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## Multivariable calculus

### 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? • 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?
• 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)? • 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? • You have to remember that it is not the direction that the vectors are pointing that matters, but rather the change in their P and Q values. Partial Q with respect to partial x (dQ/dx) represents the change in the vectors' Q value as you move in the positive direction along the input x-axis. It is true that the vectors point downward for both positive and negative x inputs along the P-axis. However, when x < 0, the Q values are increasing (the vectors are getting shorter, or less negative in the Q direction). When x > 0, the Q values are decreasing (the vectors are getting longer, or more negative in the Q direction).
(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? • 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?  