# 1: Curves

- Page ID
- 91891

We are now going to study vector-valued functions of one real variable. That is, we are going to study functions that assign to each real number \(t\) (typically in some interval) a vector^{ 1} \(\vecs{r} (t)\text{.}\) For example

\[ \vecs{r} (t) = \big( x(t), y(t), z(t)\big) \nonumber \]

might be the position of a particle at time \(t\text{.}\) As \(t\) varies, \(\vecs{r} (t)\) sweeps out a curve.

While in some applications \(t\) will indeed be “time”, it does not have to be. It can be simply a parameter that is used to label the different points on the curve that \(\vecs{r} (t)\) sweeps out. We then say that \(\vecs{r} (t)\) provides a parameterization of the curve.

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.

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 \((x-h)^2+(y-k)^2=a^2\text{.}\)

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.

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

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

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.

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

\[ x=\cos(7-y) \nonumber \]

Curves often arise as the intersection of two surfaces. For example, the intersection of the ellipsoid \(x^2+\frac{y^2}{2}+\frac{z^2}{3}=1\) with the paraboloid \(z=x^2+2y^2\) is the blue curve in the figure below.

One way to parametrize such curves is to choose one of the three coordinates \(x\text{,}\) \(y\text{,}\) \(z\) as the parameter, and solve the two given equations for the remaining two coordinates, as functions of the parameter. Here are two examples.

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.

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.

- We are going to use boldface letters, like \(\vecs{r} \text{,}\) to designate vectors. When writing by hand, it is clearer to use arrows, like \(\vec r\text{,}\) instead.

- 1.1: Derivatives, Velocity, Etc.
- This being a Calculus text, one of our main operations is differentiation. We are now interested in parametrizations \(\vecs{r} (t)\text{.}\) It is very easy and natural to extend our definition of derivative to \(\vecs{r} (t)\) as follows.

- 1.2: Reparametrization
- There are invariably many ways to parametrize a given curve. Kind of trivially, one can always replace \(t\) by, for example, \(3u\text{.}\) But there are also more substantial ways to reparametrize curves.

- 1.3: Curvature
- So far, when we have wanted to approximate a complicated curve by a simple curve near some point, we drew the tangent line to the curve at the point. That's pretty crude. In particular tangent lines are straight — they don't curve. We will get a much better idea of what the complicated curve looks like if we approximate it, locally, by a very simple “curvy curve” rather than by a straight line.

- 1.4: Curves in Three Dimensions
- So far, we have developed formulae for the curvature, unit tangent vector, etc., at a point \(\vecs{r} (t)\) on a curve that lies in the \(xy\)-plane. We now extend our discussion to curves in \(\bbbr^3\text{.}\)

- 1.5: A Compendium of Curve Formula
- In the following \(\vecs{r} (t)=\big(x(t)\,,\,y(t)\,,\,z(t)\big)\) is a parametrization of some curve.

- 1.6: Integrating Along a Curve
- Suppose that we have a curve \(\mathcal{C}\) that is parametrized as \(\vecs{r} (t)\) with \(a\le t\le b\text{.}\) Suppose further that \(\mathcal{C}\) is actually a piece of wire and that the density (i.e. mass per unit length) of the wire at the point \(\vecs{r} \) is \(\rho(\vecs{r} )\text{.}\) How do we figure out the mass of \(\mathcal{C}\text{?}\) Of course we use the standard Calculus divide and conquer strategy.

- 1.7: Sliding on a Curve
- We are going to investigate the motion of a particle of mass \(m\) sliding on a frictionless, smooth curve that lies in a vertical plane. We will consider three scenarios:

- 1.8: Optional — Polar Coordinates
- So far we have always written vectors in two dimensions in terms of the basis vectors \(\hat{\pmb{\imath}}\) and \(\hat{\pmb{\jmath}}\text{.}\) This is not always convenient.

- 1.9: Optional — Central Forces
- One of the great triumphs of Newtonian mechanics was the explanation of Kepler's laws, which said

- 1.10: Optional — Planetary Motion
- We now return to the claim, made in §1.9 on central forces, that if \(\vecs{r} (t)\) obeys Newton's inverse square law

- 1.11: Optional — The Astroid
- Imagine a ball of radius \(a/4\) rolling around the inside of a circle of radius \(a\text{.}\) The curve traced by a point \(P\) painted on the inner circle (that's the blue curve in the figures below) is called an astroid. We shall find its equation.

- 1.12: Optional — Parametrizing Circles
- We now discuss a simple strategy for parametrizing circles in three dimensions, starting with the circle in the \(xy\)-plane that has radius \(\rho\) and is centred on the origin. This is easy to parametrize: