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2.3 Parametrized curves

Each of the examples we’ve seen so far have been curves, which – informally – means that they look a bit like a bit of string. However, “like a bit of string” isn’t a proper definition. We will make this precise in this section. Also, so far we treated each curve as a shape. However, sometimes there is more information included in the situation, namely when the curve is actually a record of a movement (the path traced by a bug, for example). The shape itself is then not enough, since two bugs can crawl along the same path in many different ways (speed and even direction can vary). This additional information is described very naturally by a parametrized curve. Let’s look at some examples before introducing the proper definition.

Example 2.13 a) The circle $C=\{(x,y)\in{\mathbb{R}}^{2}:x^{2}+y^{2}=1\}$ can be described as $\{(\cos t,\sin t):t\in[0,2\pi)\}$ .

b) Similarly, an ellipse $E=\{(x,y)\in{\mathbb{R}}^{2}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\}$ can be described as $\{(a\cos t,b\sin t):t\in[0,2\pi)\}$ .

c) The curve $\{(x,y)\in{\mathbb{R}}^{2}:y^{2}=x^{3}\}$ can be described as $\{(t^{2},t^{3}):t\in{\mathbb{R}}\}$ .

In each case we have expressed the point $(x,y)$ as a (vector-valued) function of a single coordinate $t$ .

Definition 2.14

A parametrized curve in ${\mathbb{R}}^{n}$ is a continuous map $\gamma:I\rightarrow{\mathbb{R}}^{n}$ , where $I$ is an interval in ${\mathbb{R}}$ . If $n=2$ , then $\gamma$ is a plane curve.

You don’t need to worry too much about what “continuous” means in this context: it merely says that the curve progresses smoothly as $t$ changes, rather than jumping rapidly from one position to another.

Example 2.15 For an example of a map which isn’t continuous, let $\gamma:{\mathbb{R}}\rightarrow{\mathbb{R}}^{2}$ ,

\gamma(t)=\left\{\begin{array}[]{cc}(t,-1)&\mbox{if $t<0$,}\\ (t,1)&\mbox{if $t\geq 0$.}\end{array}\right.

We can sketch this as follows:

Physical interpretation when $t$ is time

Let $(x(t),y(t))$ be the position of a particle in space at time $t$ . Assuming the functions $x(t)$ and $y(t)$ are (doubly) differentiable, we can determine the velocity, which is a vector: $(x^{\prime}(t),y^{\prime}(t))$ , and acceleration: $(x^{\prime\prime}(t),y^{\prime\prime}(t))$ .

For example, suppose the particle has position $(t^{2},t^{3})$ at time $t$ , where $t\geq 0$ : then the particle moves on the curve $y=x^{3/2}$ . In fact, we get all possible $x^{3/2}$ , so the particle traces out the whole curve $\{(x,x^{3/2}):x\geq 0\}$ . If we allow $t$ to be negative too, then we get $\{(x,y)\in{\mathbb{R}}^{2}:y^{2}=x^{3}\}$ . In this case, the velocity of the particle is $(2t,3t^{2})$ ; the acceleration is $(2,6t)$ .

Hence we can think of the curve $\{(x,y)\in{\mathbb{R}}^{2}:x\geq 0,y=x^{3/2}\}$ as the image of the parametrized curve $\gamma:[0,\infty)\rightarrow{\mathbb{R}}^{2}$ , $t\mapsto(t^{2},t^{3})$ . Similarly, $\{(x,y)\in{\mathbb{R}}^{2}:x^{3}=y^{2}\}$ can be described as the image of the curve $\gamma:{\mathbb{R}}\rightarrow{\mathbb{R}}^{2}$ , $t\mapsto(t^{2},t^{3})$ .

Example 2.16. Suppose a ball is thrown from a height $h$ above the ground (i.e. with position vector $(0,h)$ ), with a velocity of $(u,v)$ . We want to describe the position $(x(t),y(t))$ of the ball at time $t$ . We know that the acceleration $(x^{\prime\prime}(t),y^{\prime\prime}(t))=(0,-g)$ , where $g\approx 9.8$ m s ${}^{-2}$ . (We assume all distances are small compared to the size of the earth, so we can think of the surface of the earth as a plane, and we can assume gravity to be constant. We also ignore air resistance.)

Integrating, we obtain $(x^{\prime}(t),y^{\prime}(t))=$ for some constants $A, B$ . But $(x^{\prime}(0),y^{\prime}(0))=(u,v)$ so $A=u$ , $B=v$ . Thus $(x^{\prime}(t),y^{\prime}(t))=$ . By integrating again, we obtain $(x(t),y(t))=$ for some constants $C, D$ . Since $x(0)=0$ and $y(0)=h$ , we have:

(x(t),y(t))=\hskip 227.622047pt

Problem 1. Find the horizontal distance travelled at the time the ball hits the ground. We need to find the value of $x(t)$ when $y(t)=0$ . So first of all, we solve for $h+vt-\frac{1}{2}gt^{2}=0$ , or equivalently, for $\frac{1}{2}gt^{2}-vt-h=0$ . This is a quadratic equation, with roots

t=\hskip 227.622047pt

But we want the positive value of $t$ for which $y(t)=0$ , and $\sqrt{v^{2}+2gh}>|v|$ , so the time at which the ball hits the ground is

Now we have to calculate $x(t)$ at this time:

x(\frac{v+\sqrt{v^{2}+2gh}}{g})=\hskip 227.622047pt

Problem 2. Suppose $h=0$ . Assuming $u^{2}+v^{2}=1$ , which values of $u, v$ would you choose to throw the ball the furthest? How far can you throw it?

We have to choose $u, v$ with $u^{2}+v^{2}=1$ such that $(u\frac{v+\sqrt{v^{2}+2gh}}{g})$ is maximal; but here $h=0$ , so the distance is .

Now $u^{2}+v^{2}=1$ so the arithmetic mean $A$ of $u^{2},v^{2}$ is $(u^{2}+v^{2})/2=$ . But the geometric mean $(u^{2}v^{2})^{\frac{1}{2}}$ is less than the arithmetic mean since

So $(u^{2}v^{2})^{\frac{1}{2}}\leq\hskip 56.905512pt=\frac{1}{2}$ , with equality if and only if $u^{2}=v^{2}$ . Thus $|2uv/g|\leq 1/g$ , with equality if $u=\pm v$ . So we find that the optimal direction to throw the ball is with the direction vector $(1/\sqrt{2},1/\sqrt{2})$ ; the maximum possible distance is $1/g$ metres.

Example 2.17 A swimmer is in a river, at position $(0,0)$ at time $t=0$ . One bank of the river is at the line $x=1$ ; the stream flows with a constant current $(0,-1)$ . The swimmer swims in the direction $(\cos\theta,\sin\theta)$ for a fixed $\theta\in[0,\pi/2]$ . As he tires he slows down: his speed at time $t$ is $2e^{-t}$ m s ${}^{-1}$ . What is his position at time $t$ ?

In order to calculate the velocity of the swimmer, we have to take into account both his swimming, and the movement of the water. So his velocity at time $t$ is:

(x^{\prime}(t),y^{\prime}(t))=\hskip 227.622047pt

To determine his position at time $t$ , we integrate:

(x(t),y(t))=\hskip 227.622047pt

for some constants $A, B$ . But we have $x(0)=y(0)=0$ , so:

x(0)=\hskip 284.527559pt=0

y(0)=\hskip 284.527559pt=0

Thus $A=$ , $B=$ , that is

(x(t),y(t))=(2(1-e^{-t})\cos\theta,2(1-e^{-t})\sin\theta-t).

Problem. Show that if he is to reach the bank of the river, he must choose $\theta<\pi/3$ .

For him to reach the bank, there must be a positive time $t$ such that $x(t)=1$ . Thus he reaches the bank if and only if there is a positive solution to . If $t>0$ then $0<e^{-t}<1$ , so $1-e^{-t}\in(0,1)$ . So in order for there to be a positive solution to $2\cos\theta(1-e^{-t})=1$ , we must have $\hskip 85.358268pt\in(0,1)$ . Since $\theta\in[0,\pi/2)$ , we necessarily have $\cos\theta>0$ . So $\frac{1}{2\cos\theta}\in(0,1)$ if and only if $2\cos\theta>1$ . But this is true if and only if $\cos\theta>$ . Since $\cos\pi/3=\frac{1}{2}$ and $\cos\theta$ is strictly decreasing for $\theta\in[0,\pi/2]$ , we see that there is a solution to $x(t)=1$ if and only if $\theta\in[0,\pi/3)$ .