**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:

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?

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?

Dancer_Engineer

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7 years ago

It's a bit confusing to me.

What do

What do

*theta*and*r*value signify here?Dancer_Engineer

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7 years ago

Yes, I can make that out.

But confused which is x value and which is y value. 😔

But confused which is x value and which is y value. 😔

I am confused too but that is the answer given 😔Dancer_EngineerYes, I can make that out.

But confused which is x value and which is y value. 😔

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

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7 years ago

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

pranav_vanarp

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7 years ago

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.

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.

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??

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??

but why do we have to put x = r*cos theta?pranav_vanarptry 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

Is it valid?

pranav_vanarp

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7 years ago

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.

https://en.wikipedia.org/wiki/List_o..._Cartesian_coordinates_from_polar_coordinates

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

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.

https://en.wikipedia.org/wiki/List_o..._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_vanarpof 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.

https://en.wikipedia.org/wiki/List_o..._Cartesian_coordinates_from_polar_coordinates

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

pranav_vanarp

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7 years ago

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 ?

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 ?

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_vanarpdo 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.

pranav_vanarp

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7 years ago

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.AruwinOh 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.

Anyways this is how your plot should look for a = 1.

https://www.wolframalpha.com/input/?i=plot+(x^2+++y^2)(+x+++y)+=+2(x*y)

Oh no,wait a minute.Now what do I do with a??Can I assume a=1 always?pranav_vanarpOk. 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.

https://www.wolframalpha.com/input/?i=plot (x^2 + y^2)( x + y) = 2(x*y)

pranav_vanarp

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7 years ago

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).

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).

Praveen Kumar Purushothaman

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7 years ago

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

Hey, I have one doubt here. In the table graph,When theta=0,r=0 but according to the equationpranav_vanarpdo 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 ?

(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

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7 years ago

Hello.

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).

You seem to be ignoring sin(theta). rewrite the said equation asAruwinHey, 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??

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).

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?Deepesh BagwaleHello.

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.

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