## Monday, April 30, 2007

### Precalculus 12.3 The Tangent Line Problem

12.3 The Tangent Line Problem

The Slope of the graph of a function can be used to analyze rates of change at particular points on the graph.

The Slope of a line indicates the rate at which a line rises or falls.
- For a straight line, this rate (or slope) is the same at every point on the line.
- For other graphs than lines, the rate at which the graph rises or falls changes from point to point.

Recall: slope formula = (the change in y's)/(the change in x's) = Δy/Δx

Review: With circles, you were taught about tangent lines. A tangent line is a line that intersects a circle in exactly one point, called the point of tangency. Using this process, to determine the rate at which a graph rises or falls at a single point, you can find the slope of the tangent line at that point.

The tangent line to a graph of a function "f" at a point P(x1, y1) is the line that best approximates the slope of the graph at that point.

Example1: f(x) = x2 - 2x + 1

find the tangent line at the point (1, f(x))

f(1) = 12 -2(1) + 1 = 0
so the tangent line at (1,0) is a straight line and the equation of the tangent line at (1, f(x)) is y=0.

A more precise method then "eyeballing" the tangent line
is making use of the secant line through the point of tangency
and a second point on the line where (x, f(x)) is the first point
and (x + h, f(x + h)) is a second point on the graph of "f", the
slope of the secant line through these two points is:

Msecant = (f(x + h) - f(x))/h

Using the limit process, you can find the
exact slope of the tangent line at (x, f(x)).

Definition of the Slope of a Graph:

The slope m of the graph of "f" at the point (x, f(x)) is equal to the slope
of its tangent line at (x, f(x)) and is given by:

M =

lim msec =
h®0

lim (f(x+h) - f(x))/h
h®0

provided the limit exists!

Example 2: given f(x) = 2x + 5, find the slope of the tangent line at (-1, -3)

Msec = (f(x+h) - f(x))/h

= (f(-1 + h) - f(-1))/h
= (2(-1 + h) + 5 - (2(-1) + 5))/h
= (-2 + 2h + 5 +2 - 5)/h
= (2h)/h
= 2
The graph has a slope of 2 at the point (-1, -3)
To find the equation of the tangent line:
y = mx + b
(-3) = 2(-1) + b
-1 = b

so the equation of the tangent line at point (-1, -3) is
y = 2x -1

Example 3: given f(x) = 10x - 2x2 at point (3, 12)

Msec = [10(x+h) - 2(x + h)2 - (10x - 2x2)]/h

Msec = [10x + 10h -2(x2 +2xh + h2) -10x + 2x2]/h

Msec = [10h - 2x2 - 4xh - 2h2 + 2x2]/h

Msec = [10h - 4xh - 2h2]/h

Msec = 10 - 4x - 2h

Slope of the tangent line = M =
lim Msec =
h®0

lim 10 - 4x - 2h = 10 - 4x so at point (3, 12), M = 10 - 4(3) = -2
h®0

Therefore the slope of the tangent line at the point (3, 12) is -2.

To find the equation of the tangent line, use the point and the slope:

y = mx + b
12 = (-2)(3) + b
12 = -6 + b
18 = b

y = 2x + 18 is the equation of the tangent line at the point (3, 12)

Finding a Formula for the slope of a Graph:
Example4: g(x) = x3 at points (1, 1) and (-2, -8)

Msec = [(x + h)3 - (x3)]/h

Msec = [x3 + 3x2h + 3xh2 + h3 - x3]/h

Msec = [3x2h + 3xh2 + h3]/h

Msec = 3x2 + 3xh + h2 , where h¹0

Next take the limit of Msec as h approaches 0.

The slope of the tangent line M =

lim 3x2 + 3xh + h2 = 3x2
h®0

Now use this M = 3x2 for the slope at (1,1)

M = 3(1)2 = 3

now y = mx + b
1 = 3(1) + b
b = -2

so the equation of the tangent line at point (1,1) is y = 3x - 2

Again now at point (-2, -8)

Now use this M = 3x2 for the slope at (-2,-8)

M = 3(-2)2 = 3(4) = 12

now y = mx + b
-8 = 12(-2) + b
-8 = -24 + b
b = 16

so the equation of the tangent line at point (-2,-8) is y = 12x +16

The formula that you derived from the function f(x) = x3 and used the limit process to get M = 3x2 represents the slope of the graph of "f" at the point (x, f(x)).

The Derived function is called the derivative of f at x. It is donoted by f ' (x), which is reas as "f prime of x".

Definition of the Derivative:

The Derivative of "f" at "x" is

f ' (x) = limh®0 (f(x+h) - f(x))/h

provided the limit exists!

Example 5: f(x) = x2 - 3x + 4

f ' (x) = limh®0 [(x + h)2 - 3(x + h) + 4 - (x2 - 3x + 4)]/h

f ' (x) = limh®0 [x2 + 2xh + h2 - 3x -3h + 4 - x2 + 3x -4]/h

f ' (x) = limh®0 [2xh + h2 - 3h]/h

f ' (x) = limh®0 2x + h - 3 = 2x - 3

f '(x) = 2x - 3

So therefore the derivative of f(x) = x2 - 3x + 4 is f ' (x) = 2x - 3

NOTE that in addition to f ' (x) , other notations can be used to denoted the derivatives of y = f(x). The most common are:

(dy)/(dx)

y'

(d/(dx))[f(x)]

and Dx[y]

Using the derivative of f(x) = x3 - x, find the slope of the tangent line at point (2,6).

f'(x) = limh®0 [(x + h)3 - (x + h ) - (x3 - x)]'h

f'(x) = limh®0 [x3 + 3x2h + 3xh2 + h3 - x - h - x3 - x]/h

f'(x) = limh®0 [3x2h + 3xh2 + h3 - h]/h

f'(x) = limh®0 (3x2 + 3xh + h2 - 1)

f ' (x) = 3x2 - 1

So at the point (2, 6), the slope is
f ' (2) = 3(2)2 - 1 = 3(4) - 1 = 12 - 1 = 11

So the equation of the tangent line at point (2,6) is
6 = 11(2) + b
6 = 22 + b
b = -16

so the equation of the tangent line at point (2,6) is y = 11x - 16

Example 6: Find the derivative of f. Use the derivative to determine any points on the graph of f where the tangent line is horizontal.

f(x) = x3 - 3x

f ' (x) = limh®0 [(x + h)3 - 3(x + h) - (x3 - 3x)]/h

f ' (x) = limh®0 [x3 + 3x2h + 3xh2 + h3 - 3x -3h - x3 + 3x]/h

f ' (x) = limh®0 [3x2h+ 3xh2 + h3 -3h]/h

f ' (x) = limh®0 [3x2 + 3xh+ h2 -3]

f ' (x) = 3x2 - 3

Recall a horizontal line has a slope of zero so

f ' (x) = 0 so
0 = 3x2 - 3
3 = 3x2
1 = x2
x = ±1

putting these values in for x:

So "f" has horizontal tangents at (1, 0) adn (-1, 0)

If you need extra help, watch the Chapter 2 lesson 1 tutorial at
http://www.calculus-help.com/funstuff/phobe.html

## Thursday, April 26, 2007

### 12.2b Precalculus - One-Sided Limits

12.2b One-Sided Limits

One-Sided Limits - when a limit approaches one value from the left side of "c" and a different value from the right side of "c"

lim f(x) = L
x®c-

and

lim f(x) = K
x®c+

where if L = K, then the limit exists but if L ¹K, then the limit does not exist for the value of x.

Example 1:

lim f(x)
x®1

Given f(x) = 4 - x2 when x£1 or f(x) = 3 - x when x is greater than 1

therefore:

lim f(x) = 4 - (1)2 = 4 - 1 = 3
x®1-

and

lim f(x) = 3 - 1 = 2
x®1+

so since 3 ¹ 2, then
lim f(x) does not exist since the function approaches different values
x®1
from the left and from the right.

I. Existence of a Limit:

If "f" is a function and "c" and "L" are real numbers, then

lim f(x) = L
x®c

if and only if BOTH the left and the right limits exist and are equal to L.

Example 2:

f(x) = 3 - x when x is less than 1 or 3x - x2 when x is greater than 1

lim f(x) = 2
x ®1-

lim f(x) = 2
x ®1+

Because the one-sided limits both exist and are equal to the same number 2, it follows that

lim f(x) = 2 so the limit does exist at x = 1.
x®1

II. A limit from Calculus:

Evaluating a Limit from Calculus - the limit of a difference quotient:

lim (f(x+h) - f(x))/h where h¹0
h®0

Direct substitution into the difference quotient ALWAYS produces the indeterminate
form 0/0.

Example 3: find the difference quotient of f(x) = 5 - 6x

lim [5 - 6(x + h) - (5 - 6x)]/h
h®0

lim [5 - 6x - 6h - 5 + 6x]/h
h®0

lim -6h/h = -6
h®0

therefore:
lim [5 - 6(x + h) - (5 - 6x)]/h = -6
h®0

Example 4: find the difference quotient of f(x)= 4 - 2x - x2

lim ([4 - 2(x + h) - (x+h)2] - [4 - 2x - x2])/h =
h®0

lim ([4 - 2x -2h -(x2 + 2xh + h2] - 4 + 2x + x2)/h =
h®0

lim (4 - 2x -2h -x2 -2xh - h2 - 4 + 2x + x2)/h =
h®0

lim (- 2h - 2xh - h2)/h =
h®0

lim (- 2 -2x -h) = -2 -2x
h®0

## Wednesday, April 25, 2007

### Precalculus - 12.2a Techniques for Evaluating Limits

12.2 Techniques for Evaluating Limits

I. Limits of Polynomial and Rational Functions

1. If p is a polynomial function and "c" is a real number, then

lim p(x) = p(c)
x®c

2. If "r" is a rational function given by r(x) = p(x)/q(x) and "c" is a real number such that q(x) ¹0.

Example 1: Evaluating Limits by Direct Substitution

lim (.5x3 - 5x) = (.5)(.2)2 - 5(2) = 6
x®-2

Example 2:
lim (x2 + 1)/(x) = (32 + 1)/ 3 = (9 + 1)/ 3 = 10/3
x®3

Example 3:
lim (x2 - 1)1/3 = (32 - 1)1/3 = (9 - 1)1/3 = 81/3 = 2
x®3

II. Factoring: Dividing out Technique

Example 4:
lim (x2 - 3x)/x = (x)(x-3)/(x)
x®0

lim x - 3 = -3
x®0

Example 5:
lim (x2 - 1)/(x + 1) =
x®-1

lim(x + 1)(x - 1)/(x + 1) =
x®-1

lim (x - 1) = -2
x®-1

This technique should be applied only when direct substitution produces zero in both the numerator and the denominator. The resulting fraction 0/0, has no meaning as a real number.

It is called an indeterminate form because you cannot, from the form alone, determine the limit.

III. Rationalizing Technique

First Rationalize the numerator of the function.

Example 6:
lim (3 - Ö (x))/(x - 9)
x®9

multiply the numerator and the denominator by the conjugate of the numerator:

lim[ (3 - Ö (x))(3 + Ö(x))] /[(x - 9) (3 + Ö(x))] =
x®9

lim (9 - x) / [(x - 9 )(3 + Ö(x))] =
x®9

As you can see in the numerator, you have (9 - x) and in the denominator you have (x - 9) so if we multiply the numerator by (-1) you would have:

lim (-1)(- 9 + x) / [(x - 9 )(3 + Ö(x))] =
x®9

so now x - 9 divides out of the numerator and denominator

lim -1 / (3 + Ö(9)) = -1/ (3 + 3 ) = -1/6
x®9

IV. Using Technology:

lim (1 + 2x)1/x
x®0

Since you cannot use substitution, use a table:
(x, f(x)) = {(-.1, 9.313), (-.01, 7.540), (-.001, 7.404), (-.0001, 7.391), (-.00001, 7.3892), (-.000001, 7.38907), (.000001, 7.38904), (.00001, 7.3898), etc...}

so you can see that it is estimated about

lim (1 + 2x)1/x »7.3890561 » e2
x®0

Algebraically:
Choose 2 points that are equidistant from 0 and close to 0 and find their average:

(7.38907 + 7.38904)/2 »7.389055 » e2

Example 7:

lim (1/(2+x) - 1/2)/x =
x®0

again using the calculator when x = -.0000000001, f(x) = -1/4
and when x = .0000000001, f(x) = -1/4

so

lim (1/(2+x) - 1/2)/x = -1/4
x®0

Algebraically:

lim (1/(2+x) - 1/2)/x =
x®0

simplifying the fraction:

[(2 - (2 + x))/(2 + x)(2)]/x =

(- x)/ [(2 + x)(2)(x)] =

-1/(4 + 2x) so:

lim (1/(2+x) - 1/2)/x = -1/(4 + 2(0)) = -1/4 same answer!!
x®0

## Thursday, April 19, 2007

### 12.1 Introduction to Limits

12.1 Introduction to Limits

Watch the tutorials on lessons 1 - 3 on the following website.

http://www.calculus-help.com/funstuff/phobe.html

Lesson 1. What is a limit
Lesson 2. When does a limit exist.
Lesson 3. How do you evaluate a limit?

______________________________________________________

The Limit Concept - this notion of a limit is a fundamental concept of calculus.

Example 1: You are given 32 feet of fence and are asked to form a rectangular pen whose area is as large as possible. What is the largest area the pen would cover and what dimensions should the pen be?

If we let w be the width of the pen and l be the length of the pen.
Therefore we know that 2w + 2l = 32 for the perimeter.

Therefore if we solve for the length, l = 16 - w

then the area would be

A = lw
A = (16 - w)(w)
A = 16w - w2

Using this equation, we can experiment with different values of w to see how to obtain the maximum area.

(w, A) = {(1, 15), (2, 28), (3, 39), (4, 48), (5, 55), (6, 60), (7, 63), (8, 64), (9, 63), (10, 60), (11, 55), (12, 48), (13, 39) ...}

so as you can see it appears that (8, 64) is a maximum point so therefore when w = 8, the maximum area would be 64. The dimensions of the pen would be an 8 by 8 pen.

In limit terminology, you can say that "the limit of A as w approaches 8 is 64"
This is written as

lim A = 64
w®8

Definition of Limit:

If f(x) becomes arbitrarily close to a unique number L as x approaches c from either side, the limit of f(x) as x approaches c is L. This is written as

lim f(x) = L.
x®c

Example 2: Estimating a Limit Numerically

lim (4 - 3x)
x®3

let's plug in some values and see what happens when x approaches 3.
(x,f(x)) = {(2.9, -4.7), (2.99, -4.97), (2.999, -4.997), (3, ?), (3.001, -5.003), (3.01, -5.03), (3.1, -5.3)}

So we can conclude that as x approaches 3, f(x) approaches -5.

So lim (4 - 3x) = -5
x®3

this graph is continuous. For graphs that are not continuous, finding a limit can be more difficult.

Example 3: Estimating a Limit Numerically:

lim (x+1)/(x2-x-2)
x®-1

Again, let's look at some values for x around -1.

(x, f(x)) = {(-1.1, -.3226), (-1.01, -.3322), (-1.001, -.3332), (-1, error),
(-.999, -.3334), (-.99, -.3344), (-0.9, -.3448)}

So as you can see in the table when x = -1, f(x) = error because if you place -1 in for x in the equation, f(x) = 0/0 which is impossible to have because you cannot divide by 0.

But if you place values close to -1, f(x) ® -1/3 so

lim (x+1)/(x2-x-2) = -1/3
x®-1

Now try:

lim (x + 1)/(x2-x-2)
x®2

again f(2) = 3/0 which you cannot do so:

when we look at the value of f(x) at values of x around 2, you have two different answers.

1. as x approaches 2 from the left side:

lim (x + 1)/(x2-x - 2)
x®2-

you can see the limit approaches -¥

2. as x approaches 2 from the right side:

lim (x + 1)/(x2-x - 2)
x®2+

you can see the limit approaches +¥

so since f(x) is not bound as x approaches 2, you can conclude that the limit does not exist at x = 2.

Example 4:

lim (3x2-12)/(x-2)
x®2

Again plugging x = 2, f(2) = 0/0 which is impossible to have so

(x, f(x)) = {(0, 6), (1, 9), (2, error), (3, 15), (4, 18), (5, 21), ...} so you can see that it is suppose to be 12. we can do this if we use factoring:

(3x2 - 12)/(x - 2) =

(3(x2 - 4))/(x - 2) =

( 3 (x + 2)(x - 2))/ (x - 2) =

3 (x + 2) because the (x - 2) divide out so we can simplify the limit to look like:

lim 3(x + 2)
x®2

= 3 ( 2 + 2) = 12

Example 5:

lim (cos (x) - 1)/ x
x®0

again when x =0, f(0) = error but when we look at the graph, as x approaches 0 from the left side and the right side, f(x) = 0 so we can conclude

lim (cos (x) - 1)/ x = 0
x®0

Example 6:

lim (cos (1/x))
x®0

Try using different windows:
let x be between -2p and +2p and y between -1 and +2

Now zoom in alittle: let x be between -1 and +1 by .1

Now zoom in more: let x be between -.1 and +.1 by .01

Continue this, what do you notice?

The smaller the window, so the closer to x = 0 you get, the graph does not approach any particular number (it goes between +1 and -1).

Therefore this limit does not exist because no matter how close you are to zero, it is possible to choose values of x1 and x2 such that cos(1/x1) and cos(1/x2) that do not equal.

Example 7: Finding Limits by Direct Substitution:

lim (x + 4)/(x - 3)
x®1

Do the numerator separately from the denominator:

lim (x + 4) = 5
x®1

lim (x - 3) = -2 so
x®1

lim (x + 4)/(x - 3) = -5/2
x®1

____________________________________________________________

I. Conditions Under Which Limits Do not exist:

The Limit of f(x) as x ® c does not exist if any of the following conditions is true.

1. f(x) approaches a different number from the right side of "c" than from the left side of "c".

2. f(x) increases or decreases without bound as x approaches c.

3. f(x) oscillates between two fixed values as x approaches c.

II. Properties of Limits:

Let "b" and "c" be real numbers and let "n" be a positive integer.

1. lim b = b
x®c

2. lim x = c
x®c

3. lim xn = cn
x®c

4. lim n Ö(x) = nÖ(c) for n even and c greater than zero.
x®c

III. Operations with Limits:

Given: lim f(x) = L
x®c
and

lim g(x) = K
x®c

1. Scalar multiple:
lim [ bf(x)] = bL
x®c

2a. Sum of limits:
lim [ f(x) + g(x)] = L + K
x®c

2b. Difference of limits:
lim [ f(x) - g(x)] = L - K
x®c

3. Product:
lim [ f(x) ´ g(x)] = L ´ K
x®c

4. Quotient:
lim [ f(x) ¸ g(x)] = L ¸ K
x®c

5. Power:
lim [ f(x)]n = Ln
x®c

Example 8:

lim f(x) = 3x
x®c

and

lim g(x) = -2x + 1
x®c

Find:
lim [ f(x) + g(x)] =3(3) + (-2)(3) + 1 = 9 - 6 + 1 = 4
x®3

Find:
lim [ f(x) g(x)] = [(3)(4)][-2(4) + 1] = (12)(-8 + 1) = (12)(-7) = -84
x®4

Example 9:
lim f(x) = 3
x®c

and

lim g(x) = -2
x®c

Find:
lim [ f(x) + g(x)]2 = (3 + -2)2 = 1
x®c

Find:
lim [ 6f(x)g(x)] = 6(3)(-2) = -36
x®c

## Wednesday, April 4, 2007

### Precalculus 10.7 Graphs of Polar Equations

10.7 Graphs of Polar Equations

A. Graphing a Polar Equation by Point Plotting by:

1. Convert to Rectangular Form. Multiply both sides of the Polar Equation by "r" and convert the result to rectangular form.

Example 1:
r = 3 cos q
r2 = r 3 cos q
x2 + y2 = 3x
x2 -3x + y2 = 0

Using complete the square:

x2 - 3x + (-3/2)2 + y2 = 0

(x - 3/2)2 + y2 = 9/4

This is the graph of a circle with center point (3/2, 0) and r = 3/2

2. Use a Polar Coordinate Mode

Put your calculator in polar mode and place the equation in the r =

r = 3 cos q
Use 0 £ q £ 2p , -6 £ x £ 6, and -4 £ y £ 4

This produces the same graph.

3. Use the Parametric Mode:

The graph of the Polar Equation r = f (q ) can be written in parametric form, using t as a parameter, as follows:

x = f (t) cos t and y = f (t) sin t

r = 3 cos q

x = 3 cos t cos t

y = 3 cos t sin t

Again, this will produce the same graph.

B. Tests for Symmetry:

The graph of a polar equation is symmetric with respect to the following if the given substitution yields an equivalent equation.

1. The line q = p/2
Replace (r, q ) by (r, p-q ) or (-r, -q )

2. The polar axis:
Replace (r, q) by (r, -q) or (-r, p-q)

3. The Pole:
Replace (r, q) by (r, p+ q) or (-r, q)

Example 2: Use symmetry to sketch the graph r = 16 cos 3q
1. The line q = p /2 replace (r, q) by (r, p - q) or (-r, -q)

-r = 16 cos (3(-q))
-r = 16 cos (-3q )
-r = 16 cos 3q
Þthis is not the same equation so it is not equivalent

2. The polar axis replace (r, q) by (r, -q)
r = 16 cos 3(-q)
r = 16 cos 3q
Þthis is the same equation so it is equivalent

3. The pole replace (r, q) by (r, p + q) or (-r, q)
-r = 16 cos 3q
Þthis is not equivalent

Therefore r = 16 cos 3q is symmetric with respect to polar axis.

Plotting the points in the table and using the polar axis symmetry, you obtain the graph which is called limacon. (the c has a ' attached to the bottom of it)

The points are (r, q) = (0, 16), (p/6, 0), (p/3, -16), (p/2, 0), (2p/3, 16), (5p/6, 0), (p, -16)

Example 3: r = 4 - sin q

1. line q = p/2
r = 4 - sin (p- q )
r = 4 - (sin p cos q - cosp sin q)
r = 4 - ((0) cos q - (-1) sinq)
r = 4 - sin q
Þsame equation

2. Polar axis
r = 4 - sin (-q)
r = 4 + sin q
Þnot same equation

or -r = 4 - sin ( p- q)
-r = 4 - (sinp cos q - cos p sin q)
-4 = 4 - sin q
Þnot equivalent

3. Pole
-r = 4 - sin q not equivalent
or
r = 4 - sin ( p+ q)
r = 4 - (sinp cos q + cosp sin q)
r = 4 - (-sin q)
r = 4 + sin q
Þnot equivalent

Therefore r = 4 - sin q is symmetric with respect to q = p/2

Example 4: r = 2 csc q cos q
r = 2(1/sinq)(cos q) = 2 cot q

1. q = p /2
-4 = 2 cot (-q)
-r = -2 cot q
r = 2 cot q
Þtherefore it is the same equation or equivalent

2. Polar axis:
-r = 2 cot ( p- q)
-r = 2 cot (-q)
-r = -2 cot q
r = 2 cot q
ÞEquivalent

3. Pole:
r = 2 cot ( p+ q)
r = 2 cot q
Þ Equivalent

ÞTherefore r = 2 csc q cos q is symmetric with respect to q = /2, polar axis and pole.

B. Zeros and Maximum r - values

½ r ½ is the maximum value for r and knowing the q - values for which r = 0 are two helpful ways to sketch the graph.

Example 4: Find the maximum value of r = 6 + 12 cos q

½r ½ = ½6 + 12 cosq ½

½ r ½ = ½ 6 + 12 cos q ½ £ ½6 ½ + ½ 12 cos q½

½ r ½ = 6 + 12½ cos q½£ 18 (from the thought that 6 + 12 = 18)

so if cos q = 1 then q = 0

So when q = 0, ½ r ½ = 18

Zeros:
Let r = 0

0 = 6 + 12 cos q
-6 = 12 cos q
-1/2 = cos q

Therefore q = 2p/3, 4p/3

Now for the table:
(r , q) goes with (x, y)
(18, 0°) and (18, 0)
(6, 90°) and (0, 6)
(0 , 120°) and (0 , 0)
(-6, 180 °) and (6, 0)
(0, 240°) and (0, 0)
(6, 270°) and (0, -6)
(18, 360°) and (18, 0)

Put your calculator in polar mode and you can get the polar coordinate. Graph the equation and go to 2nd trace, value and place the angle measure in, and the x - values and y - values will be displayed.

Example 5: r = 5 sin (2q)
½ r ½ = ½ 5 sin 2q ½

½ r ½ = 5

If we let sin b = 1 so b = 90° and b = 2q so therefore q = 45° or p/4

so when q = p/4, 3p/4 , 5p /4, 7p/4

So again to get the table of values:

(r, q) goes with (x, y)
(0, 0°) and (0, 0)
(5, 45°) and (3.5, 3.5)
(0, 90°) and (0, 0)
(-5, 135°) and (3.5, 3.5)
(0, 180°) and (0, 0)
(5, 225°) and (-3.5, -3.5)
(0, 270°) and (0, 0)
(-5, 315 °) and (-3.5, 3.5)
(0, 360°) and (0, 0)
This is a rose curve.

check out this video at
http://sam.ntpi.spcollege.edu/spjc/view/eventListing.jhtml?eventid=5323&c=13169

## Tuesday, April 3, 2007

### Precalculus 10.6 Polar Coordinates

10.6 Polar Coordinates:

So far, you have been representing graphs of equations as collections of points (x, y) on the rectangular coordinate system, where x and y represent the directed distances from teh coordinate axes to the point (x, y). In this section, you will study a second system - the Polar Coordinate System.

I. To form the polar coordinate system in the plane, fix a point O, called the pole (or origin) and construct from O an initial ray called the polar axis. Then each point P in the plane can be assigned polar coodinates (r,q ) as follows:

1. r = directed distance from O to P.
2. q = directed angle, counterclockwise from polar axis to segment OP.

To find other points in the same place, do the following:

(r, q ) = (r, q ± 2np )

or

(r, q) = (-r, q ±(2n + 1)p )

Example: Given the polar coordinate (3, p/4)
= (3, p/4 + 2p) = (3, 9p/4)

OR

(-3, p/4 + (2 (0) +1)p) = (-3, 5p/4)

(-3, p/4 - p) = (-3, -3p/4)

II. Coordinate Conversion:
Because (x, y) lies on a circle of radius r, it follows that r2 = x2 + y2. Moreover, for r> 0, the definitions of the trigonometric functions imply:

tan q = y / x

cos q = x/ r

sin q = y/r

The Polar Coordinates (r, q) are related to the rectangular coordinates (x, y) as follows:

x = r cos q

y = r sin q

tan q = y/x

r2 = x2 + y2

III. Polar to Rectangular Conversion

Example 2: Convert each point to rectangular coordinates
a). (-2, 7p/6) means r = -2 and q = 7p/6

x = (-2)(cos 7p/6) = (-2)(- Ö(3)/2) = Ö3

y = (-2)(sin 7p/6) = (-2)(-1/2) = 1

Therefore the rectangular coordinate is (Ö3 , 1)

b). (8.25, 3.5) means r = 8.25 and q = 3.5 radians

x = (8.25)(cos 3.5) = -7.72576767
y = (8.25)(sin 3.5) = -2.89391628

Therefore the rectangular coordinate is (-7.72576767, -2.89391628)

IV. Rectangle to Polar Conversion

Example 3: Convert each point to polar coordinates

a) Let 0£ q £ 2p and using the rectangular coordinate (0, -5)

tan q = -5/0 = undefined, therefore q = 270° based on where the point is.

r = Ö (x2 + y2)
= Ö(02 + (-5)2)
= 5

Therefore a polar coordinate is (5, 270°) or (5, 3p/2)

and the negative would be (-5, -90°) or (-5, p/2)

b) (3, -1)

Tan q = -1/3

q = -18.43494882° or -.3217505544 radians

so q = 341.5650512° or 5.961434753 radians

r = Ö(32 + (-1)2) = Ö(9 + 1) = Ö10

Therefore the polar coordinate is (Ö10, 5.961434753 radians)

To find another coordinate for the same location: 5.961434753-p = 2.819842099

V. Equation Convertion

Recall x = r cos q and y = r sin q and x2 + y2 = r2

Given: x2 + y2 - 6x = 0 substitute in the above equations for x and y

r2 - 6(r cos q) = 0

r ( r - 6 cos q) = 0

r = 0 and r - 6 cos q = 0
r = 6 cos q

so the polar equation is r = 6 cos q

VI. Convert the polar equation to rectangular form:

Example: r = 4 cos q

r2 = 4r cos q

x2 + y2 = 4x

x2 -4x + y2 = 0 would be the rectangular equation

Example: q = 5p/3

Tan q = tan 5p/3 = -Ö3

y/x = -Ö3/1
y = -Ö3 x
y = xÖ3 = 0

Example: r = 6/(2cosq - 3 sin q)

r = 6 / (2(x/r) - 3(y/r))

r = 6r/(2x - 3y)

2x - 3y = 6r/r

2x - 3y = 6