Tyler Clarke
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3.9 KiB
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[!]
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[=post-]
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<p>
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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 <i>any</i> path between the endpoints
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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
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<i>conservative</i> 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 <i>potential function</i>
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`f` where `grad f = F`, and the integral between points A and B is `f(B) - f(A)` <i>regardless of the path</i>.
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</p>
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<p>
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Finding the potential function is a pretty complex operation. Let's start by doing it in 2d.
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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`),
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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 <i>anti</i>derivative.
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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
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clearly `xy + C(y)`.
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</p>
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<p>
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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}`.
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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
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`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`.
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</p>
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<p>
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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.
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</p>
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<p>
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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
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`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
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`x + H_y(y, z) = x`. `H_y(y, z)` is 0.
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</p>
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<p>
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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.
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</p>
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<p>
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This is where the process recurses.
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</p>
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<p>
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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
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that the answer is very obviously `H(y, z) = -z^2`. These are <i>much</i> faster to do if you can recognize some simple forms that produce a given gradient, and swap them in immediately;
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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!
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</p>
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<p>
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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 <i>possible</i> to construct
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`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,
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and just the value of the potential function at the endpoints.
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</p>
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[/]
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[=title "Multivariable Exam 3 Review: Thomas 16.3"]
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[=author "Tyler Clarke"]
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[=date "2025-4-4"]
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[=subject "Calculus"]
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[=unpub]
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[#post.html]
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