solving an equation to plot a graph

Hi everyone.Someone please help me solve this problem.I am stucked,really.This is actually part 2 of a question where part one is already solved by getting the equation needed to draw the graph.
And here is the equation:
(x^2+y^2)(x+y)=2axy
and the plotting table is as shown in the image:
[​IMG]

and a hint is that when x and y are at their higest values, the equation
(x^2+y^2)(x+y)=0 and hence,x+y=0

but how do I solve this to plot the graph?

Replies

  • Dancer_Engineer
    Dancer_Engineer
    It's a bit confusing to me.
    What do theta and r value signify here?
  • Aruwin
    Aruwin
    the angle and radius i suppose...
  • Dancer_Engineer
    Dancer_Engineer
    Yes, I can make that out.
    But confused which is x value and which is y value. 😔
  • Aruwin
    Aruwin
    Dancer_Engineer
    Yes, I can make that out.
    But confused which is x value and which is y value. 😔
    I am confused too but that is the answer given 😔
    Well, can you try and figure it out your own way without referring to the answers?Perhaps from there you may get something?
  • Dancer_Engineer
    Dancer_Engineer
    Post the complete question, both part 1 and part 2.
  • Aruwin
    Aruwin
    Dancer_Engineer
    Post the complete question, both part 1 and part 2.
    OK,but you're gonna have to wait because I have to write the solution manually cuz it is easier for me.
  • pranav_vanarp
    pranav_vanarp
    try putting x = r * cos (theta)
    y = r * sin (theta) .
    equation becomes
    r^3 (cos (theta) + sin (theta) ) = 2(r^2)acos(theta)sin(theta) ;

    => r(cos(theta) + sin(theta)) = 2a cos(theta)sin(theta)
    => 1/sin(theta) + 1/cos(theta) = 2a / r

    if we try theta = pi/4 and r = a/sq root(2) . as the image suggests ,
    the equation does seem to be satisfied

    Thus one solution is x = a/2 and y = a/2.
  • Aruwin
    Aruwin
    So here is the question and the solution of the first part.The question is actually in japanese so I had to translate it into english so you can understand.

    There is a rectangle paper ABCD and let point B be the peak as it is folded.The paper is folded at line EF.Let P be the shifted point of B. If PE+PF= a, how is the curve P drawn??

    [​IMG]
    [​IMG]
  • Aruwin
    Aruwin
    pranav_vanarp
    try putting x = r * cos (theta)
    y = r * sin (theta) .
    equation becomes
    r^3 (cos (theta) + sin (theta) ) = 2(r^2)acos(theta)sin(theta) ;

    => r(cos(theta) + sin(theta)) = 2a cos(theta)sin(theta)
    => 1/sin(theta) + 1/cos(theta) = 2a / r

    if we try theta = pi/4 and r = a/sq root(2) . as the image suggests ,
    the equation does seem to be satisfied
    but why do we have to put x = r*cos theta?
    Is it valid?
  • pranav_vanarp
    pranav_vanarp
    of course it is valid.
    in any equation you can put x = rcos theta
    y = rsin theta.

    the equation is of 2 variables. we are merely transforming the co-ordinate system.
    List Of Common Coordinate Transformations To Cartesian Coordinates From Polar Coordinates

    where did you get the table in first post from. the one containing values of r and theta.
  • Aruwin
    Aruwin
    pranav_vanarp
    of course it is valid.
    in any equation you can put x = rcos theta
    y = rsin theta.

    the equation is of 2 variables. we are merely transforming the co-ordinate system.
    List Of Common Coordinate Transformations To Cartesian Coordinates From Polar Coordinates

    where did you get the table in first post from. the one containing values of r and theta.
    the teacher showed it.So,how do I plot the points??How does the graph gonna turn out?I am so lost 😔
  • pranav_vanarp
    pranav_vanarp
    do you only want the solutions to the equation or the entire graph ?

    Plotting r-theta graph will be easier than plotting x-y graph.
    of course the shape will be same. Only procedure will be different.
    Your equation here is
    (1/sin(theta) + 1/cos(theta)) = 2a/r .

    Notice that you will have to plot r versus theta.
    Vary theta from 0 to 2pi and find what values of r you get. Then mark the various points.
    For ex in above equation at theta = 0 , r = 0. (i.e the point x=0 , y = 0)
    at theta = pi/4 r = a/sq root(2). (i. e the point x=y= a/2 ).

    You getting the drift ?
  • Aruwin
    Aruwin
    pranav_vanarp
    do you only want the solutions to the equation or the entire graph ?

    Plotting r-theta graph will be easier than plotting x-y graph.
    of course the shape will be same. Only procedure will be different.
    Your equation here is
    (1/sin(theta) + 1/cos(theta)) = 2a/r .

    Notice that you will have to plot r versus theta.
    Oh thank you thank you!!Well, I want the solutions only for the equation because when the equation is solved, then I can just plot r against theta.
  • pranav_vanarp
    pranav_vanarp
    Aruwin
    Oh thank you thank you!!Well, I want the solutions only for the equation because when the equation is solved, then I can just plot r against theta.
    Ok. Well it is good to know how to plot equations in r and theta. It simplifies matters in some cases in some cases it does not.

    Anyways this is how your plot should look for a = 1.
    plot (x^2 + y^2)( x + y) = 2(x*y) - Wolfram|Alpha
  • Aruwin
    Aruwin
    pranav_vanarp
    Ok. Well it is good to know how to plot equations in r and theta. It simplifies matters in some cases in some cases it does not.

    Anyways this is how your plot should look for a = 1.
    plot (x^2 + y^2)( x + y) = 2(x*y) - Wolfram|Alpha
    Oh no,wait a minute.Now what do I do with a??Can I assume a=1 always?
  • pranav_vanarp
    pranav_vanarp
    I think not. The equation needs an 'a'. That plot was just to illustrate for a = 1.

    But 'a' is a constant , so you don't need to worry about it too much. You have to plot in terms of 'a'.
    i.e at theta = pi/4 , r = a/sq root(2) (or x = y = a/2).
  • PraveenKumar Purushothaman
    PraveenKumar Purushothaman
    Why don't you try google graphs? It would help you plot the graph of the equation na???
  • Aruwin
    Aruwin
    pranav_vanarp
    do you only want the solutions to the equation or the entire graph ?

    Plotting r-theta graph will be easier than plotting x-y graph.
    of course the shape will be same. Only procedure will be different.
    Your equation here is
    (1/sin(theta) + 1/cos(theta)) = 2a/r .

    Notice that you will have to plot r versus theta.
    Vary theta from 0 to 2pi and find what values of r you get. Then mark the various points.
    For ex in above equation at theta = 0 , r = 0. (i.e the point x=0 , y = 0)
    at theta = pi/4 r = a/sq root(2). (i. e the point x=y= a/2 ).

    You getting the drift ?
    Hey, I have one doubt here. In the table graph,When theta=0,r=0 but according to the equation
    (1/sin(theta) + 1/cos(theta)) = 2a/r ,isn't it supposed to be that r=2a???
    Because cos(0)=1,am I right??
    Same goes to when theta=pi,why is r=0 and not 2a?
    Because sin(pi)=1

    And can you explain to me why is when theta=3pi/4,r=+/- infinity????
    How come it becomes infinity?
  • Deepesh Bagwale
    Deepesh Bagwale
    Hello.

    Aruwin
    Hey, I have one doubt here. In the table graph,When theta=0,r=0 but according to the equation
    (1/sin(theta) + 1/cos(theta)) = 2a/r ,isn't it supposed to be that r=2a???
    Because cos(0)=1,am I right??
    You seem to be ignoring sin(theta). rewrite the said equation as
    r = 2a(sin(theta)cos(theta))/(sin(theta) + cos(theta)) ----- (i)
    putting theta = 0
    you get
    r = 2a(sin(0)cos(0)/(sin(0) + cos(0)). Similarly for theta = pi use equation (i). Earlier form of equation must have confused you. There 1/sin(0) term was present ,you missed it.


    r = 2a(sin(3pi/4)cos(3pi/4))/(sin 3pi/4 + cos 3pi/4). Denominator is zero. Hence you approach infinity for r. I don't know why +/- infinity is given to you. Perhaps when you approach theta from right side of 3pi/4 you get + infinity and when approaced from left side of 3pi/4 you get -infinity (or other way around).
  • Aruwin
    Aruwin
    Deepesh Bagwale
    Hello.



    You seem to be ignoring sin(theta). rewrite the said equation as
    r = 2a(sin(theta)cos(theta))/(sin(theta) + cos(theta)) ----- (i)
    putting theta = 0
    you get
    r = 2a(sin(0)cos(0)/(sin(0) + cos(0)). Similarly for theta = pi use equation (i). Earlier form of equation must have confused you. There 1/sin(0) term was present ,you missed it.
    Oh my,you are right!Didn't realize that mistake before.Thanks.Hey,can you please help me out with my othe post about Leaf of Descartes?

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