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### Course: Multivariable calculus>Unit 2

Lesson 7: Partial derivatives of vector-valued functions

# 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?
• If you watch the video without fullscreen mode, a note comes up on the screen that corrects this mistake.
(1 vote)
• At when 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)?
• Yes. The derivative is only instantaneously zero.
• 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.
• 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?
• Around you 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?
• 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?
• 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.
• 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?
• 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?