I Need help in this and how to convert rectangular coordinates to polar coordinates and vice versa. my teacher vaguely described how to graph also. Also finding other polar representations of any point.
for example: find an equivalent way of writing (2, pi/4)?
finding polar representations of (2,pi/3?
convert (-2, 5pi/4) to rectangular coordinates.
convert (0,2) to polar coordinates, and r=5costheta to rec. coordinates.
whats the rules of that if r is negative suppose to do?
Polar coordinates are of the form $\displaystyle (r,\;\theta)$, where $\displaystyle r$ is the directed distance (i.e., it can be positive or negative) between the point and the origin, and $\displaystyle \theta$ is the angle measured counterclockwise from the polar axis to the radial line containing the point. For example, the point (-1, 0) in rectangular coordinates would be $\displaystyle (1,\;\pi)$ in polar since the distance from the origin is 1 and the angle it makes with the axis is $\displaystyle \pi$.
Look at the graph.
Two points have been plotted. For now, look at $\displaystyle \left(3,\;\frac{\pi}4\right)$. The $\displaystyle r=3$ means that the point is 3 units from the origin, and the $\displaystyle \theta=\frac{\pi}4$ means that the line intersecting the point and origin makes an angle of $\displaystyle \frac{\pi}4 = 45^\circ$ with the polar axis.
So how would we convert this to rectangular coordinates? Notice that I have drawn a perpendicular from the point to the axis. This gives us a right triangle. You can see that the two legs of the triangle are of length $\displaystyle x$ and $\displaystyle y$ (the distance from the $\displaystyle y$- and $\displaystyle x$-axis respectively), and the hypotenuse is $\displaystyle r$. Thus, by the Pythagorean theorem, we have
$\displaystyle r^2 = x^2 + y^2$
Notice also that we can take the tangent of $\displaystyle \theta$ to get
$\displaystyle \tan\theta = \frac yx$
These two formulas make it pretty easy to convert from rectangular to polar coordinates. What about from polar to rectangular? Well, taking the sine and cosine of the angle, we get
$\displaystyle \sin\theta = \frac yr\Rightarrow y = r\sin\theta$
and
$\displaystyle \cos\theta = \frac xr\Rightarrow x = r\cos\theta$
These four formulas should be all you need for conversions. For our point $\displaystyle \left(3,\;\frac{\pi}4\right)$, we have $\displaystyle r = 3$ and $\displaystyle \theta = \frac{\pi}4$. Thus
$\displaystyle x = r\cos\theta = 3\cos\left(\frac{\pi}4\right) = \frac{3\sqrt2}2$
$\displaystyle y = r\sin\theta = 3\sin\left(\frac{\pi}4\right) = \frac{3\sqrt2}2$
and our point is $\displaystyle \left(\frac{3\sqrt2}2,\;\frac{3\sqrt2}2\right)$ in rectangular coordinates.
Now, notice that $\displaystyle \theta$ is not unique. For example, an angle of $\displaystyle 15^\circ$ is really the same as $\displaystyle 375^\circ$. Also, notice that we can make $\displaystyle r$ negative, and as long as we flip the angle around, we get the same point (look at the other point on the graph). So, any point $\displaystyle (r,\;\theta),\;r\neq0$ can be written as
$\displaystyle (r,\;\theta + 2n\pi),\;n\in\mathbb{Z}$
or
$\displaystyle \left(-r,\;\theta + (2n - 1)\pi\right),\;n\in\mathbb{Z}$
When graphing, you can do things in two ways. Either convert the polar equations to rectangular ones, or just try to plot some points directly. For an example of a polar graph of a curve, this is a graph of a curve called a limaçon (its equation is of the form $\displaystyle r = a + b\cos\theta$).
Does that help?
Find another angle that is the same as $\displaystyle \frac{\pi}4$.
See my post above.
Hint: multiply both sides by $\displaystyle r$ to get $\displaystyle r^2 = 5r\cos\theta = 5\left(r\cos\theta\right)$.
$\displaystyle r$ is the directed distance between the point and the origin. If $\displaystyle r$ is positive, it indicates that $\displaystyle (r,\;\theta)$ is $\displaystyle r$ units from the origin at the angle $\displaystyle \theta$, while if $\displaystyle r$ is negative, it means that the point is $\displaystyle \lvert r\rvert = -r$ units from the origin, at the angle opposite of $\displaystyle \theta$. So $\displaystyle (2,\;0^\circ) = (-2,\;180^\circ)$.