CLP-4 Vector Calculus

That is,

is a parametrization of the circle $x^2+y^2=a^2\text{.}$ Just looking at the figure above, it is clear that, as $\theta$ runs from $0$ to $2\pi \text{,}$ $\mathbf{r}(\theta )$ traces out the full circle.

However beware that just knowing that $\mathbf{r}(t)$ lies on a specified curve does not guarantee that, as $t$ varies, $\mathbf{r}(t)$ covers the entire curve. For example, as $t$ runs over the whole real line, $\frac{2}{\pi }\arctan (t)$ runs over the interval $(-1,1)\text{.}$ For all $t\text{,}$

is well-defined and obeys $x(t{)}^2+y(t{)}^2=a^2\text{.}$ But this $\mathbf{r}(t)$ does not cover the entire circle because $y(t)$ is always positive.

We see from the sketch that

is a parametrization of the circle $(x-h{)}^2+(y-k{)}^2=a^2\text{.}$

A second way to come up with this parametrization is to observe that we can turn the trig identity ${\cos }^{2}t+{\sin }^{2}t=1$ into the equation $(x-h{)}^2+(y-k{)}^2=a^2$ of the circle by

We can build parametrizations of the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^{2/3}+y^{2/3}=a^{2/3}$ from the trig identity ${\cos }^{2}t+{\sin }^{2}t=1\text{,}$ like we did in the second part of the last example.

So

give parametrizations of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^{2/3}+y^{2/3}=a^{2/3}\text{,}$ respectively. To see that running $t$ from $0$ to $2\pi$ runs $\mathbf{r}(t)$ once around the curve, look at the figures below.

A very easy method that can often create parametrizations for a curve is to use $x$ or $y$ as a parameter. Because we can solve $e^y=1+x^2$ for $y$ as a function of $x\text{,}$ namely $y=\ln (1+x^2)\text{,}$ we can use $x$ as the parameter simply by setting $t=x\text{.}$ This gives the parametrization

It is also quite common that one can use either $x$ or $y$ to parametrize part of, but not all of, a curve. A simple example is the circle $x^2+y^2=a^2\text{.}$ For each $-a<x<a\text{,}$ there are two points on the circle with that value of $x\text{.}$ So one cannot use $x$ to parametrize the whole circle. Similarly, for each $-a<y<a\text{,}$ there are two points on the circle with that value of $y\text{.}$ So one cannot use $y$ to parametrize the whole circle. On the other hand

provide parametrizations of the top half and bottom half, respectively, of the circle using $x$ as the parameter, and

provide parametrizations of the right half and left half, respectively, of the circle using $y$ as the parameter.

In this example, we will undo the parametrization $\mathbf{r}(t)=(\cos t,7-t)$ and find the Cartesian equation of the curve in question. We may rewrite the parametrization as

Note that we can eliminate the parameter $t$ simply by using the second equation to solve for $t$ as a function of $y\text{.}$ Namely $t=7-y\text{.}$ Substituting this into the first equation gives us the Cartesian equation

The set of all $(x,y,z)$ obeying

is a curve. We can choose to use $y$ as the parameter and think of

as a system of two equations for the two unknowns $x$ and $z\text{,}$ with $y$ being treated as a given constant, rather than as an unknown. We can now solve the first equation for $x\text{,}$ substitute the result into the second equation, and finally solve for $z\text{.}$

So

is a parametrization for the given curve.

The previous example was rigged so that it was easy to solve for $x$ and $z$ as functions of $y\text{.}$ In practice it is not always easy, or even possible, to do so. A more realistic example is the set of all $(x,y,z)$ obeying

which is the blue curve in the figure above. Substituting $x^2=z-2y^2$ (from the second equation) into the first equation gives

or, completing the square,

If, for example, we are interested in points $(x,y,z)$ on the curve with $y\ge 0\text{,}$ this can be solved to give $y$ as a function of $z\text{.}$

Then $x^2=z-2y^2$ also gives $x$ as a function of $z\text{.}$ If $x\ge 0\text{,}$

The other signs of $x$ and $y$ can be gotten by using the appropriate square roots. In this example, $(x,y,z)$ is on the curve, i.e. satisfies the two original equations, if and only if all of $(\pm x,\pm y,z)$ are also on the curve.