Class 12 - Parametrization and Arc length (9/29/2017, Friday)

September 25, 2017

Class 12 - Parametrization and Arc length (9/29/2017, Friday)

2nd Gateway

The second Gateway will start on Sept 26, Tuesday and you will earn extra 10% quiz points if you pass it by Sept 29, Friday. (Of course, 20% total if you passed 1st Gateway early.)

Webwork

8.4 due 10/2 Mon

Today's posting will roughly cover 4.8 and (part of) 8.2 of your book.

Parametrization: case study

Consider our favorite example

$y = x^{2}$ . We may ask consider the curve from

$(0, 0)$ to

$(1, 1)$ . Naturally, you think about a particle moving along the curve. We may consider the particle to be

$(x(t), y(t)) = (t, t^{2})$ because then it will satisfy

$y(t) = x(t)^{2}$ . Notice that

$y'(t)/x'(t) = 2t = f'(t),$

where

$f(t) = t^{2}.$

In general, when

$y$ is a function in

$x$ and

$x$ is a function in

$x$ so that

$y = y(x)$ ;
$x = x(t)$ .

Then we have

$y(t) = y(x(t))$ , so

$y'(t) = y'(x(t))x'(t)$ by Chain Rule. Hence, we get

$y'(t)/x'(t) = y'(x(t)).$

In the above example, we had

$x(t) = t$ and

$y(x) = x^{2}.$ The above can be remembered as:

$\frac{dy/dt}{dx/dt} = \frac{dy}{dx},$

but beware that this is just a trick to memorize. However, this gives a nice way to take derivatives.

Example. Consider the implicit function

$y(x)$ that is given by the unit circle

$x^{2} + y^{2} = 1.$ We can locally compute

$y'(x)$ as follows: first, consider the parametrization given by

$x(t) = \cos(t)$ and
$y(t) = \sin(t)$ .

Then

$y'(\cos(t)) = y'(x(t)) = x'(t)/y'(t) = -\sin(t)/\cos(t) = -y(t)/x(t),$ so you can see that

$y'(x) = -y/x$ when

$x \neq 0$ .

Arc length: case study

You know the length from

$(0,0)$ to

$(1,1)$ along the line

$y = x$ . Can you compute the length from

$(0,0)$ to

$(1,1)$ along the curve

$y = x^{2}$ ?

Write

$f(x) = x^{2}$ . To compute the length, let's break the interval

$[0, 1]$ into pieces. Each subinterval would have length

$\Delta x$ and you can estimate the length you want by summing up lengths of line segments

$\sqrt{(\Delta x)^{2} +(f(x + \Delta x) - f(x))^{2}}.$ Note that

$\sqrt{(\Delta x)^{2} +(f(x + \Delta x) - f(x))^{2}} = \sqrt{1 + \left(\dfrac{f(x + \Delta x) - f(x)}{\Delta x}\right)^{2}}\Delta x,$

so when you sum them up and take

$\Delta x \rightarrow 0$ , you should get

$\textbf{Arc length} = \int_{0}^{1}\sqrt{1 + f'(x)^{2}}dx = \int_{0}^{1}\sqrt{1 + 4x^{2}} dx.$

In general, if you have a smooth function

$y = f(x)$ , then

$\textbf{Arc length from } x = a \textbf{ to } x = b \text{ is } \int_{a}^{b}\sqrt{1 + f'(x)^{2}}dx.$

If you have a parametrization

$(x(t), y(t)),$ again, thinking

$y(t) = y(x(t))$ , you may apply the change of variable in the above formula. That is, you consider

$x = x(t)$ so that

$dx = x'(t)dt$ . Find

$t = c$ such that

$x(c) = a$ and

$t = d$ such that

$x(d) = b$ , you get

$\int_{c}^{d}\sqrt{1 + (y'(t)/x'(t))^{2}}x'(t)dt = \int_{c}^{d}\sqrt{x'(t)^{2} + y'(t)^{2}}dt.$

You can consider the above as the arc length of the particle

$(x(t), y(t))$ from time

$t = c$ to

$t = d$ .

Memorization trick. Given a particle

$(x(t), y(t))$ , its velocity is

$(x'(t), y'(t))$ and its speed is the size of the velocity, which is

$\sqrt{x'(t)^{2} + y'(t)^{2}}$ . You notice that:

Arc length is the integral of speed!

Exercise. Using arc length of the parametrization

$(2\cos(t), 2\sin(t))$ of the circle of radius

$2$ for

$0 \leq t \leq 2\pi$ , derive the usual fact that its circumference is equal to

$4\pi$ .

Search This Blog

Math 116 - Section 024, Fall 2017