Unit 1
HomeRelevant topics: parametrics, polar form, conic sections
This unit was about parametrics, polar equations, and conic sections. It is sensical in many ways to cover these concepts in the same unit, especially the former two because they are extensions beyond the traditional Cartesian coordinate system. Parametric and polar form are are partiuclarly useful because they allow you to express curves that aren't expressable in the Cartesian form.
Problem 1: parametrics
Let's take a look at a parametrics problem first. An example is the cycloid curve:
To solve for the position of a point on the wheel as a function of time, I first described the motion in the vertical direction. If we ignore the horizontal movement, the vertical motion is simply sinusoidal, shifted up by \(r\) (since the center of the wheel is a height of \(r\) above the ground). The cosine function is also negated, because the motion of the wheel is clockwise.
\[ y(t) = r - r\cos(t) \]
From there, the horizontal motion consists of two components: a constant velocity to the right and a sinusoidal term. The center of the wheel moves to the right at a steady velocity \(r t\), while the rotation of the point introduces the oscillation:
\[ x(t) = r t - r\sin(t) \]
This technique of using a parametric description is useful not only because we can separate the horizontal and vertical components of the motion, but also because in problems like this it's literally required. There isn't a clean way to write the Cartesian equation, since \(t\) appears in both polynomial and trigonometric forms.
Problem 2: polar
Now let's see a problem with polar equations. I have this problem about arc length in polar form:
\[ \text{Find the length of the cardioid} \quad r = 1 + \sin(\theta). \]
We can use the pythagorean theorem to find the length of a curve: \(\sqrt{(dx)^2+(dy)^2}\). If we switch to polar coordinates, where every point is described by \((r,\theta)\), the same idea works but the formula changes slightly. After simplifying, the result is:
\[ L = \int_{\theta_1}^{\theta_2} \sqrt{\,r(\theta)^2 + \left(\frac{dr}{d\theta}\right)^2\,}\,d\theta \]
For our curve \(r = 1 + \sin(\theta)\), we first find the derivative:
\[ \frac{dr}{d\theta} = \cos(\theta) \]
And now substitute
\[ L = \int_0^{2\pi} \sqrt{(1+\sin\theta)^2 + (\cos\theta)^2}\,d\theta \\[6pt] = \int_0^{2\pi} \sqrt{1+2\sin(\theta)+\sin(\theta)^2 + \cos(\theta)^2}\,d\theta \\[6pt] = \int_0^{2\pi} \sqrt{2+2\sin\theta}\,d\theta \]
I evaluated the integral using a graphing calculator. The result is \(L=8\).
Again, polar coordinates are very useful in this scenario because the curve is tedious to describe with Cartesian coordinates. In general, I find that most problems containing polar form require very similar methodology to cartesian form. Since we know how to convert between the forms, we can rewrite many of the formulas too.
Conic sections cheat sheet
This is all of the info about conic sections, both cartesian and polar, in one place:
First in carteisan form:
| Conic | Equation | Focus / Foci | Directrix |
|---|---|---|---|
| Circle | \(\small{(x-h)^2 + (y-k)^2 = r^2}\) | - | - |
| Ellipse (horizontal) | \(\small{\tfrac{(x-h)^2}{a^2} + \tfrac{(y-k)^2}{b^2} = 1}\) | \((h \pm c, k)\) | \(x = h \pm \tfrac{a^2}{c}\) |
| Ellipse (vertical) | \(\small{\tfrac{(x-h)^2}{b^2} + \tfrac{(y-k)^2}{a^2} = 1}\) | \((h, k \pm c)\) | \(y = k \pm \tfrac{a^2}{c}\) |
| Hyperbola (horizontal) | \(\small{\tfrac{(x-h)^2}{a^2} - \tfrac{(y-k)^2}{b^2} = 1}\) | \((h \pm c, k)\) | \(x = h \pm \tfrac{a^2}{c}\) |
| Hyperbola (vertical) | \(\small{\tfrac{(y-k)^2}{a^2} - \tfrac{(x-h)^2}{b^2} = 1}\) | \((h, k \pm c)\) | \(y = k \pm \tfrac{a^2}{c}\) |
| Parabola (horizontal) | \(\small{(y-k)^2 = 4p(x-h)}\) | \((h+p, k)\) | \(x = h-p\) |
| Parabola (vertical) | \(\small{(x-h)^2 = 4p(y-k)}\) | \((h, k+p)\) | \(y = k-p\) |
Where \(a\) is the major axis, \(b\) is the minor axis, and:
For an elipse: \(c^2\) = \(a^2 + b^2\)
For a hyperbola: \(c^2\) = \(a^2 - b^2\)
Also consider the asymptotes for a hyperbola:
For horizontal: \[ y - k = \pm \frac{b}{a}(x - h) \]
For vertical: \[ y - k = \pm \frac{a}{b}(x - h) \]
Reflection
I liked this unit because I understood the connection between polar and cartesian formulas like area and arc length. Like I wrote above. many of the same techniques were applicable both in polar, parametric, and cartesian form. In particular, I found it interesting to derive the arc length formula in all 3 coordinate forms. However, it was challenging to derive the area formula for polar form. In part this was because a polar curve repeats over some period, which added another step to finding the bounds for the integral. Aside from these topics, I also found it hard to memorize the rules and features of the conic sections. I hope that we will cover the derivations for these equations soon when we cover 3-dimensional coordinate systems. In unit 11, my goal is to understand more of the calculus derivaitons behind where these formulas come from.