1: Curves

Example 1.0.1. Parametrization of $x^2+y^2=a^2$

While we will often use $t$ as the parameter in a parametrized curve $\vecs{r} (t)\text{,}$ there is no need to call it $t\text{.}$ Sometimes it is natural to use a different name for the parameter. For example, consider the circle $x^2+y^2=a^2\text{.}$ It is natural to use the angle $\theta$ in the sketch below to label the point $\big(a\cos\theta\,,\,a\sin\theta\big)$ on the circle.

That is,

$$\vecs{r} (\theta) = \big(a\cos\theta\,,\,a\sin\theta\big)\qquad 0\le \theta\lt 2\pi \nonumber$$

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 $2\pi\text{,}$ $\vecs{r} (\theta)$ traces out the full circle.

However beware that just knowing that $\vecs{r} (t)$ lies on a specified curve does not guarantee that, as $t$ varies, $\vecs{r} (t)$ 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 $t\text{,}$

$$\vecs{r} (t) = \big(x(t),y(t)\big) = a\left(\frac{2}{\pi}\arctan(t)\,,\, \sqrt{1-\frac{4}{\pi^2}\arctan^2(t)}\,\right) \nonumber$$

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

Example 1.0.2. Parametrization of $(x-h)^2+(y-k)^2=a^2$

We can tweak the parametrization of Example 1.0.1 to get a parametrization of the circle of radius $a$ that is centred on $(h,k)\text{.}$ One way to do so is to redraw the sketch of Example 1.0.1 with the circle translated so that its centre is at $(h,k)\text{.}$

We see from the sketch that

$$\vecs{r} (\theta) = \big(h+a\cos\theta\,,\,k+a\sin\theta\big)\qquad 0\le \theta\lt 2\pi \nonumber$$

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

• multiplying the trig identity by $a^2$ to get $(a\cos t)^2 +(a\sin t)^2 =a^2$ and then
• setting $\ a\cos t=x-h\ $ and $\ a\sin t=y-k\ \text{,}$ which turns $(a\cos t)^2 +(a\sin t)^2 =a^2$ into $(x-h)^2+(y-k)^2=a^2\text{.}$

Example 1.0.3. Parametrization of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and of $x^{2/3}+y^{2/3}=a^{2/3}$

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.

• Setting $\ \cos t=\frac{x}{a}\ $ and $\ \sin t=\frac{y}{b}\ $ turns $\cos^2 t +\sin^2 t =1$ into $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{.}$
• Setting $\ \cos t= \big(\frac{x}{a}\big)^{\frac{1}{3}}\ $ and $\ \sin t=\big(\frac{y}{a}\big)^{\frac{1}{3}}\ $ turns $\cos^2 t +\sin^2 t =1$ into $\frac{x^{2/3}}{a^{2/3}}+\frac{y^{2/3}}{a^{2/3}}=1\text{.}$

So

$$\begin{alignat*}{2} \vecs{r} (t) &= \big(a\cos t\,,\,b\sin t\big)\qquad &0\le t\lt 2\pi\\ \vecs{r} (t) &= \big(a\cos^3 t\,,\,a\sin^3 t\big) &0\le t\lt 2\pi \end{alignat*}$$

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 $\vecs{r} (t)$ once around the curve, look at the figures below.

The curve $x^{2/3}+y^{2/3}=a^{2/3}$ is called an astroid. From its equation, we would expect its sketch to look like a deformed circle. But it is probably not so obvious that it would have the pointy bits of the right hand figure. We will not explain here why they arise. The astroid is studied in some detail in Example 1.1.9. In particular, the above sketch is carefully developed there.

Example 1.0.4. Parametrization of $e^y=1+x^2$

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 $y=\ln\big(1+x^2\big)\text{,}$ we can use $x$ as the parameter simply by setting $t=x\text{.}$ This gives the parametrization

$$\vecs{r} (t) = \big(t\,,\,\ln(1+t^2)\big)\qquad -\infty\lt t\lt \infty \nonumber$$

Example 1.0.5. Parametrization of $x^2+y^2=a^2\text{,}$ again

It is also quite common that one can use either $x$ or $y$ to parametrize part of, but all of, a curve. A simple example is the circle $x^2+y^2=a^2\text{.}$ For each $-a\lt x\lt a\text{,}$ 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 $-a\lt y\lt a\text{,}$ 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

$$\begin{alignat*}{2} \vecs{r} (t) &= \big(t\,,\,\sqrt{a^2-t^2}\big)\qquad &-a\lt t\lt a \\ \vecs{r} (t) &= \big(t\,,\,-\sqrt{a^2-t^2}\big)\qquad &-a\lt t\lt a \end{alignat*}$$

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

$$\begin{alignat*}{2} \vecs{r} (t) &= \big(\sqrt{a^2-t^2}\,,\,t\big)\qquad &-a\lt t\lt a \\ \vecs{r} (t) &= \big(-\sqrt{a^2-t^2}\,,\,t\big)\qquad &-a\lt t\lt a \end{alignat*}$$

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

Example 1.0.6. Unparametrization of $\vecs{r} (t)=(\cos t, 7-t)$

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

$$\begin{align*} x&=\cos t \\ y&=7-t \end{align*}$$

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

$$x=\cos(7-y) \nonumber$$

Example 1.0.7

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

$$\begin{alignat*}{2} x^3&-e^{3y} &&=0\\ x^2&-e^{y} +z &&=0 \end{alignat*}$$

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

$$\begin{alignat*}{2} x^3& &&=e^{3y}\\ x^2&+z &&=e^{y} \end{alignat*}$$

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 $z\text{.}$

$$\begin{alignat*}{4} x^3& &&=e^{3y} &&\implies x=e^y\\ x^2&+z &&=e^{y} && &&\implies e^{2y}+z=e^y \implies z=e^y-e^{2y} \end{alignat*}$$

So

$$\vecs{r} (y) = \big(e^y\,,\,y\,,\,e^y-e^{2y}\big) \nonumber$$

is a parametrization for the given curve.

Example 1.0.8

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

$$\begin{alignat*}{1} x^2+\frac{y^2}{2}+\frac{z^2}{3}&=1\\ x^2+2y^2&=z \end{alignat*}$$

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

$$-\frac{3}{2}y^2+z+\frac{z^2}{3}=1 \nonumber$$

or, completing the square,

$$-\frac{3}{2}y^2 + \frac{1}{3}\Big(z+\frac{3}{2}\Big)^2 = \frac{7}{4} \nonumber$$

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 $z\text{.}$

$$y=\sqrt{\frac{2}{9}\Big(z+\frac{3}{2}\Big)^2-\frac{14}{12}} \nonumber$$

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

$$\begin{align*} x&=\sqrt{z-\frac{4}{9}\Big(z+\frac{3}{2}\Big)^2+\frac{14}{6}}\\ &=\sqrt{\frac{4}{3}-\frac{4}{9}z^2-\frac{1}{3}z} \end{align*}$$

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.