Expert: Scott A Wilson - 8/1/2009

Question
QUESTION: 3r= sin A, r and A represent polar coordinates. Write each polar equation as an rectangular coordinates (x,y)

I have learned this stuff today only..iam kinda confused please help !

ANSWER: Think about a triangle with the center at the origin.

The point in question is the point at the end.

The (x,y) coordinates would be measured with the x on the x-axis and the y on the y-axis.

The polar coordinates of that point would be the angle the line to the point makes with the positive x axis and the r value would be the length of that line.

In this way, if the angle made with the positive x axis is Θ,
then tan(Θ) = y/x and r = √(x²+y²).


So given that 3r = sinA, to transfor them, we would need to find the value of sinA.
We know that tanA is the opposite side (which is y) over the hypotenuse (which is r).

Thus, we have 3√(x²+y²) = y/√(x²+y²).

Multiplying both sides by √(x²+y²) gives us 3(x²+y²) = y.

Multiplying this out gives us 3x² + 3y² = y.

This can be solved for x, but not for y.

Sovling for x gives us x = ±√[(y-3y²)/3].

Since we multiplied by √(x²+y²), we need to look for where this is 0.
It is only 0 at the origin.  Looking back at the problem, when r is zero, is doesn't really matter what value of A is chosen - it is still the point in the center of the graph.


---------- FOLLOW-UP ----------

QUESTION: Sorry to bother you, so the (x,y) coordinates are (0,0) right ?

thanks !!

Answer
If we look at the function x = ±√[(y-3y²)/3],
we need 3y²<y for what's under the √ to be positive.

We know that ² terms are always positive, so y is greater than or equal to 0.
Since we know y =0 would give x=0, we can now say that (0,0) is a solution.

To find the rest of the points, we know that y is positive,
so we can divide 3y²<y by y and get 3y  < 1, which means y < 1/3.

Given that 0<= y <= 1/3, the x values when y=0 and when y=1/3 are zero.

The curve looks symmetrix, so the maximum value for x is most likely at y=1/6.

I have made a graph of the solution points and will send that along.

The solution is a graph.  Here is the data in terms of r, Θ:
0   0
0.01   0.030004502
0.02   0.060036058
0.03   0.090121945
0.04   0.120289882
0.05   0.150568273
0.06   0.180986451
0.07   0.21157496
0.08   0.242365851
0.09   0.273393031
0.1   0.304692654
0.11   0.336303575
0.12   0.368267893
0.13   0.400631593
0.14   0.43344532
0.15   0.466765339
0.16   0.500654712
0.17   0.53518479
0.18   0.570437109
0.19   0.606505855
0.2   0.643501109
0.21   0.681553212
0.22   0.720818761
0.23   0.761489053
0.24   0.803802319
0.25   0.848062079
0.26   0.894665817
0.27   0.944152115
0.28   0.997283222
0.29   1.055202321
0.3   1.119769515
0.31   1.194412844
0.32   1.287002218
0.33   1.429256853
0.333333333   1.570796327

Here it is converted to (x,y)
0.000000000   0
0.009995499   0.0003
0.019963968   0.0012
0.029878253   0.0027
0.039710956   0.0048
0.0494343   0.0075
0.059019997   0.0108
0.068439097   0.0147
0.077661831   0.0192
0.086657429   0.0243
0.09539392   0.03
0.103837903   0.0363
0.111954276   0.0432
0.119705931   0.0507
0.127053375   0.0588
0.133954283   0.0675
0.140362958   0.0768
0.146229648   0.0867
0.151499703   0.0972
0.156112491   0.1083
0.160000000   0.12
0.163084978   0.1323
0.165278432   0.1452
0.166476154   0.1587
0.166553775   0.1728
0.165359457   0.1875
0.162702674   0.2028
0.158336067   0.2187
0.151924192   0.2352
0.142984999   0.2523
0.130766968   0.27
0.113943451   0.2883
0.089600000   0.3072
0.046552229   0.3267
0.000000000   0.333333333

As can be seen, when y is 0 or a 3rd, x is 0.  In this graph, x is a function of y.
The ineverse is not true, since for almost every x between 0 and 1/6 there are two y values.

So what I'm trying to say is there is more in the solution than just (x,y) = (0,0),
but not much.