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### Course: Multivariable calculus > Unit 2

Lesson 7: Partial derivatives of vector-valued functions- Computing the partial derivative of a vector-valued function
- Visual parametric surfaces
- Partial derivative of a parametric surface, part 1
- Partial derivative of a parametric surface, part 2
- Partial derivatives of vector valued functions
- Partial derivatives of vector fields
- Partial derivatives of vector fields, component by component

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# Partial derivatives of vector fields, component by component

Here we step through each partial derivative of each component in a vector field, and understand what each means geometrically. Created by Grant Sanderson.

## Want to join the conversation?

- Should partial Q with respect to x at pt (2,0) be -4, not -2 as written?(55 votes)
- If you watch the video without fullscreen mode, a note comes up on the screen that corrects this mistake.(1 vote)

- At6:15when Grant starts to look at the partial of Q with respect to y, how is the magnitude of y not changing? It sure looks like it is. Is this just a consequence of Grant evaluating to far away from the original input (x,y) coordinate of (2,0)?(12 votes)
- Yes. The derivative is only instantaneously zero.(15 votes)

- this question may not be in tnis context. but what is the difference between position vector valued function and the vector field function. position vector valued function seems pretty decent to understand because it is vector value at each point, usually interpreted as beautifully when taking the gradient of scalar temperature function. But when the function is already defined to be vector field equation, whats its gradient going to be. Also how do we measure the directional directive in each of the cases.

mathmatically, suppose we have scalar function of temperature T to be ... f(x,y,z), that returns scalar function at each point in 3D. taking gradient gives us position vector valued function. and we can proceed to find the divergence and so the source and sink and so forth. pretty decent.

but when we intially have a vector valued function as f(x,y,z) =x(t)i+y(t)j+z(t)k. is this a position vector valued function or is this a function of magnitude of vector in corresponding direction. for instance for a function, f(v) =xi+yj+zk. its magnitude when x,y and z =1; is 1. and when x,y and z=2, magnitude is sqrt (12). but is still in the same direction.(5 votes)- Position vector-valued functions have a one-dimensional input (usually thought of as time), and a multidimensional output (the vector itself).

Vector fields have a multidimensional input (e.g. the xy-plane), and a multidimensional output (vectors in that same space).

The gradient only applies to functions with a multidimensional input, and a scalar-valued output, or in other words, a one-dimensional output.

Does that help?(31 votes)

- Around5:45you show that the vectors' y-value decrease as the x increases, which seems correct since dQ/dx = -2x, but this equation also indicates that the y-value of the vectors should increase when x < 0. Which is contradicted by the fact that every vector on the graph, when y = 0 (and thus irrelevant for the vectors' y-value), point downward, even when x < 0.

Is the visual representation faulty or am I missing something here?(5 votes)- Choose any two vectors that start on the x-axis. Draw a new vector from the tip of the left vector to the tip of the right vector. For vectors along the negative x-axis, this connecting vector will be positive. Notice how the x-component can be thought of as dx, while the y-component can be thought of as dQ, thus you can see see geometrically that the vector representing dQ/dx has positive slope along the negative x-axis. Similarly, doing this along the positive x-axis gives a negative dQ/dx, as you would expect. I had the same confusion, but remember that you move left to right when analyzing how y changes with x, so your change should be pointing from left to right, with the sign determining if you then move up or down.(1 vote)

- dQ/dy explanation does not make sense. It is changing in the graph, but he is saying it is not and it is zero? really?(3 votes)
- That's just the nature of the function. At that single point, it appears that the output vector's y-coordinate is not changing at all, even though it is in the midst of changing its value. Essentially, it's the same thing in single variable calculus -- if we think of g(x)=x^2 as a transformation, it would appear that at x=0, g(x) is not really changing if we nudge it -- even though it is. Similarly, if we nudge v(x, y) in the y direction at (2, 0), it would appear that there is no change in the y-value.

The example Grant uses might be a bit confusing, since for del(P)/del(x), y stays at 0 when x is changed and hence the vectors in the field really don't change at all. However, this doesn't apply to del(Q)/del(y), since the y-value is changing and is causing del(Q)/del(y) to also change.

Something that helped me understand a bit is the fact that the vector field isn't necessarily showing you the partial derivatives; it only gives you mere feels for what the output vectors are relative to other inputs. It shows outputs, not slopes.(4 votes)

- This was asked years ago but never received a real answer. Why is it that with dQ/dx, Q is continuing to get more and more negative as x < 0? The derivative shows the change in Q should be increasingly more positive. Is this a bad graphical representation?(4 votes)
- Choose any two vectors that start on the x-axis. Draw a new vector from the tip of the left vector to the tip of the right vector. For vectors along the negative x-axis, this connecting vector will be positive. Notice how the x-component can be thought of as dx, while the y-component can be thought of as dQ, thus you can see see geometrically that the vector representing dQ/dx has positive slope along the negative x-axis. Similarly, doing this along the positive x-axis gives a negative dQ/dx, as you would expect. I had the same confusion, but remember that you move left to right when analyzing how y changes with x, so your change should be pointing from left to right, with the sign determining if you then move up or down.(1 vote)

- Hi. Are the values for the partial derivatives like 0, -2 -4 (-2 in the video) and 2 signify the magnitude or direction?(2 votes)
- The partial derivatives (0,-2,-4,2) represent the matrix values of the partial derivatives, they represent neither the magnitude or direction.(3 votes)

- In navier stokes, the equation given for the change in vector V (x,y,z,t), dv = (pV/px) dx + (pV/py) dy + (pV/pz) dz + (pV/pt) dt, where p is a partial. This makes sense, but my question is this.

We try to find the "material derivative" of V with respect to time.

We divide that original equation giving dv by dt, meaning total derivative. This means each dx, dy, dz, and dt is divided by dt, but shouldn't this be a partial derivative? Or is this just bad notation?

Thanks for any help, Im sure this isn't the right place to ask too, but hopefully someone knows.(1 vote)

## Video transcript

- [Voiceover] Let's
continue thinking about partial derivatives of vector fields. This is one of those things
that's pretty good practice for some important concepts coming up in multivariable calc,
and it's also just good to sit down and take a complicated thing and kind of break it down piece by piece. So a vector field like
the one I just showed is represented by a
vector-valued function, and since it's two-dimensional, it'll have some kind of two-dimensional input. And the output will be a vector, each of whose components
is some kind of function of x and y, right? So I'll just write P of
x, y for that x-component and Q of x, y for that y-component. And each of these are just
scalar-valued functions. It's actually quite common to
use P and Q for these values. It's one of those things where
sometimes you'll even see a theorem about vector calculus in terms of just P and Q, kind of leaving it understood
to the reader that, yeah, P and Q always refer
to the x and y components of the output of a vector field. And in this specific case,
the function that I chose, it's actually the one that
I used in the last video. P is equal to x times y, and Q is equal to y
squared minus x squared. And in the last video, I was talking about interpreting
the partial derivative of v, the vector-valued function with respect to one of the variables, which has its merits, and
I think it's a good way to understand vector-valued
functions in general. But here that's not what I'm gonna do. It's actually, another useful skill is to just think in terms
of each specific component. So if we just think of P and Q, we have four possible partial derivatives at our disposal here, two
of them with respect to P. So you can think about
the partial derivative of P with respect to x, or the partial derivative
of P with respect to y. And then similarly, Q, you could think about
partial derivative of Q with respect to x. This should be a partial. Or the partial derivative of Q with respect to y. So four different values
that you could be looking at and considering and
understanding how they influence the change of the vector field as a whole. And in this specific example, let's actually compute these. So derivative of P with respect to x. P is this first component. We're taking the partial
of this with respect to x. y looks like a constant. Constant times x. Derivative is just that constant. If we took the derivative
with respect to y, the roles have reversed, and its partial derivative is x, 'cause x looks like that constant. But Q, its partial derivative with respect to x, y
looks like a constant, negative x squared goes to negative 2x. But then when you're taking
it with respect to y, y squared now looks like a function whose derivative is 2y, and negative x squared
looks like the constant. So these are the four
possible partial derivatives. But let's actually see
if we can understand how they influence the
function as a whole. What it means in terms of the picture that we're looking at up here. And, in particular,
let's focus on a point, a specific point, and let's do this one here. So it's something that's
sitting on the x-axis. So this is where y equals zero and x is something positive. So this is probably when x
is around two-ish, let's say. So the value we wanna look at is x, y when x is two and y is zero. So if we start plugging that in here, what that would mean,
this guy goes to zero, this guy goes to two, this guy, negative two times x, is gonna be negative two. And then negative two
times y is gonna be zero. And let's start by just looking
at the partial derivative of P with respect to x. So what that means is
that we're looking for how the x-component of
these vectors change as you move in the x direction. For example, around this
point, we're kind of thinking of moving in the x direction, vaguely. So we wanna look at the
two neighboring vectors and consider what's going
on with the x direction. But these vectors, this
one points purely down. This one also points purely down, and so does this one. So no change is happening when it comes to the
x-component of these vectors, which makes sense because the
value at that point is zero. The partial derivative of P
with respect to x is zero, so we wouldn't expect a change. But on the other hand, on the other hand, if we're
looking at partial derivative of P with respect to y, this should be positive. So this should suggest that
the change in the x-component as you move in the y
direction is positive. So we go up here and now we're not looking at change in the x-direction. Not looking at change in the x-direction. But instead we're wondering what happens as we move generally upwards. So we're gonna kind of
compare it to these two guys. And in that case, the
x-component of this one is a little bit to the left. The one below it, it's a
little bit to the left. Then we get to our main
guy here and it's zero. The x-component is zero 'cause
it's pointing purely down. And up here it's pointing
a little bit to the right. So as y increases, the x-component of these
vectors also increases. And again, that makes sense because this partial
derivative is positive. This two suggests that
as you're changing y, the value of p, the
x-component of our function, should probably keep that on screen. You know, the x-component of
our vector-valued function is increasing 'cause that's positive. For contrast, let's say we look at the Q component over here. So what this is doing, we're looking at changes in the x, and we're wondering what the
y value of the vector does. So we go up here. And now, we're not looking at
changes in the y direction. But instead, we're going
back to considering what happens as we change x, as we're kind of moving in
the horizontal direction here. So again, we look at
these neighboring guys. And now, the y-component
starts off small, but negative, then gets a little bit more negative, then gets even more negative. And if we kind of keep looking at these, the y-component is getting more,
and more, and more negative so it's decreasing. The value of Q, the
y-component of these vectors should be decreasing. And that lines up 'cause
the partial derivative here was negative two, so that's telling us, given that it started negative,
it's getting more negative. If it started positive, they would've been kind of getting shorter as vectors as their y-component got smaller. And then finally, just to close things off nice and simply here, if
we look at partial of Q with respect to y, so now if we start looking at
changes in the y direction, and we start considering
how, as you move below, and then starting to go up, what happens to the y-component, here, the y-component is a
little bit negative, right? It's pointing down and to the left. So down, it's a little bit negative. Here, the y-component is
also a little bit negative. Over here, well, it remains
a little bit negative. And you know, from our heuristic look, there's no discernible change. Maybe it's changing a little bit, and we don't have a fine
enough vision of these vectors to see that, but if we actually go back to the analysis, and see what we computed, in fact, it is zero. That fact that it looked like there was not too much change in the y-component of
each one of these vectors corresponds with the fact
that the partial derivative of that y-component with respect to y, with respect to vertical
movements, is zero. So this kind of analysis
should give a better feel for how we understand the
four different possible partial derivatives and what they indicate about the vector field,
and you'll get plenty of chance to practice that understanding as we learn about divergence and curl, and try to understand
why each one of those represents the thing
that it's supposed to. And you'll see what I mean by
that in just a couple videos.