Now that we’re a bit more familiar with parametric curves, we’ll think about them in the context of calculus. There are only a few results that we’ll build, but all of them should end up being relatively intuitive, as long as we refer back to the results in our standard context.
The visual above is helpful in building the parametric definition of the cycloid. We start with a circle. Due to clockwise rotation (instead of the typical counter-clockwise rotation) as well as the point beginning at the bottom of the circle, we parameterize it as:
\begin{align*}
x \amp = -\sin(t)\\
y \amp = -\cos(t)
\end{align*}
In order to shift the circle up to be centered at \((0,1)\text{,}\) we can add \(1\) to the \(y\)-value and use the parametric curve:
\begin{align*}
x \amp = -\sin(t)\\
y \amp = 1-\cos(t)
\end{align*}
Then, in order to shift the circle to the right as the time changes, we add \(t\) to the \(x\)-value:
\begin{align*}
x \amp = t-\sin(t)\\
y \amp = 1-\cos(t)
\end{align*}
If we want to scale the circle to have a generic radius, \(r\text{,}\) we can scale the whole thing!
Definition10.3.1The Cycloid.
The cycloid passing through the origin and generated by a circle of radius \(r\) rolling along the \(x\)-axis from left to right is parameterized as:
\begin{align*}
x \amp = r(t-\sin(t))\\
y \amp = r(1-\cos(t))
\end{align*}
Before we begin, let’s remind ourselves of a key idea when we were introducing parametric curves: Both \(x\) and \(y\) are independent functions of some third variable (or parameter), \(t\text{.}\)
First, make a conjecture: where do you think the cycloid will have horizontal tangent lines? What about vertical tangent lines, or other points where the derivative doesn’t exist?
Use these values of \(t\) to find the points on the cycloid where there are horizontal tangent lines or points where the derivative doesn’t exist. Do they match what you conjectured?
In order to find the area under a curve, we will still use the basic idea of a Riemann Sum: we want to multiply a height (\(y\)) by a width (\(dx\)) on small and specific intervals, and add them up to construct an integral.
Since our height variable, \(y\text{,}\) is a function of \(t\text{,}\) the integral will end up with an input variable that doesn’t match the differential. We can use a change of variables (like in \(u\)-Substitution) to change the differential from \(dx\) to \(dt\text{.}\)
\begin{equation*}
dx = x'(t) \;dt
\end{equation*}
This changes our differential, allowing us to integrate with regard to \(t\text{,}\) the input variable.
\begin{align*}
x \amp = x(t) \\
y \amp = y(t)
\end{align*}
on the interval where \(x(t)\) and \(y(t)\) are continuous on the interval \(a\leq t\leq b\) and \(x(t)\) is differentiable on \(a\lt t \lt b\text{,}\) then the signed area bounded between the parametric curve and the \(x\)-axis from from \(t=a\) until \(t=b\) is
We have built an arc length formula already in this textbook! Remember: Definition 6.5.7 Length of a Curve? Let’s rebuild this integral formula in a way that makes sense for a parametric curve.
Let’s drop back in to that derivation with some small changes. Because we’re dealing with a parametric curve, we’ll use \(t\) as our input variable, and so we’ll eventually need a \(\Delta t\) to turn into a differential \(dt\) in the integral. This means that we’ll have non-uniform \(\Delta x\) distances, since they’ll be based on the changes in the input, \(t\text{.}\) So we’ll actually use \(\Delta x_k\) to represent the change in the \(x\)-variable on the \(k\)th subinterval.
In order to build the arc length formula in Section 6.5, we factored out \(\sqrt{\Delta x^2}\) in order to end up with a \(\Delta x\) that turned into the differential \(dx\) in the integral. This time, factor out a \(\Delta t^2\) under the square root.
Now, create a Riemann sum and let \(n\to \infty\) (and, correspondingly, \(\Delta t\to 0\)). Note that you should end up with some differentials instead of deltas!
