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Finding dy/dx for a circle

  1. Jul 4, 2017 #1

    MPZ

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    1. The problem statement, all variables and given/known data
    Hello
    I have this circle with the equation :

    (x-a)^2+(y-b)^2=r^2
    I want to find dy/dx for it

    2. Relevant equations
    (x-a)^2+(y-b)^2=r^2


    3. The attempt at a solution
    I am looking on the internet and it appears that I should use what is called "Implicit differentiation", can anyone use "Implicit differentiation for this circle please, thanks!
     
  2. jcsd
  3. Jul 4, 2017 #2

    BvU

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    Hi,
    Keep it simple and set a=b=0 . If you differentiate ##x^2+y^2 = r^2## you get ##2xdx+2ydy=0## so ##\displaystyle {{dy\over dx}= -{x\over y}}## which can be checked easily.

    PS is this homework ?
     
  4. Jul 4, 2017 #3

    MPZ

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    this is not a homework, i am trying to use a mathematical software to draw images using mathematical equations :) Can you please tell me a way to not "keep it simple" since i need the values of a and b since I can't draw it that small!
     
  5. Jul 4, 2017 #4

    BvU

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    Fair enough.
    The idea was to make it easier to understand, so that you would be able to complicate things on your own. Same differentiation on ##(x-a)^2+(y-b)^2 = r^2## gives you ##2(x-a)dx+2(y-b)dy=0## which is not very surprising if you see it as a translation of the origin to ##(a,b)##.

    Can you elaborate? Give an example why you need ##dy\over dx## ?
     
  6. Jul 4, 2017 #5

    MPZ

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    are you a detective? LOOL. I need the derivative because from it I can get the slope of the normal at any point since I want to find the equation of multiple normal lines at different points to draw "the hair" of the thing I am drawing. No more questions please with this sort
     
  7. Jul 4, 2017 #6

    LCKurtz

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    You hardly need calculus and derivatives for that. The normal lines to a circle are radius lines. Straight lines through the center.
     
  8. Jul 4, 2017 #7

    LCKurtz

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    Here's a sample written in Maple:
    > restart; with(plots);
    > circle := plot([cos(theta), sin(theta), theta = 0 .. 2*Pi], color = black, axes = none, thickness = 2):
    > hair := seq(plot([r*cos((1/3)*Pi+(1/36)*k*Pi), r*sin((1/3)*Pi+(1/36)*k*Pi), r = 1 .. 1.2], thickness = 2), k = 0 .. 12);
    > display({hair}, circle);
    HairyCircle.jpg
    Maybe that will give you some ideas.
     
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