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+[!]
+[=post-]
+<p>
+    Hello again, <i>mes amis</i>! We've only four sections left in the chapter, and this is one of them. In this section, we're taking a bit of a trip down memory lane,
+    revisiting some old concepts with new insight: namely, <i>parametric vector forms</i>. If you don't already know how to handle parametric forms with one free variable (a line or path),
+    you should probably learn about those externally; we're covering parametric forms with <i>two</i> free variables. Surfaces!
+</p>
+<p>
+    The general idea is that, just like you can represent any line or path with a function `r(u) = \[x(u), y(u), z(u)]`, you can represent any surface with a function
+    `r(u, v) = \[x(u, v), y(u, v), z(u, v)]`. `u` and `v` might be bounded, or might not. That's all there is to it!
+</p>
+<p>
+    For example, say we're given the plane `x + y + z = 1`. How can this be vector parameterized? Obviously doing this in one free variable won't work; it's not possible to represent
+    a plane with a line, no matter how bendy. In two variables, however, it's quite possible. There are many, many possible ways to find the paramterization, but I'm going to use the
+    most obvious one: set `x = u`, `y = v`, and we have `u + v + z = 1`: algebra turns this into `z = 1 - u - v`. So now we have three equations, relating each dimension
+    to <i>two</i> variables. It's a parameterization! Sub'ing in gives us `r(u, v) = \[u, v, 1 - u - v]`.
+</p>
+<p>
+    What about a more complicated shape? It turns out that most shapes are pretty intuitive: for instance, a cylinder with radius `rho` could be parameterized as `r(u, v) = \[rho cos(u), rho sin(u), v]`
+    where `v` is bounded to the height of the cylinder and `0 <= u <= 2 pi`. It's critical to remember, when parameterizing, the bounds of `u` and `v` are <i>your choice</i> - unless you've
+    actually been provided bounds, you can set them to whatever you want; this means cylindrical (or even spherical!) coordinates are an option.
+</p>
+<p>
+    It's often useful to find the <i>tangent plane</i> at a given point on a surface. This is much, much easier if you have a parametric vector function. If we have a function `r(u, v)`
+    describing a curve, and a point `(u, v)` where we want to construct the tangent plane (note: if you're given a point in `(x, y, z)`, you'll have to do some algebra to find `u` and `v`),
+    we immediately know that the initial point `p` is `r(u, v)`. The normal vector is a bit harder to find: the function for it is `n = frac {dr} {du} times frac {dr} {dv}`. To understand why
+    this works, you'll need to visualize `frac {dr} {du}` and `frac {dr} {dv}` as being directional vectors along the surface: assuming they both exist and are nonequal, both are
+    aligned with the surface, and so the cross product between them <i>must</i> point directly out of or into the surface. Once we know `n` and `p`, of course, the plane is simply
+    `n cdot \[x, y, z] = n cdot p`.
+</p>
+<p>
+    Let's do an example. We've the sphere `r(u, v) = \[4cos(u)sin(v), 4sin(u)sin(v), 4cos(v)]`, `0 <= u <= 2pi`, `0 <= v <= pi`. What's the tangent plane at the point
+    `(2, 2, 2sqrt(2))`? (I <i>might</i> be a bit too lazy with these examples). We know that `4cos(u)sin(v) = 2`, `4sin(u)sin(v) = 2`, and `4cos(v) = 2sqrt(2)`. The first
+    two are hairy, but that last one works out nicely to `v = frac {pi} 4`. Because we know `sin(frac {pi} 4) = frac {sqrt(2)} 2`, the equation for y becomes
+    `4sin(u) = frac {4} {sqrt(2)}` - `sin(u) = frac {sqrt(2)} 2`. Obviously, then, `u = frac {pi} 4`. 
+</p>
+<p>
+    Our initial point `p` was helpfully provided, so we don't have to do any work to find that. Now onto the normal vector! `frac {dr} {du} = \[-4sin(u)sin(v), 4cos(u)sin(v), 0]`,
+    and `frac {dr} {dv} = \[4cos(u)cos(v), 4sin(u)cos(v), -4sin(v)]`. Calculating the cross product of those is a good exercise, but ultimately fairly simple; I recommend you do it,
+    but if you don't want to, the answer is `\[-16 cos(u) sin^2 (v), -16 sin(u) sin^2 (v), -16 sin(v) cos(v)]`. This is... pretty awful. Fortunately, we're almost done: all we have
+    to do is plug in those values for `u` and `v`. Final result: `n = \[-4sqrt(2), -4sqrt(2), -8]`. And we're done! Constructing the plane equation is really very easy; we end up with
+    `-4sqrt(2)x - 4sqrt(2)y - 8z = -32sqrt(2)`, or, simplified, `sqrt(2)x + sqrt(2)y + 2z = 8sqrt(2)`. 
+</p>
+<p>
+    Finally: after parameterizing a surface, it becomes quite simple to find the area. Plug in your bounds for `u` and `v` into a double integral, and simply apply the area element!
+    For a general parameterized function, the area element is actually just `abs {frac {dr} {du} times frac {dr} {dv}}` - the magnitude of the normal vector to the tangent plane.
+    In our sphere example above, that would be `sqrt((-16 cos(u) sin^2 (v))^2 + (-16 sin(u) sin^2 (v))^2 + (-16 sin(v) cos(v))^2)` - which simplifies to `16sin(v)`, magically enough.
+    The integral is thus `int_0^{2pi} int_0^{pi} 16sin(v) dv du = 64pi`. This matches up perfectly with the formula for the surface area of a sphere, `4pi r^2`. Great!
+</p>
+<p>
+    I've intentionally ignored implicit surfaces because they don't seem to be showing up in this course. If you actually have the Thomas textbook, I'd recommend reading through
+    that part, for interestingness value if nothing else. Until 16.6, <i>arrivederci</i> and good luck!
+</p>
+[/]
+[=title "Multivariable Exam 3 Review: Thomas 16.5"]
+[=subject "Calculus"]
+[=author "Tyler Clarke"]
+[=date "2025-4-10"]
+[#post.html]