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### Course: Multivariable calculus>Unit 2

Lesson 7: Partial derivatives of vector-valued functions

# Partial derivatives of vector fields

How do you intepret the partial derivatives of the function which defines a vector field? Created by Grant Sanderson.

## Want to join the conversation?

• At , since v1+dv=v2, shouldn't the tip of the second pink vector touch the tip of the differential blue vector?
• If they were in the vector space, yes. If the initial points of v1 and v2 were the same and dv started at the tip of v1, the tips of dv and v2 would be touching. In the xy plane, however, the initial points of the vectors are shifted to the xy coordinates where they are evaluated - v1 and v2 start at different points.
• I just cannot get it what is the difference between a vector field, which is in this video, and the previous example where we had a parametrically defined plane?
• "Vector field" is just a representation for a function, just like a regular map represents Earth. We can represent the Earth with different kinds of maps: spherical - it's the most accurate (because of how the Earth is shaped) but it's less practical to inspect because you need to rotate it and manufacturing it costs more. Same for multivariable functions. If you have n inputs (variables) and m outputs as a vector:
f(a_1, a_2, ... a_n) = (b_1, b_2, ... b_m)
then you need (n + m) dimensional space in order to represent it accurately. If we have here n=2 and m=2 then we need 4 dimensional space to imagine how the function looks which obviously is already "impossible" as we can comprehend up to 3 dimensions. For the vector field representation we only need 2 dimensions (half less!). In general the optimal representation for a multivariable function can be categorized as follows:
n > m (more inputs) -> regular graph
n = m -> vector field
m > n -> (more outputs) -> parametric (curve or surface...)
• The mathematical formalism looks like the same as a parametric surface, so every time we do this in "real life" we need to precise "it is a vector field" or "it is a parametric surface", is not it?
• Why do the dimensions of the input and output space need to match? Couldn't we have a vector field that associates an input in R^3 with vectors that only point along the xy-plane f(x,y,z)=[f1(x,y,z),f2(x,y,z)]?
(1 vote)
• You are asking about a vector field in a 2D plane based upon 3D inputs. A R^2 vector will always be f=<a,b>, where a is the change along one axis and b is the change along another axis. Is it possible then that you could have some vector f=<a(x,y,z), b(x,y,z)>?

I imagine there's nothing saying you can't... but graphically it would be difficult to interpret. If thinking of a function as a transformation, we are essentially bending and compressing every point in a 3 dimensional space into a 2 dimensional plane. So... you could map a vector field onto the input space, but what would a 2D vector f(a,b) represent in a 3D space? You'd have to set (a,b) in the (x,y) plane and have the vectors graphically drawn as <a,b,0> (or <0,a,b> or <a,0,b> et al...), but then that graphically would imply a and b as changes in their respective axes.

If you chose to plot the vectors on a 2D plot, a vector with 3D inputs and a 2D output also may not have every point on a 2D plot in its domain, which means the vector field is not zero but undefined.
(1 vote)
• Why this is different to a transformation presented in parametric surface from previous videos? 'I think they are similar':
we have two inputs (x,y) and we get two outputs (P,Q) is also a relocation of point from a plane to another different plane.
(1 vote)