Review from Calculus II

What do parametric equations describe?

The notion of parametric equations in analytic geometry was developed to describe objects moving along curves (or surfaces, even!).  Why is this useful?  Well, imagine that we build a track shaped like a parabola:

Now, with a little work, you could figure out that the equations describing the shape of this curve is $x=y^2$.  But now, let's imagine that we put a mouse on the track at the point $(0,0)$.  If it were a mouse with a one-track mind, it might just start at the point where we placed it, and begin walking in one direction up the curve.  Keep track of the mouse's $x$ and $y$ coordinates at various times, and you might have some data that looks something like this:

Let's plot this data in two different ways.  First, we plot each of the coordinates $x$ and $y$ as functions of time $t$.  Then, we will plot the points $(x,y)$.

Okay, nothing unexpected so far.  But what if our mouse turns around at some point?

So, it is clear that the motion is different than the motion of the first mouse.  But what happens in the $(x,y)$ plane?  Let's plot our new $(x,y)$ coordinates in red, together with our old $(x,y)$ coordinates in black.

So (as expected) the two mice are traveling along exactly the same track!  So, what are parametric equations useful for?  Equations $f(x,y)=0$ are sufficient for describing shapes; in our mouse example, the shape of the track is described by$x-y^2=0$Whenever we need to describe a shape together with a motion along that shape, we need something extra (a time variable).  The time variable allows us to describe the location along a shape a particular object is at different times.  In the mouse example, the mice are both moving along the same, parabolic track (seen in the $(x,y)$ coordinate plots) but they are moving in very different ways (as seen in the plots of the coordinates $x$ and $y$ against time $t$).

Parametric equations as descriptions of motion along shapes are crucial for the study of differential equations (and physics, and economics, and just about anything that involves quantities that change over time!).

Notation for parametric equations:

You have probably seen parametric equations described something like$x=f(t),\ \ y=g(t),\ \ \text{for\ }a\leq t\leq b$In differential equations, we use this notation, but sometimes we additionally "compactify" it.  We often gather the individual coordinates into a vector, as$Y(t)=\left[\begin{array}{c}f(t)\\ g(t)\end{array}\right]$For instance, the equations$x=t^2,\ \ y=t$can be written as$Y(t)=\left[\begin{array}{c}t^2\\ t\end{array}\right]$

This notation can take a little while to get used to!  When we compute the derivative vector along a parametric curve, we do not usually write$\frac{dx}{dt}=f'(t),\ \ \frac{dy}{dt}=g'(t)$but instead we like to write$\frac{dY}{dt}=\left[\begin{array}{c}dx/dt\\ dy/dt\end{array}\right]$So, what takes getting used to?  We just wrote the derivatives in a column; what's hard about that?  Well, nothing... but it's a little harder to remember that $dY/dt$ is a tangent vector (not a number) when we don't explicitly write out the derivatives in the coordinates.

The geometry of parametric curves

As in Calculus II, we can compute the slope of the tangent line to the shape of the curve that the object is traveling along by using the following formula (derived from the Chain Rule):$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

It can be challenging to find the intersection points of two parametric curves.  Why?  Well, a parametric curve is a description of the motion of an object along a shape.  So, what do we mean by "intersection"?  If we mean that the objects moving along the shapes collide, then the objects must be at the same place, at the same time.  If we mean that the shapes intersect, then the objects need to be at the same place, but may be there at different times.  Cars driving along different roads behave the same way; if by "intersect" we mean that the cars collide, then they must be at the same crossing at the same time, but if we simply mean that the roads intersect, the cars need not ever collide.

For example, consider$X(t)=\left[\begin{array}{c}t-2\\ t-2\end{array}\right],\ \ \ Y(t)=\left[\begin{array}{c}t\\ t^2\end{array}\right],\ \ \ Z(t)=\left[\begin{array}{c}4t^2\\ 2t\end{array}\right]$We can plot these curves as follows ($X$ red, $Y$ blue, and $Z$ green).

If we are only concerned that the "roads" intersect, then we can see from these graphs that the three roads do indeed intersect at two points, $(x,y)=(0,0)$ and $(1,1)$.  But what if we are interested in whether the objects represented by the parametric equations collide?

Let's try to find out whether the objects represented by $X(t)$ and $Y(t)$ collide first.  We need to solve $X(t)=Y(t)$, that is,$t-2= t\ \ and\ \ t-2=t^2$

If we are only concerned that the "roads" intersect, then we can see from these graphs that the three roads do indeed intersect at two points, $(x,y)=(0,0)$ and $(1,1)$.  But what if we are interested in whether the objects represented by the parametric equations collide? Let's try to find out whether the objects represented by $X(t)$ and $Y(t)$ collide first.  We need to solve $X(t)=Y(t)$, that is,$t-2= t\ \ \text{and}\ \ t-2=t^2$But the first equation, $t-2=t$, tells us that we have no solutions.  Thus, the objects modeled by $X$ and $Y$ do not collide.

Next, let's try to find out whether the objects represented by $Y$ and $Z$ collide.  We need to solve$t=4t^2\ \ \text{and}\ \ t^2=2t$The first equation has solutions $t=0$ and $t=1/4$.  Since $t=0$ solves the second equation but $t=1/4$ does not, we know that $t=0$ is the only time at which these objects collide.  So, the objects represented by $Y$ and $Z$ collide at time $t=0$ only, and this occurs at the point $Y(0)=(0,0)$.

Next let's see how to find out where the curves intersect (but where collisions might not occur).  Let's look for intersections of $X$ and $Y$.  We need to solve something like $X(t)=Y(t)$... but this is not right! Why not?  Well, as we saw just above, the equation $X(t)=Y(t)$ finds collisions of objects, but not intersections of the curves.  The problem is that the times being plugged into $X$ and $Y$ are the same... but intersections are locations that both objects pass through, but possibly at different times.  So, we need to ask:  are there times $t$ and $s$ such that $X(t)=Y(s)$?  Well, let's try to solve these equations!$t-2=s\ \ \text{and}\ \ t-2=s^2$Taking the two equations together, the left hand sides are the same, so the right hand sides must be equal: $s=s^2$.  This leads us to solutions $s=0$ or $1$, and $t=s+2$.  So the curves intersect at the points $Y(s=0)=(0,0)$ and $Y(s=1)=(1,1)$ (which is what our graph tells us).  (I call this technique the method of the cloned parameter since it was necessary to copy our parameter $t$ but give it a new name, $s$.)

Exercises:

1. Suppose that we have the parametric equations$Y(t)=\left[\begin{array}{c}\cos(t)\\ \sin(t)\end{array}\right]$Compute$\left.\frac{dY}{dt}\right|_{t=\pi/4}$Is the particle being modeled moving up or down at time $t=\pi/4$?  Right or left?
2. Suppose that we have the parametric equations$Y(t)=\left[\begin{array}{c}e^{-t/10}\cos(t)\\ e^{-t/10}\sin(t)\end{array}\right]$As time $t$ increases from $0$ towards $\infty$, what is happening to the particle being modeled?  The following plot may help you to formulate an explanation, but try to connect the observed behavior to the terms in the parametric equations.

1. Suppose that$\frac{dY}{dt}=\left[\begin{array}{c}6t^5-3t^2\\ 3t^2\end{array}\right]\ \ \ \text{and}\ \ \ Y(0)=0$Find $Y(t)$.  Use this to show that the path of the parametric equations lies on the curve$x=y^3-y$in the $(x,y)$-plane.  (For bonus karma points, plot the path of the parametric curve for $-4\leq t\leq 4$ in the $(x,y)$-plane.)
2. A particle is traveling with parametric equations$Y(t)=\left[\begin{array}{c}\cos(t)\\ 2\sin(t)\end{array}\right]$At time $t=\pi/2+0.1$, the particle is released and flung away from the curve, tangential to its path along the curve at that time.  Does the particle hit the circle of radius 1 centered at $(-3,1)$?  Explain.
3. Consider the parametric curve given by$Y(t)=\left[\begin{array}{c}t^5-4t^3\\ t^2\end{array}\right]$At the point $(x,y)=(0,4)$, something "weird" happens (see the graph below).  Find the slope(s) of the tangent line(s) to the parametric curve at $(x,y)=(0,4)$.  (Hint: You were not given a time at which the object passes through $(0,4)$.  You will need to begin by finding the time(s) at which this happens.  There should be two.)

1. Suppose that $f(x,y)=x^3-5x^2-y$, $x(t)=t^2$, and $y(t)=e^t$.  First, write out $F(t)=f(x(t),y(t))$.  Then, compute the quantities$\frac{dF}{dt},\ \ \frac{dx}{dt},\ \ \frac{dy}{dt},\ \ \frac{\partial f}{\partial x},\ \ \frac{\partial f}{\partial y}$Show that in this example,$\frac{dF}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$(This is one version of the Chain Rule for multivariable functions.)
2. Suppose I hand you the vector $(a,b)$, where at least one of $a$ and $b$ is not zero.  Show that $(b,-a)$ is always perpendicular to $(a,b)$.  (Try this for a few examples.  But, your argument has to be written generically; that is, you have to convince your reader that this fact always holds!  Try writing the slopes of the lines from $(0,0)$ to $(a,b)$ and from $(0,0)$ to $(b,-a)$ and see what happens.  Also, if you have seen something called the "dot product" before, remember that you have not done anything with it in this class.  So, you can use it in this problem, but if you do, you first have to explain why it has any properties that you need it to have.)