Chapter 9: Parametric and Polar Equations

[Pages:52]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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It should be mentioned that the fact that c1 and c2 equaled the initial values is due to the equations being polynomials and the initial t = 0. The constants would not equal the initial values if this had not been true.

Ex 2 Given that an object in motion has v(t ) = sin3 t, cos2 t and the initial

position -1, 3 , find the position at t = 2 .

It this case, we cannot take the anti-derivatives of the velocity, because these functions cannot be integrated by any techniques we have. We can find the definite integrals by out calculator, though, and these would be displacements from the initial point.

x(t) =

-1 +

2 0

sin

3

t

dt

=

.059

y(t)

=

3+

2 0

cos2

t

dt

=

3.811

Ex 3 BC 2006 #3

Again, velocity is a vector, so displacement is a vector also, as we saw in the last

( ) ( ) two examples. But v(t) is the speed and v(t) = x'(t) 2 + y'(t) 2 . The distance

traveled--which, in most cases, is the Arc Length of the curve--is:

Arc Length/Distance Formulae:

Function:

b

L = a

1

+

dy dx

2

dt

( ) ( ) Parametric: L = x'(t) 2 + y'(t) 2 dt

Note that the distance traveled is not always equal to the Arc Length. The particle can trace over the part or all of the path traveled.

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Ex 4 Find the distance traveled by the object in Ex 2.

( ) ( ) Distance = 02 sin3 t 2 + cos2 t 2 dt

= 1.651

Ex 5 Find the arc length of x = cost and y = sint for t 0,4 and compare it to the distance traveled.

This problem highlights the difference between Arc Length and Distance.

( ) D = 04 (-sint )2 + cos(t ) 2 dt

=

4 0

1

dt

= t 04

= 4

But we saw that that this curve is the unit circle, with the motion traced over the circle twice. The distance traveled is 4 , but the arc length is actually half this distance, or 2 . This should make sense, since the circumference of the Unit Circle is 2 .

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9.1: Homework

Find (a) the velocity vector, (b) the speed, and (c) acceleration vector for each of the following parametric equations.

1. x(t) = t , y(t) = cos t

2. x(t) = 5sint , y(t) = t2

3. x(t) = t2 - 5t , y(t) = -4t2 +1

4. x = sec , y = tan

Find the position at t = 5 of an object moving according to the given equation with the given initial position.

5. x'(t) = t - t2 , y'(t) = 5 t23, x(1) = -2, y(1) = 3

3

6. x'(t ) = 3 - t2 , y'(t ) = t 12 - t-12 , x(0) = -2, y(0) = 3 7. x'(t) = e-t , y'(t) = 3e3t , x(-2) = 0, y(-2) = -1 8. x'(t) = t cost2 , y'(t) = 3sint , x(1) = -2, y(1) = 3 9. x'(t) = t lnt , y'(t) = tan3 t , x(0) = -3, y(0) = 1 10. x'(t) = coset , y'(t) = cot2 (t +1) , x(3) = 0, y(3) = 0

Find the distance traveled by each object below. Determine if the distance is equal to the arc length.

11.

x(t) = t

- t2 ,

y(t) =

4

t

3 2

,

3

t

1,

2

12. x(t) = t sint , y(t) = t cost , t -3, 3

13. x(t ) = 3t - t 3 , y(t ) = 3t2 , t 0, 14. x(t ) = et sint , y(t ) = et cost , t 0,

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