Monday, April 26, 2010

Stephen's Reflection

Ok so this week we did more stuff in the study guides for trig exam and yea..anyway im going to show you the half and double angle formulas for yall...

Half-Angle and Double Angle Formulas:

sin(2alpha) = 2sin(alpha)cos(alpha)
cos(2alpha) = cos^2(alpha)-sin^2(alpha)=1-2sin^2(alpha)=2cos^2
(alpha)-1
tan(2alpha) = 2tan(alpha)/1-tan^2(alpha)
sin(alpha/2)= +- sqrt(1-cos(alpha)/2)
cos(alpha/2)= +- sqrt(1+cos(alpha)/2)
tan(alpha/2)= +- sqrt(1-cos(alpha)/1+cos(alpha))=sin(alpha)/1+cos
(alpha)=1-cos(alpha)/sin(alpha)

Ok so like i said in class, everything with graphs is gonnneee. I need help working on conics and graphing them so if anyone can help i would be grateful

2 comments:

  1. Conics

    Ellipses

    Steps:

    1. find the center
    2. determine the major axis
    3. find the vertex (± √big denom)
    4. find the other intercept ( ± √small denom)
    5. find the focus (c^2 = a^2 + b^2)
    6. determine the length of the major axis (2√big denom)
    7. find the length of the minor axis (2√small denom)
    8. finally graph

    Example 1: Graph the following ellipse. Find its major intercepts, length of the major axis, minor intercepts, length of the minor axis, and foci.

    x^2/4 + y^2/9 = 1

    This ellipse is centered at (0, 0). Since the larger denominator is with the y variable, the major axis lies along the y-axis.

    Since a^2 = 9 then a = 3 & Since b^2 = 4 then b = 2

    Major intercepts: (0, 3), (0, –3)

    Length of major axis: 2 √9 = 6

    Minor intercepts: (2, 0), (–2, 0)

    Length of minor axis: 2√4 = 4

    c^2 = a^2 + b^2

    = 9 - 4

    = 5

    = √5

    Foci: (0, √5) , (0, -√5)

    then you graph your points..

    Parabolas:

    how to find the axis of symmetry, vertex, focus, & directrx??

    1.) to find the axis of symmetry: x = -b/2a

    2.) for the vertex: (-b/2a, f(-b/2a)) or use complete the square:

    y = (x+a)^2 + b.....a & b are numbers and (-a,b) = vertex

    3.) to find the focus: 1/4p= the coefficient of x^2 and then add p

    Note:

    *If opens up, add to y value from vertex, if opens down, subtract

    *If opens right, add to x value to vertex, if opens left, subtract)

    4.) directrix: is p units behind the vertex

    Note:

    *If opens up, subtract; if opens down, add from y-value of vertex.
    *If opens right, subtract x-value
    *If opens left, add x-value

    Example: x^2 + 1

    ~vertex:

    x = -b/2a

    x = 0/2(1) = 0

    0^2 + 1 = 1

    (0,1)

    ~Focus:

    1/4p = 1

    4p = 1

    p = 1/4

    (0, 1 + 1/4)

    (0, 5/4)

    ~directrix:

    y = 1 - 1/4

    y = 3/4

    CIRCLES
    The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)

    If the equation is not in standard form, you must complete the square to put it in standard form.

    If you are given a center and a point, you can use the distance formula to find the radius.

    To find the intersection of a line and a circle:

    1. solve the linear eqn for y.
    2. substitute in the circle eqn.
    3. solve for x.
    4. plug the x value in to get the y value.

    ***Reminder. If your x value is imaginary, then there is no point of intersection.

    EX: find the center and radius.

    (x-3)^2+(y+7)^2=19 c:(h,k)

    center: (3,-7) radius: square root of 19

    EX: find the eqn of the circle with the center (1,4) through (3,7)

    in the problem you are given a center and a point so you would plug into the distance formula.

    square root of (3-1)^2+(7-4)^2= square root of 4+9=square root of 13. **13 has no root.

    Your answer should be (x-1)^2+(y-4)^2=13

    Hyperbola

    (x^2-h/a^2)+(y^2-k/b^2)=1

    * Your center is (h,k)
    * your major axis has the larger denominator

    ReplyDelete
  2. Graphing Conics

    CIRCLES
    -the equation of a circle in standard form
    (x-h)^2+(y-k)^2=r^2
    where the center is (h,k) and radius=r
    -if not in standard form, you must complete the square for both x and y to put in standard form
    -give nthe center and point, you can use the distance formula to find the radius
    TO FIND THE INTERSECTION OF A LINE AND CIRCLE
    -solve the linear equation of a line
    -substitute in circle equation
    -solve for x
    -plug x value in to get y value
    *if your x value is imaginary, there is no point of intersection
    TO GRAPH A CIRCLE
    -find the center and radius
    -draw your circle

    ELLIPSES
    ((x-H)^2/(length of X/2)^2)+((Y-K)^2/(length of Y/2)^2)=1
    -(h,k)=center
    -major axis is the larger denominator
    -vertex is on major axis
    -focus> smaller#^2=larger#^2-focus^2
    focus is on major axis
    1. center
    2. major axis- bigger denominator (x or y)
    3. vertex +/- squareroot larger denominator
    4. other intercepts +/- squareroot smaller denominator
    5. focus
    6. length of major axis 2squareroot larger denominator
    7. length of minor axis 2squareroot smaller denominator
    8. graph

    HYPERBOLAS
    ((x-h)^2/(length/2)^2)-((y-k)^2/(length/2)^2)=1
    -center (h/k)
    -major axis is the non-negative denominator
    -vertex> +/- squareroot non-negative denominator
    -asymptotes> y=+/- (squareroot y/ squareroot x)X
    -focus^2= x denom+ y denom
    or focus^2= vertex^2+ other denom
    TO SKETCH
    1. shape
    2. center
    3. major
    4. minor
    5. other intercepts (none)
    6. focus
    7. asymptotes
    8. vertex
    9. sketch

    PARABOLAS
    - x=(-b/2a)
    -two ways for find the vertex
    ((-b/2a), f(-b/2a))or complete the square to get in vertex form y=(x+a)^2 +/- b >(-a,b)
    -focus (1/4p)=coefficient of x^2
    -directrix is P units behind vertex
    if opens up, add y value to vertex, if down, subtract y value from vertex
    if upens < add x value to vertex, if > subtract x value from the vertex

    ReplyDelete