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Homework Help: Circle Transformations

  1. Jan 19, 2010 #1
    If someone could explain to me how to Transform Circles ? i know how to cransform curves and such.

    for example there is a question asking me to transform a circle with Origin (57,8.5),r: 0.5
    to a circle with orgin (57,8.5) r:6

    and anther type which asks me to transform circle with orgin (57,8.5) with r:0.5 to a circle with origin (-57,-8.5) r: 0.5


    if someone could explain to me the process. there is no need to solve those question,
     
  2. jcsd
  3. Jan 19, 2010 #2

    Mark44

    Staff: Mentor

    I don't think I know what you mean by "transforming circles." I do know how to find the equations of all four of the circles you have presented. Is that what you're asking?
     
  4. Jan 19, 2010 #3
    Am sorry that i wasn't clear

    i will copy the exact word from paper

    1) Referring to lines parrallel to xaxis and y axis, describe a set of transformation that would transform the innermost circle of House I ( (57,8.5),r: 0.5 ) to the outermost circle of House I ((57,8.5),r: 0.5 )

    The diagram in nothing buy the X and y axis, with these circles on them
     
  5. Jan 19, 2010 #4

    Mark44

    Staff: Mentor

    Is this in the context of matrices that perform transformations? If so, for problem 1 you need a matrix that scales distances by a factor of 17.

    For the second problem you need a matrix that reflects points across the x-axis and another that reflects points across the y-axis.
     
  6. Jan 19, 2010 #5
    No it has nothing to do with linear algebra . it comes as a review for the Conics chapter.. am fairly sure he wants me to use the Standard form of circle ( X-h)2 + (y-k)2 = r2 and the general formula.

    Is there a way for me to upload A scanned sheet ?
     
  7. Jan 19, 2010 #6

    Mark44

    Staff: Mentor

    In the first problem, the only thing that changes is the radius, from r = .5 to r = 8.5. In the second problem, the only thing that changes is the location of the center.

    Do you know how to write the equation of a circle given its center and radius?
     
  8. Jan 19, 2010 #7
    It's ( X-h)2 + (y-k)2 = r2
    where
    h= the x cordinate of the center
    k= the y cordinate of the center
    and r is the radius of the circle

    i also would like to knopw if there is a way to show you the paper. i mean, my English ain't that good and i don;t think i explained it correct .

    Thanks Mark for your help so far
     
  9. Jan 19, 2010 #8

    Mark44

    Staff: Mentor

    People scan their problems all the time and upload them, so give it a try.
     
  10. Jan 19, 2010 #9
    Last edited by a moderator: May 4, 2017
  11. Jan 19, 2010 #10

    Mark44

    Staff: Mentor

    It's pretty straightforward to get the equations of the innermost and outermost circles in House 1, and in fact, all of the circles in all four Houses. For House 1, the equation of the inner circle is (x - 57)2 + (y - 37/4)2 = 1/4. The equation of the outer circle in this House is (x - 57)2 + (y - 37/4)2 = 36.

    I'm not sure what they're looking for as far as a transformation is concerned. A worked, similar example from your textbook would be very helpful.
     
  12. Jan 20, 2010 #11
    This is like the Province project ... i found no examples like it in the textbook.

    if you don't mind, Can you explain to me how you got the equations ?

    thanks alot for your help Mark
     
  13. Jan 20, 2010 #12

    Mark44

    Staff: Mentor

    I don't know anything about the Province project.

    As was given in the problem, the equation of a circle with center at (h, k) and radius r is (x -h)2 + (y - k)2 = r2.

    All of the circles in House 1 have their centers at (57, 37/4), in feet. All of the distances come from the drawing or the description. The radius of the smallest circle is 1/2 ft. The radius of the largest circle is 6 ft.
     
  14. Jan 20, 2010 #13
    how did you get the 37/4 ?
     
    Last edited: Jan 20, 2010
  15. Jan 20, 2010 #14

    Mark44

    Staff: Mentor

    9.25, the y-value at the center of the circle. The centerline of the circles in Houses 1 and 2 is 1.5 ft + 7.75 ft = 9.25 ft above the origin.

    I figured out the rest of it, and the key is to be working with functions. The circle equations aren't functions, but each equation can be turned into two functions: one for the upper half of the circle, and one for the lower half.

    I think that the way to approach this is to start with a circle centered at (0,0) of radius 1, and do whatever we need to make it larger or smaller, then translate it to the right or left.

    This circle's equation is x2 + y2 = 1. Solving for y, we get
    [tex]y = f(x) = +\sqrt{1 - x^2}[/tex]

    This function represents the upper half of the circle. The lower half is almost identical, with the difference being we have the negative square root.

    To get the smallest circle in House 1, we have to transform this function with these operations, in this order:
    1. Compress the graph of the function toward the y-axis and toward the x-axis, both by factors of 2. This will give us a circle (upper half) of radius 1/2.
    2. Translate the graph of the previous step so that its center is at (57, 9.25).

    For the first step, we want y = .5f(2x) = (1/2)sqrt(1 - 4x2). If you work this out, you get 2y = sqrt(1 - 4x2) or 4x2 + 4y2 = 1, or x2 + y2 = 1/4.

    For the second step, translate the graph of the function in the previous step to the right by 57 units and up 9.25 units. This would be y = .5f(2(x - 57)) + 9.25. I'll leave it to you to verify that we get the upper half of the smallest circle in House 1. You need a similar set of transformations to get the lower half of the same circle.
     
  16. Jan 20, 2010 #15
    But when you solve the y = .5f(2(x - 57)) + 9.25. the number are so huge,

    2y=[tex]\sqrt{1- (2x - 57)^2}[/tex] + 9.5
    2y=[tex]\sqrt{1- (4x^2 -228x +3249)}[/tex] -9.5

    4y2 = 1-(4x2-228x+3249)+90.25
    4y2+4x2-228x = -3157.75


    am thinking maybe he wants us to say how to translate those circles only theoretically just like you did on the last replay
    am sure your solution was right no dout, but i think we never did sucha thing (in the class or in workbooks ), so i don't know how he expect me to solve it .
     
  17. Jan 20, 2010 #16

    Mark44

    Staff: Mentor

    There's a mistake in the line below. If f(x) = sqrt(1 - x2), f(2(x-57)) = +sqrt(1 - [2(x - 57)]2). I realize I have omitted some of what you had above, but it was only to make a point. There is no need to square things out underneath the radical.
    Talk to your professor/instructor to get clarification on how you should proceed.
     
  18. Feb 7, 2010 #17
    Hey..

    Am still working on this "project" and am stuck on number 6 on the 2nd page that i uploaded.

    am not sure how to calculate the standard deviation "that would allow 80%"

    i know that standard daviation is calculated by getting the mean, then subtracting the mean from the list of numbers, then square the new result, then add them, then divided by one less than the number of items in the list,then get the square root of total number.
     
  19. Feb 7, 2010 #18

    Mark44

    Staff: Mentor

    What do you know about the standard normal distribution, and how to calculate probabilities with it?
     
  20. Feb 8, 2010 #19
    If you gave me The :
    - mean x
    - Standard deviation S
    - And a number. h
    ~The symbols are imaginary~

    I can use (h-x)/s to get the Z score.
    and from the Z-score chart i can get the probabilty of you getting this number.


    But i couldn't work around this question..
    the question gave me the mean ( -4.75) and "the other number" (80%) But seems like am lacking 2 numbers, the Zscore and the Standard Deviation.. i need either one of them to get the other.


    And Thank you really for all the help :)
     
  21. Feb 8, 2010 #20

    Mark44

    Staff: Mentor

    You can find the z-score from a table of z-scores so that the probability in the right tail is .10. (The probability in the left tail is also .10.) Then you can find sigma.
     
  22. Feb 8, 2010 #21
    Ahaaa, i got it now.. so, to cover 80% of the Curve, we will drop 10% from the right and 10 % from the left (we will take the middle 80%) , an

    It will be something like 0.27=(0.80-(-4.75))/X

    Where x is the standard Deviation.
    X= 1.50=Standard Deviation.

    Edit:
    Do i keep 80% in this form, or do i divided it by 100 ?
     
  23. Feb 8, 2010 #22

    Mark44

    Staff: Mentor

    Is this formula just a "for instance?"
    The 0.80 you have in there shouldn't have anything to do with the 80% probability. Maybe this is just a coincidence that you have used this number.

    In your calculations, percent figures should be changed to decimal fractions.
     
  24. Feb 8, 2010 #23
    am confused again

    Okay, Here is the formula i know

    Z= (x - [tex]\mu[/tex])/[tex]\sigma[/tex]

    Where
    Z= I check forr 0.10 in the Z-score chart and it was 0.27
    x= i thought it to be 0.80 because the question said (the probabilty of "80%" )
    [tex]\mu[/tex] = Mean = -4.75
    [tex]\sigma[/tex] = Standard Deviation, and thats what we gonna solve for


    But you said x has nothing to do with the 80% probability
     
  25. Feb 8, 2010 #24

    Mark44

    Staff: Mentor

    I don't know what you did to get 0.27 for .10.

    I looked in a z-score table and found a z-score of about 1.28 for which the area (probability) to the left of that z-score was .90, which means there is an area (probability) of .10 in the right-hand tail.

    In the table I looked at, for a z-score of 1.28, the area was .8997. For a z-score of 1.29, the area was .9015. Since .8997 is a lot closer to .9 than is .9015, I'll stick with 1.28 and not both interpolating to get a closer value.

    At this point you should have two bell-shaped graphs: one is of the ice temps on the curling rinks, and the other is the standard normal curve.

    On the first graph, the high point (mean) is at -4.75 and on either side of it are -5 and -4.5. We need to figure out sigma so that the area under the bell curve between -5 and -4.5 is .80.

    On the second graph, the high point is at 0, and on either side are -1.28 and + 1.28. On this graph, the area between -1.28 and 1.28 is (close to) .80.

    On the standard normal curve, the two endpoints are 1.28 standard deviations away from the mean. (Do you follow this? For the standard normal curve, the mean is 0 and the st. dev. is 1.)

    This means that on the ice temp graph, the values -5 and -4.5 are 1.28 standard deviations away from -4.75. I.e., (-4.5 - (-4.75)) = 1.28*sigma.

    I know you can find sigma from this equation! I have faith in you...
     
  26. Feb 8, 2010 #25
    Okay i pretty much understood all what you said .

    But there is one little thing that i don't understand yet .... I have to Limits for the graph, -4.5 and -5 .... which one do i use to calculate the standard deviation.

    For -4.5 SD is 0.20
    For-5 it is -0.20

    am sorry, but i had a long hard day, and am really really exhusted, so am sorry if i sound stupid or lazy
     
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