Theory

Theory 1

We can use $x(t)$ and $y(t)$ data to compute the slope of a parametric curve in terms of $t$.

Slope formula

Given a parametric curve $(x(t),y(t))$, its slope satisfies:

$$\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}(\text{where }{x}^{\prime }(t)\ne 0)$$

Concavity formula

Given a parametric curve $(x(t),y(t))$, its concavity satisfies the formula:

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)\cdot \frac{1}{x^{\prime }(t)}(\text{where }{x}^{\prime }(t)\ne 0)$

Extra - Derivation of slope and concavity formulas

For both derivations, it is necessary to view $t$ as a function of $x$ through the inverse parameter function. For example if $x=f(t)$ is the parametrization, then $t=f^{-1}(x)$ is the inverse parameter function.

We will need the derivative $\frac{dt}{dx}$ in terms of $t$. For this we use the formula for derivative of inverse functions:

$$\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}$$

Given all this, both formulas are simple applications of the chain rule.

For the slope:

$$\begin{array}{c}\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}\gg \gg y^{\prime }(t)\cdot \frac{1}{dx/dt}\\ \\ \gg \gg \frac{y^{\prime }(t)}{x^{\prime }(t)}\end{array}$$

For the concavity:

$$\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\gg \gg \frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{dt}{dx}\\ \\ \gg \gg \frac{d}{dt}\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)\cdot \frac{1}{x^{\prime }(t)}\end{array}$$

(In the second step we inserted the formula for $\frac{dy}{dx}$ from the slope.)

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Pure vertical, Pure horizontal movement

In view of the formula $\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}$, we see:

• Pure vertical: when $x^{\prime }(t)=0$ and yet $y^{\prime }(t)\ne 0$
• Pure horizontal: when $y^{\prime }(t)=0$ and yet $x^{\prime }(t)\ne 0$

When $x^{\prime }(t_0)=y^{\prime }(t_0)=0$ for the same $t=t_0$, we have a stationary point, which might subsequently progress into pure vertical, pure horizontal, or neither.

Theory 2

Arclength formula

The arclength of a parametric curve with coordinate functions $x(t)$ and $y(t)$ is:

$$L={\int }_a^b\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt$$

This formula assumes the curve is traversed one time as $t$ increases from $a$ to $b$.

Counts total traversal

This formula applies when the curve image is traversed one time by the moving point.

Sometimes a parametric curve traverses its image with repetitions. The arclength formula would add length from each repetition!

Extra - Derivation of arclength formula

The arclength of a parametric curve is calculated by integrating the infinitesimal arc element:

$$\begin{array}{c}ds=\sqrt{d{x}^2+d{y}^2}\\ \\ L={\int }_a^{b}ds\end{array}$$

In order to integrate $ds$ in the $t$ variable, as we must for parametric curves, we convert $ds$ to a function of $t$:

$$\begin{array}{c}ds=\sqrt{d{x}^2+d{y}^2}\gg \gg \sqrt{\frac{1}{d{t}^2}\cdot (d{x}^2+d{y}^2)\cdot d{t}^2}\\ \\ \gg \gg \sqrt{\frac{d{x}^2}{d{t}^2}+\frac{d{y}^2}{d{t}^2}}\cdot \sqrt{d{t}^2}\gg \gg \sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt\\ \\ \gg \gg ds=\sqrt{x^{\prime }{(t)}^2+y^{\prime }{(t)}^2}dt\end{array}$$

So we obtain $ds=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt$ and the arclength formula follows from this:

$$L={\int }_a^b\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt$$

Theory 3

Distance function

The distance function $s(t)$ returns the total distance traveled by the particle from a chosen starting time $t_0$ up to the (input) time $t$:

$$s(t)={\int }_{t_0}^{t}ds={\int }_{t_0}^t\sqrt{x^{\prime }{(u)}^2+y^{\prime }{(u)}^2}du$$

We need the dummy variable $u$ so that the integration process does not conflict with $t$ in the upper bound.

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Speed function

The speed of a moving particle is the rate of change of distance:

$$v(t)=s^{\prime }(t)=\sqrt{x^{\prime }{(t)}^2+y^{\prime }{(t)}^2}$$

This formula can be explained in either of two ways:

1. Apply the Fundamental Theorem of Calculus to the integral formula for $s(t)$.
2. Consider $ds=\sqrt{x^{\prime }{(t)}^2+y^{\prime }{(t)}^2}dt$ for a small change $dt$: so the rate of change of arclength is $\frac{ds}{dt}$, in other words $s^{\prime }(t)$.

Theory 4

Surface area of a surface of revolution: thin bands

Suppose a parametric curve $(x(t),y(t))$ is revolved around the $x$-axis or the $y$-axis.

The surface area is:

$$A={\int }_a^{b}2\pi R(t)\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt$$

The radius $R(t)$ should be the distance to the axis:

$$\begin{array}{cc}R(t)& =y(t)\text{revolution about }x\text{-axis}\\ R(t)& =x(t)\text{revolution about }y\text{-axis}\end{array}$$

This formulas adds the areas of thin bands, but the bands are demarcated using parametric functions instead of input values of a graphed function.

The formula assumes that the curve is traversed one time as $t$ increases from $a$ to $b$.