0
$\begingroup$

$ r_1 \ = \ 4 \sin(3 \theta) \ $ and $ \ r_2 \ = \ 3 \cos(3\theta) \ $

a) find the solutions to the system using polar coordinates

I was able to solve this by setting $ \ r_1 \ $ and $ \ r_2 \ $ to be equal to each other. I got up to the part where $ \ \theta \ = \ \frac{tan^{-1}(4/3)}{3} \ + \ \frac{ nπ}{3} $ and plugged in 0,1,2,3 for $ \ n \ $ . I don't know what to do with the theta values that I found. How can i make it into a solution?

b) find the rectangular coordinates of the solution in the 2nd quadrant

If I knew the answer to a, I think I would be able to solve this but I am not sure.

c) find the rectangular coordinates of the points were the graphs appear to intersect but whose polar coordinates are not solutions

I am completely lost with this question

$\endgroup$
2
  • $\begingroup$ What are we solving for in part a? Do you mean to solve the similar system of equations: $$x=4\sin(3\theta)$$$$x=3\cos(3\theta)$$Solve for $x$ in terms of $\theta$?$${}$$Also note that systems of equations of the form $x=\sin(\dots)$ don't have solutions for $x$ in terms of elementary functions. $\endgroup$ Jun 6, 2016 at 23:19
  • $\begingroup$ You are right: the question, as you report it, is nonsense. "Find the values of $\theta$ for which $r_1=r_2$" is what the question should say, and you have a good idea for the answer to (b). For (c), I don't want to ruin your education and destroy the whole point of the exercise. Draw the graphs. Mark your ($r$,$\theta$) solutions. Notice that there are other intersections, and work out why they exist. $\endgroup$ Jun 7, 2016 at 6:35

2 Answers 2

0
$\begingroup$

Since you are simply looking for the values of $ \ \theta \ $ which satisfy both equations simultaneously, you are correct in setting $ \ 4 \ \sin 3 \theta \ = \ 3 \ \cos 3 \theta \ $ . However, you should have from this

$$ \tan \ 3 \theta \ = \ \frac{3}{4} \ \ \Rightarrow \ \ 3 \theta \ = \ \tan^{-1} \left( \frac{3}{4} \right) \ + \ n \ \pi $$

$$ \Rightarrow \ \ \theta \ = \ \frac{1}{3} \ \tan^{-1} \left( \frac{3}{4} \right) \ + \ n \ \frac{\pi}{3} \ \ . $$

You will not get "nice" angles out of this, but that is not actually important to solve the rest of your problem. The first solution in the "principal circle" is $ \ \theta \ \approx \ 0.2145 \ ( \approx \ 12.3º ) $ , and because the tangent function has a period of $ \ \pi \ $ (as you noted) , there are six solutions.

[If you are working with these equations as descriptions of polar curves, these six "collapse" to three intersection points because we have equations of two "three-petal" rosettes, which are retraced over a cycle of $ \ 2 \ \pi \ $ radians, so each intersection point corresponds to one angle solution with a positive radius and one with a negative radius.

Further, if we were concerned with polar curves (especially for purposes of arclength or area integration), we would also have to consider the values of $ \ \theta \ $ at which either $ \ r_1 \ = \ 0 \ $ or $ \ r_2 \ = \ 0 \ $ . This may be what is being asked for in part (c) .]

You will not need the exact value for these angles, so we can just designate $ \ \Theta_1 \ = \ \frac{1}{3} \ \tan^{-1} \left( \frac{3}{4} \right) \ $ and observe that the other solutions are successively $ \frac{\pi}{3} \ = \ \ 60º \ $ further counter-clockwise from there. This places $ \ \Theta_1 \ $ and $ \ \Theta_2 \ $ in Quadrant I, $ \ \Theta_3 \ $ in Quadrant II, $ \ \Theta_4 \ $ and $ \ \Theta_5 \ $ in Quadrant III, and $ \ \Theta_6 \ $ in Quadrant IV.

To work with these in part (b), we want to return to $ \ \ 3 \theta \ = \ \tan^{-1} \left( \frac{3}{4} \right) \ $ , from which we can extract

$$ \sin \ 3 \theta \ = \ \frac{3}{\sqrt{3^2 + 4^2}} \ = \ \frac{3}{5} \ \ , \ \ \cos \ 3 \theta \ = \ \frac{4}{\sqrt{3^2 + 4^2}} \ = \ \frac{4}{5} \ \ . $$

Since the values of $ \ 3 \Theta \ $ are separated by $ \ \pi \ = \ 180º \ $ , we have either

$$ \ r_1 \ = \ 4 \ \sin( 3 \Theta_{1,3,5} ) \ = \ r_2 \ = \ 3 \ \cos( 3 \Theta_{1,3,5} ) \ = \ \frac{12}{5} $$ or $$ \ r_1 \ = \ 4 \ \sin( 3 \Theta_{2,4,6} ) \ = \ r_2 \ = \ 3 \ \cos( 3 \Theta_{2,4,6} ) \ = \ -\frac{12}{5} \ \ . $$

As mentioned earlier, there are three "positive radius" solutions and three for "negative radius". These last three are redundant, as we shall see (and which you may already be aware of, if your course has discussed the interpretation of negative radius).

We will not be able to obtain exact values for the intersection point coordinates at all easily, but we can get approximate values by setting up

$$ \ x_1 \ = \ \frac{12}{5} \ \cos \Theta_1 \ \approx \ 2.4 \ \cos (12.29º) \ \approx \ 2.34 \ \ , $$ $$ y_1 \ = \ \frac{12}{5} \ \sin \Theta_1 \ \ \approx \ 2.4 \ \sin (12.29º) \ \approx \ 0.51 \ \ , $$

and we proceed similarly for

$$ \ x_3 \ = \ \frac{12}{5} \ \cos \Theta_3 \ \approx \ 2.4 \ \cos (132.29º) \ \ , \ \ y_3 \ = \ \frac{12}{5} \ \sin \Theta_3 \ \ \approx \ 2.4 \ \sin (132.29º) $$ and $$ \ x_5 \ = \ \frac{12}{5} \ \cos \Theta_5 \ \approx \ 2.4 \ \cos (252.29º) \ \ , \ \ y_5 \ = \ \frac{12}{5} \ \sin \Theta_5 \ \ \approx \ 2.4 \ \sin (252.29º) \ \ . $$

If we look at the negative-radius solution for $ \ \Theta_4 \ $ , we find

$$ \ x_4 \ = \ -\frac{12}{5} \ \cos \Theta_4 \ \approx \ -2.4 \ \cos (192.29º) \ \approx \ 2.34 \ \ , $$ $$ y_4 \ = \ -\frac{12}{5} \ \sin \Theta_4 \ \ \approx \ -2.4 \ \sin (192.29º) \ \approx \ 0.51 \ \ ; $$

the other two negative-radius solutions have the correspondence $ \ \Theta_2 \ \rightarrow \ \Theta_5 \ $ and $ \ \Theta_6 \ \rightarrow \ \Theta_3 \ $ .

I am not clear as to whether these are the points being referred to in part (c) -- that they are apparent additional intersections arising from the analysis of the functions, but are redundant on the graph -- or whether it is referring to the origin, in which case the answer is just $ \ ( 0, \ 0) \ $ . As mentioned before, we would solve

$$ r_1 \ = \ 4 \ \sin \ 3\theta \ = \ 0 \ \ \Rightarrow \ \ \theta \ = \ \frac{n \ \pi}{3} \ \ , $$ $$ r_2 \ = \ 3 \ \cos \ 3\theta \ = \ 0 \ \ \Rightarrow \ \ \theta \ = \ \frac{(2n+1) \ \pi}{6} \ \ , $$

but we would find that there are no values of $ \ \theta \ $ which solve both curve equations simultaneously. While both rosettes pass through the origin, they do not do so at the same values of the angle-parameter. Below is a graph of the two rosettes.

[pic to be added shortly]

Note that while we said the values of $ \ \Theta_{2,4,6} \ $ lie in Quadrants I, III, and IV, respective, the points these correspond to fall into diametrically-opposite positions in Quadrants III, I, and II.

$\endgroup$
0
$\begingroup$

Divide first equation by the second and expand triple angle of tan.

$$ \dfrac {3t-t^3}{1-3t^2} = \dfrac{3}{4} $$

$$ 4t^3 -9t^2-12t+3=0$$

Can be solved numerically. Three real roots, one real root near $12^0$ and two others $120^0 $ apart due to same periodicity of both roses.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .