Tyler Clarke
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2.8 KiB
HTML
[!]
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[=post-]
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<p>
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Welcome back! We continue chapter 16 with a discussion of <i>vector fields</i>. Essentially, a vector field is a function with the same input dimensions as output dimensions. In `R^2` and `R^3`,
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we visualize this as a bunch of vectors permeating all of space. If you've taken physics, you've probably encountered vector fields - the electric and magnetic fields are in fact
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vector fields! A way to think about vector fields is as the <i>velocity</i> of a given infinitesimal region of a fluid: it's at a position, and the velocity vector is pointing in some direction
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with some magnitude. This is actually a very good way to represent a flowing fluid, for our purposes.
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</p>
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<p>
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An example of a vector field we've used extensively thus far is the gradient function of a surface! At every point along the surface,
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it outputs a vector representing the slope at that point.
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</p>
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<p>
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One of the most useful operations we can perform on a vector field is a line integral (remember 16.1?). The integral of a vector field over a line is a fairly abstract concept,
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but it has real-world uses: imagine we have a function for the force at any given point, and we want to know the total force over a line. Vector fields! The form for a vector
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field integral is simply the dot product of the vector field and the tangent vector to the point on our line: given a line `r(t)` in a vector field `f(x, y) = [u, v]`, we have
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`int f(x, y) * dr`. This is one of those very elegant things that has interesting consequences.
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</p>
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<p>
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Let's do an example. Straight from the textbook, we have `f(x, y, z) = \[z, xy, -y^2]` and `r(t) = \[t^2, t, sqrt(t)]`, where `a <= t <= b`. Getting `f(x, y, z)` in terms of `t` is simple enough:
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just substitute the components of `r`: `f(t) = \[sqrt(t), t^3, -t^2]`. `dr` is going to be `frac {dr} {dt} * dt`, so let's find `frac {dr} {dt}`:
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`frac d {dt} \[t^2, t, sqrt(t)] = \[2t, 1, frac 1 {2sqrt(t)}]`.
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So our integral is `int_a^b \[sqrt(t), t^3, -t^2] * \[2t, 1, frac 1 {2sqrt(t)}] dt`. Evaluating the dot product gives us `int_a^b 2t^{3/2} + t^3 - frac {t^{3/2}} {2} dt`.
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I'll leave further simplification as an exercise to the reader.
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</p>
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<p>
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This section is cut rather short, as the introduced idea is simple and it's too late at night to try making it more complex. I might expand it
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with more examples and applications later. Section 16.3 will go into quite a bit more detail, as we'll be dealing with the problem of conservative vs
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nonconservative paths. For now, da svidanya and bonne nuit!
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</p>
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<p>
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P.S: If you notice broken or missing math, it's probably due to my template engine freaking out about the square brackets. Will fix in a bit!
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</p>
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[/]
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[=title "Multivariable Exam 3 Review: Thomas 16.2"]
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[=author "Tyler Clarke"]
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[=date "2025-4-3"]
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[=subject "Calculus"]
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[#post.html]
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