Polar coordinates

M

Mathygirl

New member
Joined
Dec 17, 2020
Messages
5
Hello

I’m stuck with something that i know is an easy problem.

I have the following equation: 4(x^2 + y^2 — ax)^3 = 27a^2(x^2+y^2)^2

And i need to put the equation in polar coordinates so i can show that it can be written as: r = 4a• cos^3(theta/3) .

I tried replacing x^2+y^2 by r^2, x by r•sin(theta) and y by r•cos(theta) but i don’t see how it can be equal tor = 4a• cos^3(theta/3) .

Thank you for your help.
 
Mathygirl said:
Hello

I’m stuck with something that i know is an easy problem.

I have the following equation: 4(x^2 + y^2 — ax)^3 = 27a^2(x^2+y^2)^2

And i need to put the equation in polar coordinates so i can show that it can be written as: r = 4a• cos^3(theta/3) .

I tried replacing x^2+y^2 by r^2, x by r•sin(theta) and y by r•cos(theta) but i don’t see how it can be equal tor = 4a• cos^3(theta/3) .

Thank you for your help.
Click to expand...
This involves nasty algebra.

Please share your work.
 
I converted the equation to the following:
4(-ar•sin(theta)+r^2)^3 = 27a^2r^4.
Then i’m stuck because i don’t know how to go further.
 
Mathygirl said:
I’m stuck with something that i know is an easy problem.

I have the following equation: 4(x^2 + y^2 — ax)^3 = 27a^2(x^2+y^2)^2

And i need to put the equation in polar coordinates so i can show that it can be written as: r = 4a• cos^3(theta/3) .

I tried replacing x^2+y^2 by r^2, x by r•sin(theta) and y by r•cos(theta) but i don’t see how it can be equal tor = 4a• cos^3(theta/3) .
Click to expand...
Clearly you'll need a formula for either cos(alpha/3) or sin(3 alpha), which you can put into either the original form or the target form, and try to manipulate the two equations into the same form.

But how do you "know" it's easy? It doesn't look that way to me, even if it may be straightforward.
 
Try going from r = 4a• cos^3(theta/3) to 4(x^2 + y^2 — ax)^3 = 27a^2(x^2+y^2)^2 .
If you can do that then read your work from the last line to the first line and there will be your solution. Study the solution to see where you got stuck.
 
Dr.Peterson said:
Clearly you'll need a formula for either cos(alpha/3) or sin(3 alpha), which you can put into either the original form or the target form, and try to manipulate the two equations into the same form.

But how do you "know" it's easy? It doesn't look that way to me, even if it may be straightforward.
Click to expand...
I really don’t know how i can get a formula for one of those. This kind of questions are always really hard for me.

Because my teacher said it was, but he th
 
H
Jomo said:
Try going from r = 4a• cos^3(theta/3) to 4(x^2 + y^2 — ax)^3 = 27a^2(x^2+y^2)^2 .
If you can do that then read your work from the last line to the first line and there will be your solution. Study the solution to see where you got stuck.
Click to expand...

So going from polar to carthesian coordinates?
 
Dr.Peterson said:
Clearly you'll need a formula for either cos(alpha/3) or sin(3 alpha), which you can put into either the original form or the target form, and try to manipulate the two equations into the same form.

But how do you "know" it's easy? It doesn't look that way to me, even if it may be straightforward.
Click to expand...

I went from this
BD924B80-5BCB-41D5-86AD-94D31890A6EA.jpeg
To this
5C39FAB8-2F34-4962-8531-37C17019B9F0.jpeg
How can i find a formula for what you said?
 
Mathygirl said:
I really don’t know how i can get a formula for one of those.
Click to expand...
I expected that either you would have learned a formula for trig functions of theta/3 or 3theta (more likely the latter), or else would have been taught enough that you could try writing sin(3theta) = sin(2theta + theta) and use a double-angle formula.

Alternatively, you can look it up, either in your book or online. (Or check post #10)
 
You must log in or register to reply here.
Top