Lecture 5 Parametric Curves

Text References: Course notes pp. 10-22 & Rogawski 14.4, 11.1, 14.6

5.1 Recap

Last time, we discussed the equation of tangent planes and how to use them to approximate values of a function at a given point.

Exercise 5.1 You measure a cylinder that has a height of $3.34$ meters and a radius of $2.77$ meters. If the measurement error of the height is $2$cm and the measurement error of the radius is $3$cm, what is the measurement error on the volume of the cylinder?

Solution. We know that the volume of the cylinder is given by $V(r, h)=\pi r^2 h$. Using the differential form of the linear approximation, we have \[\begin{align*} dV &= V_r dr+ V_h dh \\ &= 2\pi r h dr + \pi r^2 dh\\ &= 2\pi(2.77)(3.34)(0.03)+\pi(2.77)^2(0.02)\\ &=2.23 m^3 \end{align*}\]

5.2 Learning Objectives

Calculate the speed, velocity, and acceleration of a particle whose movement is described by a vector function.
Use vector functions to describe curves and paths in parametric form.

5.3 Vector Functions

In single-variable calculus, we learned how to parametrize curves by describing a point $(x,y)$ using the variable $t$ and thinking of $x$ and $y$ as functions of $t$. This parametric description allows us to describe a broader set of relations between $x$ and $y$. Not only that, but we now get additional information about the direction of the curve and its speed. It’s common to distinguish between the curve and its path. the curve is what gets ‘traced’ on paper; the path describes how that ‘tracing’ is done. As you might imagine, there are infinitely many ways to parametrize a given curve!

The interactive applet illustrates this difference. Use the checkboxes to select one of the two parametric curves and advance the slider for $t\in [0, 2\pi]$ to view the path along which the curve is drawn. Notice how both curves are the same, but their paths are different.

Given that we’re thinking of both $x$ and $y$ as functions of $t$, we can think of the corresponding parametric equations as vector functions which take a scalar as input and output a vector, that is, $\vec{r}: \mathbb{R}\to\mathbb{R}^n$. The vector $\vec{r}(t)$ is interpreted as starting from the origin so that its tip traces out the path described by the parametric curves.

Given a vector function $\vec{r}(t)=\begin{bmatrix}x(t)\\y(t)\end{bmatrix}$, we define its derivative as $\vec{r}'(t)=\begin{bmatrix}x'(t)\\y'(t)\end{bmatrix}$. If we interpret $\vec{r}(t)$ as describing the motion of a point, $\vec{r}'(t)$ gives its velocity, $\Vert \vec{r}'(t)\Vert$ gives its speed, and $\vec{r}''(t)$ gives the acceleration.

Exercise 5.2 Consider a particle moving according to the vector function $\vec{r}(t)=\begin{bmatrix}\cos(2t)\sin(t)\\ \sin(3t)\end{bmatrix}$. Determine

the velocity;
the speed; and
the acceleration

of the particle.

Solution. We have $\vec{r}(t)=\begin{bmatrix}x(t)\\y(t)\end{bmatrix}$ where $x(t)=\cos(2t)t$ and $y(t)=\sin(3t)$. Taking the derivative of each component, we find the velocity: \[\vec{v}(t)=\vec{r}'(t)=\begin{bmatrix}\cos(2t)-2t\sin(2t)\\ 3\cos(3t) \end{bmatrix}\] The speed is therefore \[\Vert\vec{v}(t)\Vert= \sqrt{(\cos(2t)-2t\sin(2t))^2+(3\cos(3t))^2} = \sqrt{9 \cos^2(3 t) + (\cos(2 t) - 2 t \sin(2 t))^2}\]

Finally, taking derivatives once more, we find the acceleration: \[\vec{a}(t)=\vec{r}''(t)=\begin{bmatrix}-4 (\sin(2 t) + t \cos(2 t)) \\ -27\cos(3t)\end{bmatrix}\]

5.4 Parametrization of Curves

5.4.1 Circles

So far, we’ve seen how to parametrize circles; a standard for parametrizing a circle of radius $a$ is \[\vec{r}(t)=\begin{bmatrix}a\cos(\omega t)\\ a\sin(\omega t)\end{bmatrix} \quad \mbox{where $\omega$ is the angular frequency}\]

The interactive applet illustrates this standard parametrization of a circle. Start by tracing the curve by moving the slider for $t$ when $a=1$ and $\omega=1$. See how the curve and path change as the values of $a$ and $\omega$ change.

Let’s take a look at a few other types of curves.

5.4.2 Ellipses

Related to circles are ellipses; in fact, circles can be thought of as a sub-case of ellipses.

Exercise 5.3 Give parametric equations for the ellipse with equation $9x^2+y^2=4$?

Solution. Note that $-\frac{2}{3}\leq x\leq \frac{2}{3}$ and $-2\leq y \leq 2$. We can find these bounds by setting $y=0$ and $x=0$, respectively.

Working from the parametrization of a circle and taking the above constraints into account, we find \[\vec{r}(t)=\begin{bmatrix}\frac{2}{3}\cos(\omega t) \\ 2\sin(\omega t)\end{bmatrix}, \quad t\in [0, 2\pi/\omega]\]

What if we want to shift the centre of the ellipse to a point other than $(0,0)$?

Exercise 5.4 Give parametric equations for the ellipse with equation $9(x+1)^2+(y-3)^2=4$?

Solution. Note that this is the same ellipse as the previous exercise, but centred at $(-1, 3)$. We need to apply these shifts to the parametric equations:

\[\vec{r}(t)=\begin{bmatrix}\frac{2}{3}\cos(\omega t) -1 \\ 2\sin(\omega t)\end{bmatrix}+3, \quad t\in [0, 2\pi/\omega]\]

In the interactive applet, you can enter your own parametrizations and observe the behaviour of the curve. Try inputting the parametrizations from the previous two examples to verify your work.

5.4.3 Lines

A common curve to parametrize is a straight line passing that starts from the point $(x_0,y_0)$ and ends at the point $x_1, y_1$. We can parametrize as follows: \[\vec{r}(t)=\begin{bmatrix}x_0+(x_1-x_0)t \\ y_0 + (y_1-y_0)t\end{bmatrix}, t\in[0,1]\]

Exercise 5.5 Give parametric equations for the straight line segment from $(-1,3)$ to $(2, 9$.

Solution. We have $(x_0, y_0)=(-1,3)$ and $(x_1, y_1)=(2,9)$. Therefore a parametrization is given by \[\vec{r}(t)=\begin{bmatrix}-1+3t \\ 3+ 6t\end{bmatrix}, \quad t\in[0,1]\]

5.4.4 Curves of the form $y=f(x)$

Finally, we can also parametrize curves of the form $y=f(x)$; in fact, the parametrization has already been done for us in this case with $x$ as the parameter. Using $t$ instead, we get \[\vec{r}(t)=\begin{bmatrix}t \\f(t)\end{bmatrix}\]

MATH 119 Course Notes