and not from the origin of the graph (0,0)?
    (2 votes)
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    • leaf green style avatar for user MBCory
      Sal has the vector start at (1,1) to better visualize the movement of the particle at that specific point. Yes, we usually start vectors at the origin, but we don't have to, and in this case it wouldn't make sense to talk about the movement of a particle at (1,1) and not put our vector (which represents that movement) at the specific point we are talking about. Hope this helps!
      (1 vote)
  • piceratops ultimate style avatar for user Nazariy Olegovich Vavryk
    For the second r vector function. The graphical representation of the slope vector does not mirror the notation I believe. If we using I hat and j hat, then those are unit vectors. And since the scale of the graph is different to the first one, when Sal is moving 2 units to the right and 4 units up, he is really moving 1 whole i hat unit to the right and 2 whole j hat units up. I believe.
    (1 vote)
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    • ohnoes default style avatar for user Tejas
      The graphs and the notation agree. There is nothing wrong with using î and ĵ to represent a slope vector. You can use a linear combination of î and ĵ to represent any two-dimensional vector, including slope vectors. Also, while Sal did not go all the way and give the slope vector the full length, it was definitely larger than 1î + 2ĵ.
      (2 votes)
  • female robot amelia style avatar for user neamesis
    lots of people are getting confused and asking if the "scale" of the second graph is correct or not. I don't think they realize that Sal said t = 1/2 at , they think t ϵ {0, 1, 2} in the second graph when it's actually t ϵ {0, 1/2, 1}
    (1 vote)
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  • piceratops tree style avatar for user khc1031
    Is it safe to say that the magnitude of a velocity vector is synonymous as the slope in 2-dimensional analysis?
    (1 vote)
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Video transcript

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.