Parametrics and Polar curves (1 Viewer)

leehuan · Apr 24, 2016

Starting with a dumb question because I think my lecturer forgot to explicitly say how to.

I know what the graph looks like already
$r=6\sin{\theta}$

But how do I convert this to Cartesian form lol

InteGrand · Apr 24, 2016

leehuan said:
Starting with a dumb question because I think my lecturer forgot to explicitly say how to.

I know what the graph looks like already
$r=6\sin{\theta}$

But how do I convert this to Cartesian form lol

That's a circle with diameter 6 on the y-axis. Centre is (0, 3), radius is 3.

To derive this, you could replace sin(theta) with y/r, so r = a*(y/r) (in this case a = 6), so r^2 = ay, that is, x^2 + y^2 = ay. Now just moves things around and complete the square to get the circle's Cartesian equation.

We assumed r was non-zero above, since we had r in a denominator. For the r = 0 case, we can get r = 0 at theta = 0 in the polar equation. So the Cartesian point (0, 0) (i.e. the place where r = 0) is part of the curve.

leehuan · Apr 24, 2016

Oh fair enough, sin(theta)=y/r was what I needed to see.
___________________________________

$r=1+\sin{\theta}}$ represents a cardiod. It has a cusp at (0,0)

$r=2+\sin{\theta}}$ doesn't look like it has a cusp though.

Is there any way to deduce by inspection of the equation what's going on? So far I've been relying on my lecturer (for the former) and a graphing software.

InteGrand · Apr 24, 2016

Say the curve is r = a + sin(theta) = f(theta), say.

Now, y = r.sin(theta) = f(theta).sin(theta) and similarly x = f(theta).cos(theta)

Now, dy/d(theta) = f'(theta).sin(theta) + f(theta).cos(theta) = cos(theta).sin(theta) + (a + sin(theta)).cos(theta) = a.cos(theta) + sin(2*theta).

Also, dx/d(theta) = f'(theta).cos(theta) - f(theta).sin(theta) = cos(theta).cos(theta) - (a + sin(theta)).sin(theta) = cos(2*theta) - a.sin(theta).

Now, dy/dx = (dy/d(theta))/(dx/d(theta)) = [a.cos(theta) + sin(2*theta)]/[cos(2*theta) - a.sin(theta)].

At theta = -pi/2, we see that dy/dx is perfectly well-defined if a =/= 1, as it equals [a*0 + 0]/[-1 + a] = 0. So we have a horizontal slope when theta = pi/2 for a =/= 1.

If a = 1, the denominator is 0 at theta = -pi/2 however and we need further investigation, and it turns out we end up with a cusp.

To find where dy/dx may be undefined for arbitrary given a, we can set the denominator of dy/dx to 0 and try and solve for sin(theta) (it'll be a quadratic in sin(theta)) and hence theta.

If you solve the quadratic we get, the solutions should be

sin(theta) = [-a +/- sqrt(a^2 + 8)]/4. For real solutions, we'd need the RHS to be between -1 and 1. You can investigate for which a this will happen. You'll find that at least one of these will always be between -1 and 1 for any real a. These'll usually give places where the tangents are vertical. You can try finding for which a it is possible to have both numerator and denominator of dy/dx be 0 simultaneously. You should find this can only happen if a = +/- 1. For a cusp, we'd need both numerator and denominator to be 0 at a point theta₀, and at least one of these (x'(theta) and y'(theta)) should change sign here, which is what happens at -pi/2 for a = +/- 1 cases.

leehuan · Jun 3, 2016

Are there any techniques you could employ to sketch this weird parametric curve?

$\\x=\sin{t}\sin{2t}\\ y=\cos{t}\sin{2t}$

Cause I couldn't see anything obvious and was basically gonna bombard it with a table of values

InteGrand · Jun 3, 2016

leehuan said:
Are there any techniques you could employ to sketch this weird parametric curve?

$\\x=\sin{t}\sin{2t}\\ y=\cos{t}\sin{2t}$

Cause I couldn't see anything obvious and was basically gonna bombard it with a table of values

$\noindent One thing to note is that if we write $\theta = t$, then squaring and adding, $x^2 + y^2 = \sin^2 2\theta \Rightarrow r \equiv f(\theta)= \left |\sin 2\theta\right |$. This may be a more familiar (polar) curve for you. Anyway, note $f(\theta) = f(-\theta)$, so the curve is symmetric about $\theta = 0$ (the $x$-axis), and $f(\pi - \theta) = f(\theta)$, so the curve is symmetric about $\theta =\frac{\pi}{2}$ (the $y$-axis). In fact $f\left(\frac{\pi}{2}-\theta \right) = f(\theta)$, so the curve is also symmetric about the line $\theta = \frac{\pi}{4}$ (the 45-degree line $y=x$), and since $f\left(-\frac{\pi}{2}-\theta \right)=f(\theta)$, it's also symmetric about the line $\theta = -\frac{\pi}{2}$, i.e. the line $y=-x$. So you can use this symmetry to help sketch the curve.$

These symmetries mean if you were to use a table of values, you'd only need to do so for theta between 0 and pi/4. This gets the image for one 'octant' ('first' octant), and then to complete the image in the first quadrant, reflect it about he 45 degree line y = x. Then reflect this first-quadrant image about the x- and y-axes to get the fourth and second quadrants respectively. Then finally another reflection gives a similar image in the third quadrant.

seanieg89 · Jun 3, 2016

leehuan said:
Are there any techniques you could employ to sketch this weird parametric curve?

$\\x=\sin{t}\sin{2t}\\ y=\cos{t}\sin{2t}$

Cause I couldn't see anything obvious and was basically gonna bombard it with a table of values

tan(t)=x/y, use this and trig manipulations to eliminate the t in one of the equations if you want to get a cartesian form.

seanieg89 · Jun 3, 2016

(Personally I think it is easier to plot in polar form).

Parametrics and Polar curves (1 Viewer)

leehuan

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