solving an equation to plot a graph

Aruwin · 2012-01-23T03:02:50+00:00

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: 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?

solving an equation to plot a graph

Aruwin

Member

Updated: Oct 26, 2024

Views: 1.3K

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:

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?

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Dancer_Engineer

Member • Jan 23, 2012

It's a bit confusing to me.
What do theta and r value signify here?

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Aruwin

Member • Jan 23, 2012

the angle and radius i suppose...

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Dancer_Engineer

Member • Jan 23, 2012

Yes, I can make that out.
But confused which is x value and which is y value. 😔

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Aruwin

Member • Jan 23, 2012

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?

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Dancer_Engineer

Member • Jan 23, 2012

Post the complete question, both part 1 and part 2.

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Aruwin

Member • Jan 23, 2012

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.

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pranav_vanarp

Member • Jan 23, 2012

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.

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Aruwin

Member • Jan 23, 2012

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??

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Aruwin

Member • Jan 23, 2012

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?

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pranav_vanarp

Member • Jan 23, 2012

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.
<a href="https://en.wikipedia.org/wiki/List_of_common_coordinate_transformations#To_Cartesian_coordinates_from_polar_coordinates" target="_blank" rel="nofollow noopener noreferrer">List Of Common Coordinate Transformations To Cartesian Coordinates From Polar Coordinates</a>

where did you get the table in first post from. the one containing values of r and theta.

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Aruwin

Member • Jan 23, 2012

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.
<a href="https://en.wikipedia.org/wiki/List_of_common_coordinate_transformations#To_Cartesian_coordinates_from_polar_coordinates" target="_blank" rel="nofollow noopener noreferrer">List Of Common Coordinate Transformations To Cartesian Coordinates From Polar Coordinates</a>

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 😔

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pranav_vanarp

Member • Jan 23, 2012

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 ?

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Aruwin

Member • Jan 23, 2012

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.

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pranav_vanarp

Member • Jan 23, 2012

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.
<a href="https://www.wolframalpha.com/input/?i=plot+%28x%5E2+%2B+y%5E2%29%28+x+%2B+y%29+%3D+2%28x*y%29" target="_blank" rel="nofollow noopener noreferrer">plot (x^2 + y^2)( x + y) = 2(x*y) - Wolfram|Alpha</a>

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Aruwin

Member • Jan 23, 2012

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.
<a href="https://www.wolframalpha.com/input/?i=plot+%28x%5E2+%2B+y%5E2%29%28+x+%2B+y%29+%3D+2%28x*y%29" target="_blank" rel="nofollow noopener noreferrer">plot (x^2 + y^2)( x + y) = 2(x*y) - Wolfram|Alpha</a>
Oh no,wait a minute.Now what do I do with a??Can I assume a=1 always?

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pranav_vanarp

Member • Jan 23, 2012

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).

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PraveenKumar Purushothaman

Member • Jan 23, 2012

Why don't you try google graphs? It would help you plot the graph of the equation na???

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Aruwin

Member • Jan 23, 2012

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?

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Deepesh Bagwale

Member • Jan 24, 2012

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).

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Aruwin

Member • Jan 24, 2012

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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