1. ## r=2cos theta graph

Could you please tell me how to plot r=2cos theta ? Please show me the step of plotting this graph. Thanking in advance.

2. Originally Posted by kittycat
Could you please tell me how to plot r=2cos theta ? Please show me the step of plotting this graph. Thanking in advance.
Decide on a set of $\theta$ values, say,
$\theta = 0, \pi/4, \pi/2, 3\pi/4, pi, 5pi/4, 3\pi/2, 7\pi/4$ rad

Then calculate the r value for each $\theta$. They are
$r = 2, \sqrt{2}, 0, -sqrt{2}, -2, -\sqrt{2}, 0, \sqrt{2}$ (respectively)

What do you do with negative r values? Well, you make the r value positive and then add (or subtract) $\pi$ rad to the angle.

So we wish to plot the set of points:
$(r, \theta)$
$(2, 0)$
$(\sqrt{2}, \pi/4)$
$(0, \pi/2)$
$(\sqrt{2}, 5\pi/4)$
$(2, 0)$ <-- $\pi + \pi = 2\pi \to 0$
$(\sqrt{2}, \pi/4)$
$(0, 3\pi/2)$
$(\sqrt{2}, 7\pi/2)$

Then you plot the points on polar coordinate graph paper and sketch in a smooth curve between the points. I get the graph below.

-Dan

3. Originally Posted by kittycat
Could you please tell me how to plot r=2cos theta ? Please show me the step of plotting this graph. Thanking in advance.
Or you could do it the hard way:
$r = \sqrt{x^2 + y^2}$

$cos(\theta) = \frac{x}{\sqrt{x^2 + y^2}}$

$\sqrt{x^2 + y^2} = 2 \frac{x}{\sqrt{x^2 + y^2}}$

Multiplying both sides by $\sqrt{x^2 + y^2}$ gives:
$x^2 + y^2 = 2x$

$(x^2 - 2x) + y^2 = 0$

$(x^2 - 2x + 1) + y^2 = 1$

$(x - 1)^2 + y^2 = 1$

which is a circle centered on (1, 0) with radius 1.

-Dan

4. Sorry, i saw you used polar functions, my apologies. I'll delete my posts now.

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# circle of r - 2cos

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