In today's post, I will move on to discuss the topic of parametric equations, the graph of parametric equations, the function(s) -and the conditions under which they are defined- defined by a pair of parametric equations and finally I will state and prove a theorem concerning the differentiation of parametric curves and functions. Cases of applicability broadening the scope of the theorem will also be investigated.
$\bullet$ Let us consider the real functions
\begin{equation} \label{parameq1}
\begin{array}{ccc}
x(t):E \rightarrow \mathbb{R} & & y(t):E \rightarrow \mathbb{R}
\end{array}
\end{equation}
where $E \subseteq \mathbb{R}$. In such a case, a point $(x(t),y(t))$ of the Cartesian plane is mapped to any value of the variable $t$.
The $x(t)$, $y(t)$ functions are called parametric equations and the real variable $t$ will be called parameter.
The set of points of the Cartesian plane represented by the ordered pairs of the set
\begin{equation} \label{parameq2}
\mathcal{C} = \{ (x(t),y(t)) / t \in E \}
\end{equation}
will be called the graph or the curve of the parametric equations \eqref{parameq1}.
$\bullet$ Let us now consider the parametric equations
\begin{equation} \label{parameq3}
\begin{array}{ccc}
x=f(t), \ f : E \rightarrow \mathbb{R} & & y=g(t), \ g : E \rightarrow \mathbb{R}
\end{array}
\end{equation}
and let us suppose that there is $A \subseteq E \subseteq \mathbb{R}$, such that
$$
\forall x \in f(A) \ , \ \exists \ ! \ y \in g(A) \ , \ (x,y) \in \{ (f(t),g(t)) / t \in A \}
$$
It is clear now that a function $\phi: f(A) \rightarrow \mathbb{R}$ has been formed with formula $y=\phi(x)$, whose graph $\mathcal{C}_{\phi}$ is a part of the graph $\mathcal{C}$.
We will then say that the parametric equations \eqref{parameq3} define the function $\phi$.
Remark: The conditions of the above definition are obviously met in case $x=f(t)$ is an invertible (thus: bijective or $``1-1"$) function: in such a case $t=f^{-1}(x)$ and $\phi(x)=g(f^{-1}(x))$; in other words: $\phi = g \circ f^{-1}$. However, we stress the fact that this is not the only case: the conditions described earlier are much wider than demanding the existence of an inverse function for $f$. For example we can readily see that the parametric functions $x = f(t) = t^{2}$, $y = g(t) = t^{4}+1$ define a (unique) function $y = \phi(x) = x^{2}+1$. Of course in this case the $f$ function is clearly not invertible. However the conditions described above are met !
Examples:
$\boxed{1}$ If we consider the parametric equations $x(\theta) = \rho cos\theta$, $y(\theta) = \rho sin\theta$ for $\theta \in [0, 2\pi]$, then the graph (or: the curve) of these parametric equations is a circle of radius $\rho$ centered at the origin. It clearly is not a function (at least not a single!).
$\boxed{2}$ If we consider the parametric equations $x(\theta) = \rho cos\theta$, $y(\theta) = \rho sin\theta$ for $\theta \in [0, \pi]$, then the graph (or: the curve) of these parametric equations is the upper circle of radius $\rho$ centered at the origin. We say that the parametric equations define the function $y=\sqrt{1-x^{2}}$ whose graph completely coincides with the graph of the parametric equations.
$\boxed{3}$ If we consider the parametric equations $x(\theta) = \rho cos\theta$, $y(\theta) = \rho sin\theta$ for $\theta \in [\pi, 2\pi]$, then the graph (or: the curve) of these parametric equations is the lower semicircle of radius $\rho$ centered at the origin. We say that the parametric equations define the function $y=-\sqrt{1-x^{2}}$ whose graph completely coincides with the graph of the parametric equations.
It should be noted at this point that the intervals $[0, \pi]$, $[\pi, 2\pi]$ in which $[0, 2\pi]$ was divided in order for the parametric equations to define a single function each time, are exactly those intervals for which the $x(\theta) = \rho cos\theta$ function is bijective (i.e. $``1-1"$) and thus invertible.
$\boxed{4}$ The following figure displays the graph
.+parametric+function+example.png)
of the parametric functions $x(t)=t-3sint$, $y(t)=4-3sint$ for $t \in [0,10]$.
$\bullet$ The inverse of a function in parametric form: The above discussion implies that given any function $y=f(x)$, $x \in D_{f}$, this can be written in parametric form as: $x=t$, $y=f(t)$, $t \in D_{f}$.
The inverse function (in case it exists) can be written as $y=f^{-1}(x)$ (we assume that we are using a common coordinate system for both the initial and the inverse) which is equivalent to saying $x=f(y)$. This (inverse) function in turn, can be written -following exactly what we did earlier- in parametric form as: $x=f(t)$, $y=t$, $t \in f(D_{f})$, where $f(D_{f})$ is the range of $f$ or equivalently the domain of $f^{-1}$.
$\bullet$ Theorem: (derivative of a function defined in parametric form)
If:
$
x=f(t), \ f : I \rightarrow \mathbb{R} \ , \ \ y=g(t), \ g : I \rightarrow \mathbb{R}
$
where $I$ is an interval, are differentiable functions
$$
y = \phi(x) = g(f^{-1}(x)) = (g \circ f^{-1})(x): f(I) \rightarrow \mathbb{R}
$$
which is differentiable; moreover for any $x \in f(I)$
\begin{equation} \label{parameq4}
\frac{d\phi}{dx} = \frac{dy/dt}{dx/dt}
\end{equation}
or equivalently $\phi'(x) = \frac{g'(t)}{f'(t)}$.
We will provide a proof for this theorem in some subsequent post.
Remarks:
1. Two different situations may occur at points of the domain where $dx/dt=0$:
If at a given point $P$ we have $dx/dt = 0$ and $dy/dt \neq 0$, then at such a point $dy/dx \big|_{P}=\phi'(P)$ will be infinite, we will say that the slope is infinite at the given point and that the tangent to the graph of the parametric equations at $P$ will be vertical.
If at a given point we have $dx/dt = dy/dt = 0$ then the rhs of \eqref{parameq4} becomes an indeterminate form; such points are called singular points. Unfortunately, we have no general statement available for the behavior of parametric equations at singular points: they have to be investigated case-by-case.
2. In the case that we are working with parametric equations $x=f(t)$, $y=g(t)$ and the $f$ function is not invertible (i.e. $f$ is a ``many-to-one" function) we can still revert the $x=f(t)$ formula obtaining more than one functions of the form $t=f_{i}^{-1}(x)$ (for various values of $i$). This usually amounts to a suitable partition of the initial domain $D_{f}$ (see examples 1-3 earlier) into suitable domains $D_{f_{i}}$. Then the above theorem is valid for each one of the $f_{i}$ functions and since they all stem from the single function $x=f(t)$, $t \in D_{f}$ it can be applied once and for all !
The following example is supposed to shed some light in this last remark:
Example: The unit circle can be written in parametric form (see example 1) as: $x(\theta) = cos\theta$, $y(\theta) = sin\theta$ for $\theta \in [0, 2\pi]$. An arbitrary point on the circumference has coordinates $(cos\theta, sin\theta)$.
According to the previous theorem (and the last remark), the derivative of either of the two functions defined (i.e. the upper and the lower semicircle respectively, see ex.2,3) at the given point will be
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=-\frac{cos\theta}{sin\theta}$
thus the equation of the tangent at the (arbitrary) point $(cos\theta, sin\theta)$ will be:
$y-sin\theta = -\big( \frac{cos\theta}{sin\theta} \big)(x-cos\theta)$
Notice that the situation is exactly the same no matter which semicircle the point belongs at!. Thus, in accordance with the last remark earlier, the theorem has been applied once and for all, covering both the $y = f_{1} = \sqrt{1-x^{2}}$ and the $y = f_{2} = -\sqrt{1-x^{2}}$ functions defined by the initial parametric equations.
$\bullet$ Let us consider the real functions
\begin{equation} \label{parameq1}
\begin{array}{ccc}
x(t):E \rightarrow \mathbb{R} & & y(t):E \rightarrow \mathbb{R}
\end{array}
\end{equation}
where $E \subseteq \mathbb{R}$. In such a case, a point $(x(t),y(t))$ of the Cartesian plane is mapped to any value of the variable $t$.
The $x(t)$, $y(t)$ functions are called parametric equations and the real variable $t$ will be called parameter.
The set of points of the Cartesian plane represented by the ordered pairs of the set
\begin{equation} \label{parameq2}
\mathcal{C} = \{ (x(t),y(t)) / t \in E \}
\end{equation}
will be called the graph or the curve of the parametric equations \eqref{parameq1}.
$\bullet$ Let us now consider the parametric equations
\begin{equation} \label{parameq3}
\begin{array}{ccc}
x=f(t), \ f : E \rightarrow \mathbb{R} & & y=g(t), \ g : E \rightarrow \mathbb{R}
\end{array}
\end{equation}
and let us suppose that there is $A \subseteq E \subseteq \mathbb{R}$, such that
$$
\forall x \in f(A) \ , \ \exists \ ! \ y \in g(A) \ , \ (x,y) \in \{ (f(t),g(t)) / t \in A \}
$$
It is clear now that a function $\phi: f(A) \rightarrow \mathbb{R}$ has been formed with formula $y=\phi(x)$, whose graph $\mathcal{C}_{\phi}$ is a part of the graph $\mathcal{C}$.
We will then say that the parametric equations \eqref{parameq3} define the function $\phi$.
Remark: The conditions of the above definition are obviously met in case $x=f(t)$ is an invertible (thus: bijective or $``1-1"$) function: in such a case $t=f^{-1}(x)$ and $\phi(x)=g(f^{-1}(x))$; in other words: $\phi = g \circ f^{-1}$. However, we stress the fact that this is not the only case: the conditions described earlier are much wider than demanding the existence of an inverse function for $f$. For example we can readily see that the parametric functions $x = f(t) = t^{2}$, $y = g(t) = t^{4}+1$ define a (unique) function $y = \phi(x) = x^{2}+1$. Of course in this case the $f$ function is clearly not invertible. However the conditions described above are met !
Examples:
$\boxed{1}$ If we consider the parametric equations $x(\theta) = \rho cos\theta$, $y(\theta) = \rho sin\theta$ for $\theta \in [0, 2\pi]$, then the graph (or: the curve) of these parametric equations is a circle of radius $\rho$ centered at the origin. It clearly is not a function (at least not a single!).
$\boxed{2}$ If we consider the parametric equations $x(\theta) = \rho cos\theta$, $y(\theta) = \rho sin\theta$ for $\theta \in [0, \pi]$, then the graph (or: the curve) of these parametric equations is the upper circle of radius $\rho$ centered at the origin. We say that the parametric equations define the function $y=\sqrt{1-x^{2}}$ whose graph completely coincides with the graph of the parametric equations.
$\boxed{3}$ If we consider the parametric equations $x(\theta) = \rho cos\theta$, $y(\theta) = \rho sin\theta$ for $\theta \in [\pi, 2\pi]$, then the graph (or: the curve) of these parametric equations is the lower semicircle of radius $\rho$ centered at the origin. We say that the parametric equations define the function $y=-\sqrt{1-x^{2}}$ whose graph completely coincides with the graph of the parametric equations.
It should be noted at this point that the intervals $[0, \pi]$, $[\pi, 2\pi]$ in which $[0, 2\pi]$ was divided in order for the parametric equations to define a single function each time, are exactly those intervals for which the $x(\theta) = \rho cos\theta$ function is bijective (i.e. $``1-1"$) and thus invertible.
$\boxed{4}$ The following figure displays the graph
$\bullet$ The inverse of a function in parametric form: The above discussion implies that given any function $y=f(x)$, $x \in D_{f}$, this can be written in parametric form as: $x=t$, $y=f(t)$, $t \in D_{f}$.
The inverse function (in case it exists) can be written as $y=f^{-1}(x)$ (we assume that we are using a common coordinate system for both the initial and the inverse) which is equivalent to saying $x=f(y)$. This (inverse) function in turn, can be written -following exactly what we did earlier- in parametric form as: $x=f(t)$, $y=t$, $t \in f(D_{f})$, where $f(D_{f})$ is the range of $f$ or equivalently the domain of $f^{-1}$.
$\bullet$ Theorem: (derivative of a function defined in parametric form)
If:
1. the parametric functions
x=f(t), \ f : I \rightarrow \mathbb{R} \ , \ \ y=g(t), \ g : I \rightarrow \mathbb{R}
$
where $I$ is an interval, are differentiable functions
2. the function $x=f(t)$ is $``1-1"$ (and thus invertible)
3. For any $t \in I$, $f'(t) \neq 0$
then the parametric equations $x=f(t)$, $y=g(t)$, $t \in I$, define the function3. For any $t \in I$, $f'(t) \neq 0$
$$
y = \phi(x) = g(f^{-1}(x)) = (g \circ f^{-1})(x): f(I) \rightarrow \mathbb{R}
$$
which is differentiable; moreover for any $x \in f(I)$
\begin{equation} \label{parameq4}
\frac{d\phi}{dx} = \frac{dy/dt}{dx/dt}
\end{equation}
or equivalently $\phi'(x) = \frac{g'(t)}{f'(t)}$.
We will provide a proof for this theorem in some subsequent post.
Remarks:
1. Two different situations may occur at points of the domain where $dx/dt=0$:
If at a given point $P$ we have $dx/dt = 0$ and $dy/dt \neq 0$, then at such a point $dy/dx \big|_{P}=\phi'(P)$ will be infinite, we will say that the slope is infinite at the given point and that the tangent to the graph of the parametric equations at $P$ will be vertical.
If at a given point we have $dx/dt = dy/dt = 0$ then the rhs of \eqref{parameq4} becomes an indeterminate form; such points are called singular points. Unfortunately, we have no general statement available for the behavior of parametric equations at singular points: they have to be investigated case-by-case.
2. In the case that we are working with parametric equations $x=f(t)$, $y=g(t)$ and the $f$ function is not invertible (i.e. $f$ is a ``many-to-one" function) we can still revert the $x=f(t)$ formula obtaining more than one functions of the form $t=f_{i}^{-1}(x)$ (for various values of $i$). This usually amounts to a suitable partition of the initial domain $D_{f}$ (see examples 1-3 earlier) into suitable domains $D_{f_{i}}$. Then the above theorem is valid for each one of the $f_{i}$ functions and since they all stem from the single function $x=f(t)$, $t \in D_{f}$ it can be applied once and for all !
The following example is supposed to shed some light in this last remark:
According to the previous theorem (and the last remark), the derivative of either of the two functions defined (i.e. the upper and the lower semicircle respectively, see ex.2,3) at the given point will be
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=-\frac{cos\theta}{sin\theta}$
thus the equation of the tangent at the (arbitrary) point $(cos\theta, sin\theta)$ will be:
$y-sin\theta = -\big( \frac{cos\theta}{sin\theta} \big)(x-cos\theta)$
Notice that the situation is exactly the same no matter which semicircle the point belongs at!. Thus, in accordance with the last remark earlier, the theorem has been applied once and for all, covering both the $y = f_{1} = \sqrt{1-x^{2}}$ and the $y = f_{2} = -\sqrt{1-x^{2}}$ functions defined by the initial parametric equations.
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