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, 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 - 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 + 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 `int f(x, y) * dr`. This is one of those very elegant things that has interesting consequences.
@@ -21,15 +21,37 @@ 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}`: `frac d {dt} \[t^2, t, sqrt(t)] = \[2t, 1, frac 1 {2sqrt(t)}]`. 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`. - I'll leave further simplification as an exercise to the reader. + I'll leave further simplification as an exercise to the reader. In the case where `a = b`, this is also known as the circulation.
- 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 - 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 - nonconservative paths. For now, da svidanya and bonne nuit! + In some cases, we may want to know only the line integral in a single direction: in our wire-under-force + example, this would mean finding only the x-component or y-component or z-component of the force. + In these cases, it's as simple as zeroing the unused components: if your vector field is `F(x, y, z) = + \[m(x, y, z), n(x, y, z), o(x, y, z)]`, and we want to integrate over the y-axis, our vector field becomes + `F(x, y, z) = \[0, n(x, y, z), 0]`. Simple enough! If you prefer a vector solution, then taking the unit + vector `vec u` along the axis we care about, we have `F_{vec u}(x, y, z) = vec u * (F(x, y, z) * vec u)`. + Let's do an example. What is specifically the y-component of the line integral `f(x, y, z) = \[z, xy, -y^2]`? + Given that `vec y = \[0, 1, 0]`, we have `\[0, 1, 0] * (\[x, xy, -y^2] * \[0, 1, 0]) = \[0, xy, 0]`. + We already know that `dr = \[2t, 1, frac 1 {2sqrt(t)}] dt`. `int_a^b \[0, xy, 0] * \[2t, 1, frac 1 {2sqrt(t)}]` + simplifies to `int_a^b xy dt`. We've already determined `xy` = `t^3`, so this becomes simply `int_a^b t^3 dt`.
- 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! + Now onto flux. The flux across a simple closed curve (that is, a closed curve that does not cross over itself) in the xy plane is similar to the line integral, except it finds the value that crosses + the curve, rather than the value along the curve: if the line integral of a vector field is total force along the curve, then the flux over a vector field is the total force perpendicular to the curve. + Essentially, `int_a^b f(x, y, z) * dn` rather than `int_a^b f(x, y, z) * dr`, where `n` is the outward normal vector of `r`. What is an outward normal vector? For a closed curve, it's a vector pointing + perpendicular to the curve and facing away from the curve (that is, traveling in the direction of the normal vector will lead you out of the boundary of the curve). For a 2d curve, you can find this normal vector + with the cross product of tangent vector and the perpendicular axis - for instance, for a curve parallel to the xy plane, `T xx k`. Note that this may sometimes point inwards, and sometimes point outwards! + It's up to you to flip the direction, if necessary. Let's do an example: let's find the flux of the line `r(t) = \[t, 1 - t]` in the circulating vector field `f(x, y) = \[-y, x]`. + We can easily determine that `dr = \[1, -1] dt`, and that cross `k` is simply `\[1, -1]` (are the dimensions here a problem? No. To find the cross product, I briefly extended `dr` into three dimensions with a `0` + in the z-component, and then cut off the z-dimension at the end. This is a perfectly valid operation in this specific situation.) +
++ Hence, `dn = \[1, -1] dt`, meaning our integral is `int \[-y, x] * \[1, -1] dt` (no bounds necessary - we aren't going to evalute a final result). A quick substitution and a dot product gives us + `int -1 dt`, the indefinite integral of which is `-t + C`. Simple, and geometrically accurate. +
++ Section 16.3 will go into quite a bit more detail, as we'll be dealing with the problem of + conservative vs nonconservative paths. For now, da svidanya and bonne nuit!
[/] [=title "Multivariable Exam 3 Review: Thomas 16.2"] diff --git a/site/posts/multivariable-thomas-16.3.html b/site/posts/multivariable-thomas-16.3.html new file mode 100644 index 0000000..380ce73 --- /dev/null +++ b/site/posts/multivariable-thomas-16.3.html @@ -0,0 +1,50 @@ +[!] +[=post-] ++ Hello once again! Today we're covering a very exciting topic: path independence. The basic idea is that a line integral in a vector field is path independent if any path between the endpoints + would get the same result. Not a complex idea, but quite difficult to express in math! If the line integral of a given path in a vector field is path independent, then the vector field is said to be + conservative over that path. Conservative line integrals have a useful consequence: we don't have to consider the path! Given a conservative field `F`, all we need to know is a potential function + `f` where `grad f = F`, and the integral between points A and B is `f(B) - f(A)` regardless of the path. +
++ Finding the potential function is a pretty complex operation. Let's start by doing it in 2d. + Assuming we have a gradient `grad f(x, y) = \[y, x]` (this is a convenient easy one to demonstrate technique, as the answer is quite obviously `f(x, y) = xy`), + we're looking for a function for which the partial derivatives are `frac {df} {dx} = y` and `frac {df} {dy} = x`. There's already a known way to find the reverse of a derivative: an antiderivative. + The tricky part is that, for a situation like this, our constant `C` is actually a function of the left-out variable. For instance, the antiderivative `int y dx` is in this case + clearly `xy + C(y)`. +
++ Now comes a bit of trickery. Because `xy + C(y)` for some unknown function C must be equal to `f(x, y)`, we know that the derivative with respect to *y* must be equal to `frac {df}{dy}`. + The derivative `frac d {dy} xy + C(y)` is obviously `x + C'(y)`. Let's set up an equality: `x + C'(y) = frac {df} {dy} = x`. Ah-ha! Because we can subtract `x` from both sides, we get + `C'(y) = 0`. We can't say for sure if `C(y)` is 0 or constant, but we'll just assume it's 0: given the previous equation `f(x, y) = xy + C(y)`, and `C(y) = 0`, we know that `f(x, y) = xy`. +
++ This is a confusing, crazy, and magical way to reverse the gradient function, and it's a big part of why I'm enjoying this chapter so much. +
++ Let's do it in three dimensions. Say we've a gradient `grad f = \[y + z, x, x - 2z]` (derived from the function `f = xy - z^2 + xz`). Step one is to find `int y + z dx`, which works out to + `f(x, y, z) = xy + xz + H(y, z)`. The derivative of this with respect to `y` is `frac {df} {dy} = x + H_y(y, z)`, and we already know `frac {df} {dy} = x`, so we have + `x + H_y(y, z) = x`. `H_y(y, z)` is 0. +
++ The process is much the same for z. `frac {df} {dz} = x + H_z(y, z) = x - 2z`, meaning `H_z(y, z) = -2z`. We still don't have a value for H, though- just the two partial derivatives. +
++ This is where the process recurses. +
++ We know from these results that `grad H(y, z) = \[0, -2z]`. We can now use exactly the same process as the 2d version to get `H(y, z)` - except, in this case, we have the shortcut + that the answer is very obviously `H(y, z) = -z^2`. These are much faster to do if you can recognize some simple forms that produce a given gradient, and swap them in immediately; + this saves a whole bunch of effort. We've known for a while now that `f(x, y, z) = xy + xz + H(y, z)`, and thus `f(x, y, z) = xy + xz - z^2`. This is correct! +
++ An interesting property I won't get into proving (there's a proof in the Thomas book, very much worth reading!) is that gradients are conservative. Hence, if it's possible to construct + `f(x, y, z)` from the vector field function, this path is conservative! That means that, as long as you have your `f(x, y, z)`, you do not have to worry about the actual path you're dealing with, + and just the value of the potential function at the endpoints. +
+[/] +[=title "Multivariable Exam 3 Review: Thomas 16.3"] +[=author "Tyler Clarke"] +[=date "2025-4-4"] +[=subject "Calculus"] +[#post.html]