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Converting Polar to Rectangular

  • #1
1,235
1
Hello

If you have [itex] r = 2\sin(2\theta) [/itex], how would you convert it to rectangular form? I tried doing this:

[itex] \sin(2\theta) = 2\sin\theta\cos\theta [/itex] which means [itex] r = 4\sin\theta\cos\theta [/itex]. Then I know that [itex] r^{2} = x^{2} + y^{2} [/itex]. We know that [itex] x = r\cos\theta, y = r\sin\theta [/itex]. But then I have a circular argument, where I end up with [itex] y^{2} = y^{2} [/itex]. This means that I have to use another conversion factor. Should I use [itex] \sin\theta = \frac{y}{r} [/itex]?

Thanks :smile:
 

Answers and Replies

  • #2
hi, you're trying to write [itex] r = 4\sin\theta\cos\theta [/itex]
in terms of x and y now that you have [itex] \sin\theta = \frac{y}{r} [/itex]
and [itex] \cos\theta = \frac{x}{r} [/itex] just make the substitutions and you should have no problem getting the answer
 
  • #3
10
0
Thank you kindly for your response.
I 'think' I get the mechanics of it but me and the old voyage 200 are at differences about the answer when doing sample problems.

Example:
let's uses (-2,8),(3,5) as an example.

This what I get for AxB when done manually.

(3,5),(-2,8)
3^2+5^2=r^2
9+25=r^2
√34=r
5.8301=r
then:
toa: tan^(-1)⁡(3/5)=30.964^o
do the same to coordinate #2
-2^2+8^2=r^2
4+16=r^2
√20=r
4.47214=r
then:
toa: tan^(-1)⁡((-2)/8)=-14.036^o
we have:
(5.8301,30.964^o ),(4.47214,-14.036^o )
note: To cross product we need the sin of the angle between the two point. AKA, absolute value
30.964^o+14.036^o=44.73^o
For cross product A ⃗xB ⃗:
5.8301*4.47214 sin⁡44.73 n
18.3493 which is different from the calculator
**********
Voyage 200

crossP([3,5],[-2,8]) = [0. 0. 34.]

Whom is incorrect, me or the calculator?
Why are the answers so different?
 
  • #4
PeterO
Homework Helper
2,425
46
Thank you kindly for your response.
I 'think' I get the mechanics of it but me and the old voyage 200 are at differences about the answer when doing sample problems.

Example:
let's uses (-2,8),(3,5) as an example.

This what I get for AxB when done manually.

(3,5),(-2,8)
3^2+5^2=r^2
9+25=r^2
√34=r
5.8301=r
then:
toa: tan^(-1)⁡(3/5)=30.964^o
do the same to coordinate #2
-2^2+8^2=r^2
4+16=r^2
√20=r
4.47214=r
then:
toa: tan^(-1)⁡((-2)/8)=-14.036^o
we have:
(5.8301,30.964^o ),(4.47214,-14.036^o )
note: To cross product we need the sin of the angle between the two point. AKA, absolute value
30.964^o+14.036^o=44.73^o
For cross product A ⃗xB ⃗:
5.8301*4.47214 sin⁡44.73 n
18.3493 which is different from the calculator
**********
Voyage 200

crossP([3,5],[-2,8]) = [0. 0. 34.]

Whom is incorrect, me or the calculator?
Why are the answers so different?
I high lighted one line in your answer. Should that be 3/5 or 5/3 ?
 
  • #5
ehild
Homework Helper
15,396
1,803
Both the scalar product and the cross product can be calculated from the components directly. Check your notes.
If vector a is (ax, ay) and vector b is (bx,by) the magnitude of the cross product axb is axby-aybx, and it is parallel to the z axis.

You also can calculate it on that complicated way you tried. There is a mistake when calculating the magnitude of b:
-2^2+8^2=r^2
4+16=r^2
the square of 8 is 64. :tongue2:

ehild
 
  • #6
10
0
I high lighted one line in your answer. Should that be 3/5 or 5/3 ?
Well, 5 = x or your adjacent, no?
 
  • #7
10
0
Both the scalar product and the cross product can be calculated from the components directly. Check your notes.
If vector a is (ax, ay) and vector b is (bx,by) the magnitude of the cross product axb is axby-aybx, and it is parallel to the z axis.

You also can calculate it on that complicated way you tried. There is a mistake when calculating the magnitude of b:
the square of 8 is 64. :tongue2:

ehild
I do it the REALLY long way first to understand the mechanics and then jump to formulas.
But, you single handedly answered both my question and my next question which would have been the express route.

Now I understand both.

Thank you kindly!
 
  • #8
100
0
Hello

If you have [itex] r = 2\sin(2\theta) [/itex], how would you convert it to rectangular form? I tried doing this:

[itex] \sin(2\theta) = 2\sin\theta\cos\theta [/itex] which means [itex] r = 4\sin\theta\cos\theta [/itex]. Then I know that [itex] r^{2} = x^{2} + y^{2} [/itex]. We know that [itex] x = r\cos\theta, y = r\sin\theta [/itex]. But then I have a circular argument, where I end up with [itex] y^{2} = y^{2} [/itex]. This means that I have to use another conversion factor. Should I use [itex] \sin\theta = \frac{y}{r} [/itex]?

Thanks :smile:

in

r=4*sin(theta)*cos(theta)

multiply 2 sides by r**2
to get

r**3=4*x*y

r=sqrt(x**2+y**2)
r**3=(x**2+y**2)**(3/2)
 

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