Chapter 9: Parametric and Polar Equations

Chapter 9: Parametric and Polar

Equations

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Chapter 9 Overview: Parametric and Polar Coordinates As we saw briefly last year, there are axis systems other than the Cartesian System for graphing (vector coordinates, polar coordinates, rectangular coordinates--for Complex Numbers--and others). The Calculus should apply to them as well. Previously, we saw the relationship between parametric motion and vectors. We will explore this further and investigate how derivatives relate the concepts of tangent line slopes, increasing/decreasing, extremes, and concavity to vectorfunction (parametric) graphs. Polar coordinates are, at their base, drastically different from Cartesian or Vector coordinates. We will review the basics we learned last year about the conversions between polar and Cartesian coordinates and look at graphing in polar mode, with an emphasis on the calculator. As with vector-functions, we will investigate how derivatives relate the concepts of tangent line slopes, increasing/decreasing, extremes, and concavity. Further, we will see how integrals relate to area in polar mode. Finally, we will revisit motion in parametric mode and see how motion could be described in polar mode. There are two contexts within which we will consider parametric and polar functions. I. Motion II. Graphing

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9.1: Parametrics and Motion

Vocabulary Parameter--Defn: An independent variable, usually t, that determines x and y

separately from one another. Parametric Graphs--Defn: graphs of equations where the coordinates of the points

(x, y) are both dependent variables determined by the independent variable t.

In Chapter 3, we considered motion in a parametric context. At the time, we only had the derivative as a tool for analysis, but now we have the integral also.

Recall from earlier:

Remember:

Position = x(t), y(t) Velocity = x'(t), y'(t) Acceleration = x"(t), y"(t)

Speed = v(t) = (x'(t))2 + (y'(t))2

Therefore:

Position = x'(t)dt, y'(t)dt Velocity = x"(t)dt, y"(t)dt

Ex 1 Find the velocity vector for the particle whose position is described by t2 - 2t,t2 +1 . At t = 3, find the speed of the particle.

First, notice that this problem has a slightly different notation than we are used to. You may remember that chevrons (the angled brackets) are one of the ways that we write vectors. Since this is a position vector, we really have the following:

x = t2 - 2t and y = t2 + 1 Taking the derivatives, we get

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dx dt

= 2t - 2

and

dy dt

= 2t

Since the derivatives of x and y are actually velocities, we can simply write the velocity vector using chevrons.

2t - 2, 2t

Of course, we could still write vectors using the

!

xi

+

"!

yj

form, but this can

become cumbersome when x and y are equations in terms of t.

( ) ( ) In general for this particle, speed = 2t - 2 2 + 2t 2 and

( ) ( ) ( ) ( ) at t = 3, speed =

23

-2 2+

2

3

2

=

52

When dealing with derivatives of parametric equations, it is important to realize we have gotten a little sloppy by talking about "THE derivative"--as if there were only one. It was not that important before because, in Cartesian mode, x was the independent variable and y was the dependent variable. We did not deal with derivatives in terms of different variables within the same problem. That comes back to haunt us now. There is still the issue of displacement vs. distance traveled. We recall:

Key Idea #1:

Displacement =

b a

v

(t

)

dt

=

b a

x

'(t

)

dt

b a

y

'

(t

)

dt

Distance

traveled

=

b a

v(t) dt

OBJECTIVES Find the position of an object in motion in two dimensions from its velocity. Find the arc length of a curve expressed in parametric mode.

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Key Idea #2: Position can be found two ways

a) Indefinite integration of the velocity equation, with an initial value, will yield the position equation. Position can be determined by substituting the time value.

b) Definite integration (by calculator) will yield displacement. Adding displacement to the initial value will yields the position

Ex 1 Given that an object in motion has v(t ) = t2 + 2, t 3 - 1 and the initial

position -1, 3 (that is, t = 0 ), find the position at t = 2 .

v(t) =

t2 + 2, t 3 - 1

x'(t) y'(t)

= =

t2 t3

+ -

2 1

Considering these separately,

( ) x'(t) = t2 + 2 x(t) =

t2 + 2

dt

=

1t3 3

+

2t

+

c1

x

(0)

=

-1

-1

=

1 3

(

0

)3

+

2

(0)

+

c1

c1

=

-1

and

( ) y'(t) = t3 -1 y(t) =

t3 -1

dt

=

1 4

t4

-

t

+

c2

y(0)

=

3

3

=

1 4

04

-

0

+

c2

c2

=

3

So,

x(t) =

1 3

t

3

+ 2t

-1

and

y(t) =

1 t4 4

-t

+

3

and

x(2), y(2) = 17 , 5

3

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