Parametric curves and second derivatives: methods

   Let us know come to face the problem of computing the second derivative for functions defined through a pair of parametric equations:
Utilizing the results of the previous theorem on the first derivative, we can proceed -after having obtained the first derivative $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ as a function of the parameter $t$-  to the computation of the second derivative either as
\begin{equation} \label{secder1st}
\frac{d^{2}y}{dx^{2}} = \frac{dy'}{dx} = \frac{dy'/dt}{dx/dt}
\end{equation}
where $y'=\frac{dy}{dx}$, or as
\begin{equation}  \label{secder2nd}
\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\Big[ \frac{dy}{dx} \Big] = \frac{d}{dt}\Big[ \frac{dy}{dx} \Big] \cdot \frac{dt}{dx}
\end{equation}
Remarks:
1. Notice that -according to a previous remark in the computation of the first derivative- if we wish to apply \eqref{secder2nd}, a suitable partition of the domain $E \subseteq \mathbb{R}$ of $x=f(t)$, must be considered in order for $x=f(t)$ to be bijective ("1-1"). Thus, we will have $t=f_{i}^{-1}(x)$ (in the corresponding interval in $E$'s partition) and the correct formula for $\frac{dt}{dx}$ must be replaced.
2. Notice that in general $$\frac{d^{2}y}{dx^{2}} \neq \frac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}$$

Let us now procceed in a couple of clarifying examples:

Example1: Let us consider the pair of parametric equations $\left\{%
                               \begin{array}{l}
                               x = sint  \\
                               y = cos2t
                               \end{array}
                                \right. \ \ $, for $\ t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

It is clear that $x=sint$ is invertible in the domain $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and thus we can directly apply \eqref{secder2nd}:

   Since $\frac{dx}{dt}=cost$ and $\frac{dy}{dt}=-2sin2t$, we have:
$$
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2sin2t}{cost} = \frac{-4 sint cost}{cost} = -4 sint
$$
therefore:
$$
\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\Big[ \frac{dy}{dx} \Big] = \frac{d}{dx}(-4 sint) =  \frac{d}{dt}(-4 sint)  \cdot \frac{dt}{dx} = (-4 cost) \cdot (\frac{1}{cost}) = -4
$$
where we have made use of the fact that $x=sint \Leftrightarrow t=arcsinx$ for $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and thus $\frac{dt}{dx} = (arcsinx)' = \frac{1}{cost} = \frac{1}{\sqrt{1-x^{2}}}$ for $t \in (-\frac{\pi}{2}, \frac{\pi}{2})$. (see previous post on the derivatives of the inverse trigonometric functions).
   Notice that $\frac{dx}{dt} \Big|_{\pm \frac{\pi}{2}} = \frac{dy}{dt} \Big|_{\pm \frac{\pi}{2}}=0$ so these points are singular points and the above result does not apply at these points. Can you figure out what is happening at these points ? (plotting a graph of the parametric equations will probably help you understand the behavior at these singular points).

Example2: The curve given by the parametric equations $\left\{%
                               \begin{array}{l}
                               x = t^{2}  \\
                               y = t^{3}
                               \end{array}
                                \right. \ \ $, for $\ t \in [-\infty, \infty]$ is called a semicubic parabola.

   This curve (and the functions defined by it) can be equivalently described in implicit form by the equation $y^{2}=x^{3}$ (square $y(t)$ and cube $x(t)$ to see this!).
   The graph of these parametric equations consists of two branches: the upper branch corresponding to the function $y=x^{3/2}$ while the lower branch is the graph of the function $y=-x^{3/2}$. These two branches meet at the origin, which corresponds to the value $t=0$.
   Since $\frac{dx}{dt}=2t$ and $\frac{dy}{dt}=3t^{2}$ we can clearly see that the origin corresponds to a singular point of the graph.
   We can readily apply the theorem to compute the first derivative (for either branch):
$$
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{d}{dt}(t^{3})}{\frac{d}{dt}(t^{2})}  = \frac{3t^{2}}{2t} = \frac{3}{2}t
$$
while we will follow \eqref{secder1st} to compute the second derivative:
$$
\frac{d^{2}y}{dx^{2}} = \frac{dy'}{dx} = \frac{dy'/dt}{dx/dt} = \frac{\frac{d}{dt}(3t/2)}{\frac{d}{dt}(t^{2})} = \frac{3}{4t}
$$
   Can you apply \eqref{secder2nd}, after suitably dividing the domain $(-\infty, \infty)$, to obtain the same results?
   Can you figure out what is happening at the singular point $O(0,0)$ ?

Parametric curves: some theory and a few examples

   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

 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:

        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 function
$$
y = \phi(x) = g(f^{-1}(x)) = (g \circ f^{-1})(x):  f(I) \rightarrow \mathbb{R}
$$
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. 

Implicit functions: some remarks

$\bullet$ Let us consider the equation
\begin{equation} \label{eq1}
F(x,y)=0
\end{equation}
and the set
\begin{equation}
A=\{(x,y)/ x,y \in \mathbb{R} \ \ and \ \ F(x,y)=0 \} \subset  \mathbb{R}^{2}
\end{equation}
of real solutions of \eqref{eq1}.
   The set of points of the Cartesian plane represented by the ordered pairs of $A$ is called the graph of  \eqref{eq1}.
$\bullet$ If there is a set $E \subset  \mathbb{R}$ such that
$$
\begin{array}{ccc}
 \forall x \in E,  & \exists \ \ y  \in \mathbb{R}, &   F(x,y)=0
\end{array}
$$
then we can map for any $x \in E$ a single $y=f(x)$ such that
$$
F(x,f(x))=0
$$
In this way a function $f:E \rightarrow \mathbb{R}$ with formula $y=f(x)$ is defined and this function will be called an implicit function defined by \eqref{eq1}.
$\bullet$ If this is the case, the graph of the implicit function $y=f(x)$, $f:E \rightarrow \mathbb{R}$ defined by \eqref{eq1} will be a part of the graph of \eqref{eq1}.
Remarks: 
1. It is interesting to note that an implicit function may be defined through an equation even if there is no way to obtain an analytic expression for the function's formula by solving the equation. This is the case (why?) for example for the equation
\begin{equation}
2y-2x-siny=0
\end{equation}
2. Even in cases in which the formula is easy to handle, many "unexpected" functions may be defined implicitly through an equation; most of them may not even be continuous: this is the case for example with the equation of a circle
\begin{equation}
x^{2}+y^{2}=1
\end{equation}
when it is solved with respect to $y$ a host of functions arise; some of them are continuous:
$$
\begin{array}{ccc}
 \begin{array}{l}
y=f_{1}(x)=\sqrt{1-x^{2}}  \\
[-1,1] \rightarrow \mathbb{R}
 \end{array}   &  \begin{array}{l}
                            y=f_{2}(x)=-\sqrt{1-x^{2}}  \\
                            [-1,1] \rightarrow \mathbb{R}
                            \end{array}
\end{array}
$$

while others are not, either at a point:



$
 y=f_{3}(x)= \left\{%
                               \begin{array}{l}
                               \sqrt{1-x^{2}}, \ \ x \in [-1,0) \\
                              -\sqrt{1-x^{2}}, \ \ x \in [0,1]
                               \end{array}
                                \right.$

or nowhere in their domain:
$$
 y=f_{4}(x)=\left\{%
                            \begin{array}{l}
                            \sqrt{1-x^{2}}, \ \ x \in [-1,1],  x: \ rational \\
                           -\sqrt{1-x^{2}}, \ \ x \in [-1,1], x: \ irrational
                            \end{array}
                            \right.
$$
Note that for all the above functions the domain is common $D_{f_{i}}=[-1,1]$, $i=1,2,3,4$. 

Theoretical Remarks #2 - some insight into the function $\int_{a}^{x}f(t)dt$

   Let us first recall the proposition mentioned (without proof) in yesterday's post:
Proposition: Let a real function $f$, with domain $D_{f}$ and continuous on an interval $\Delta \subseteq D_{f} \subseteq  \mathbb{R}$ and let $a \in \Delta$ be a fixed point of $\Delta$. Then the function $F(x)=\int_{a}^{x}f(t)dt$ is an antiderivative function of $f$. In other words:
$$
F'(x) = \big( \int_{a}^{x}f(t)dt \big)' = f(x)
$$
for all $x \in \Delta$.

   If $x \in \Delta$ is a variable point,
then the value of the definite  integral  $F(x)= \int_{a}^{x}f(t)dt$ represents the area of the shaded  region shown in the figure:
Remark: Notice that the function $F(x)$ is defined on any interval $\Delta \subseteq D_{f}$ in which:

• $a \in \Delta$
• f is continuous on $\Delta$

and (according to the preceding proposition) is differentiable on that $\Delta$.

   Considering $h$ to be infinitesimal, we can now compute
$$
\begin{array}{c}
\Delta F(x) = F(x+h)-F(x) = \int_{a}^{x+h}f(t)dt - \int_{a}^{x}f(t)dt = \\
     \\
 = \int_{x}^{x+h}f(t)dt = E(\Omega) \approx h \cdot  f(x)
\end{array}
$$
In the above, the change $\Delta F(x)$ in the value of the function $F(x)=\int_{a}^{x}f(t)dt$ when $x$ changes to $x+h$ is denoted by $E(\Omega)$ and equals the area of the shaded strip $\Omega$, displayed in the next figure:

Consequently, $f(x) \approx \frac{E(\Omega)}{h}$ and since $h$ is considered to be infinitesimal, we can write:
$$
f(x)=\lim_{h \rightarrow 0}\frac{F(x+h)-F(x)}{h}=F'(x) = \frac{d \big(\int_{a}^{x}f(t)dt \big)}{dx}
$$
Remarks:
(1). The above should not be considered to be a rigorous proof. It should rather be taken as an intuitive  line of thinking, aiming to shed some light into "what is really going on" inside the function $\int_{a}^{x}f(t)dt$
(2). Let a real function $f$, continuous on an interval $\Delta \subseteq  \mathbb{R}$ and let $a \in \Delta$ be a fixed point of $\Delta$. Let another real function $g(x)$ with domain $D_{g}$ and differentiable (and thus: continuous) on an interval $\Delta_{1} \subseteq D_{g}$. We can then consider the composite function $G=F \circ g$:
$$
G(x) = \int_{a}^{g(x)}f(t)dt
$$
The domain $D_{G}$ of $G$ will be:
$$
D_{G} = \{ x \in D_{g} \ , \ g(x) \in \Delta \}
$$
   The composite function $G=F \circ g$ is differentiable on $\Delta_{2} = \Delta_{1} \cap D_{G}$. Its derivative can be computed by combining the above differentiation rule with the chain rule of differentiation (for differentiating composite functions). We readily get the following formula:
$$
\Big( \int_{a}^{g(x)}f(t)dt \Big)' = f\big( g(x) \big) \cdot g'(x)
$$
for all $x \in \Delta_{2}$.

Theoretical Remarks #1 - indefinite integrals, antiderivatives and the function $\int_{a}^{x}f(t)dt$

   Suppose we are given a continuous, real function $f(x)$ defined on an interval $\Delta \subseteq \mathbb{R}$ and let $a \in \Delta$ be a fixed point.
   Any other function $F(x)$, with domain $D_{F}= \Delta$ will be called an antiderivative function of $f$ if
$$
F'(x)=f(x)
$$
(notice that $F$ is by definition differentiable (and thus continuous) in $\Delta$.)
   The above definition implies that: the antiderivative function is not uniquely determined, but rather there is a family of functions satisfying the above relation. Actually, any other function $G(x)$, $D_{G}= \Delta$ with the property
\begin{equation} \notag
G'(x)=F'(x)=f(x)
\end{equation}
will also be an antiderivative function. In such a case $F(x)$ and $G(x)$ will differ by a constant:
\begin{equation} \notag
G(x)=F(x)+c
\end{equation}
for some $c \in \mathbb{R}$. (this comes from a well known theorem of elementary calculus). We can thus now lay the following
   Definition: We will call antiderivative or indefinite integral of $f$, and we will denote it by $\int f(x)dx$ the set of all functions satisfying the above property, thus:
\begin{equation} \notag
\begin{array}{r}
\int f(x)dx = \{F | F'(x)=f(x), \ x \in \Delta  \} = \\
     \\
= \{G(x)+c |\textrm{for all } c \in \mathbb{R} \} \ \ \ \ \ \ \ \
\end{array}
\end{equation}
where in the last equality $G$ is an antiderivative function (actually any antiderivative function) of $f$.

   Now it can be proved (the proof can be found on standard calculus texts and i will -hopefully- post it here later) that:
Proposition: one of these functions belonging in the above set (thus, one of the antiderivative functions of $f$ or equivalently: one of the indefinite integrals of $f$) is the function
\begin{equation} \notag
\int_{a}^{x} f(t)dt
\end{equation}
In other words: $(\int_{a}^{x} f(t)dt )'=f(x)$ for all $x \in \Delta$.

Remarks:
(1). Notice that the above proposition readily implies that $\int_{a}^{x} f(t)dt$ is differentiable (and thus continuous) for any $x \in \Delta$. Of course the $\ ' \ $ symbol indicates differentiation with respect to the variable $x$.
(2). Thus: the definite integral $\int_{a}^{x} f(t)dt $ with variable upper limit of integration, is an antiderivative function of $f$. Consequently, we can write
\begin{equation} \notag
\int f(x)dx = \int_{a}^{x} f(t)dt + c
\end{equation}
for all $c \in \mathbb{R}$.
(3). What the above proposition actually tells us is that: any function $f$ which is continuous on an interval $\Delta \subseteq \mathbb{R}$, has an antiderivative function given by $\int_{a}^{x} f(t)dt $ for $a,x \in \Delta$.
(4). It is worth noticing the meaning of the number $a \in \Delta$: Varying the value of $a \in \Delta$ produces different antiderivative functions (because the variation of $a \in \Delta$ simply alters the value of the constant of integration $c$). However, we cannot hope that varying the value of $a \in \Delta$ "covers" all possible antiderivatives of a given (continuous function $f$). In other words, this means that: although the above theorem tells us that  $\int_{a}^{x} f(t)dt$ is an antiderivative function of $f$, not all antiderivative functions of $f$ can necessarily be expressed as $\int_{a}^{x} f(t)dt$ for some $a \in \Delta$. This can be clearly seen in the following example:
   If $f(x)=2x$ and $\Delta = \mathbb{R}$, then for  any real $a$ we have
\begin{equation} \notag
\int_{a}^{x} 2tdt = [t^{2}]_{a}^{x}=x^{2}-a^{2}
\end{equation}
But, the family of functions $\{F(x)=x^{2}-a^{2}, \ x \in \mathbb{R} | \textrm{for all } a \in  \mathbb{R} \}$ does not include for example the function $G(x)=x^{2}+1$, which is an obvious antiderivative function of $f(x)=2x$.

Revision Exercises (with solutions) on IB Math SL material

   Here are some revision exercises (most of them come with their solutions) on the material of IB Math SL (mainly). They are of course suitable for HL courses as well.
   You can practise on

• Algebra, 
• Calculus, 
• Trigonometry, 
• Functions, 
• Matrices, 
• Statistics and Probability, 
• Vectors.

I hope that these will be of help, particularly in view of the upcoming midterms.

Marry Christmas to everybody !!

Quadratics: notes and some homework

Here are some notes on quadratics and some homework as well.
Enjoy!
 

Homework 18 (on functions)

Homework 17 (on functions)

Functions - class notes