Parametric graphing and calculus

**TenaciousE** · August 28th 2012, 03:09 PM

Problem:
x = cos(t) + 2cos(2t)
y = sin(t) + 2sin(2t)

graph the curve to discover where it crosses itself and find both tangent lines at that point.

I understand how to find the tangents after I know the point I'm looking for, but I don't know how to relate the two equations to make the curve graphable. I could use Mathematica to help me graph it, but I won't have access to that on my exam. I'm not necessarily looking for someone to give me the answer, but pointing me in the right direction would be nice.

**Prove It** · August 28th 2012, 06:41 PM

Originally Posted by TenaciousE

Problem:
x = cos(t) + 2cos(2t)
y = sin(t) + 2sin(2t)

graph the curve to discover where it crosses itself and find both tangent lines at that point.

I understand how to find the tangents after I know the point I'm looking for, but I don't know how to relate the two equations to make the curve graphable. I could use Mathematica to help me graph it, but I won't have access to that on my exam. I'm not necessarily looking for someone to give me the answer, but pointing me in the right direction would be nice.

This isn't going to be pretty...

$\displaystyle \begin{align*} x &= \cos{t} + 2\cos{2t} \\ y &= \sin{t} + 2\sin{2t} \\ \\ x &= \cos{t} + 2\left( 2\cos^2{t} - 1 \right) \\ y &= \sin{t} + 4\sin{t}\cos{t} \\ \\ x &= 4\cos^2{t} + \cos{t} - 2 \\ y &= \sin{t} \left( 1 + 4\cos{t} \right) \\ \\ \frac{x}{4} &= \cos^2{t} + \frac{1}{4}\cos{t} - \frac{1}{2} \\ y &= \pm \sqrt{ 1 - \cos^2{t} } \left( 1 + 4\cos{t} \right) \\ \\ \frac{x}{4} &= \cos^2{t} + \frac{1}{4}\cos{t} + \left(\frac{1}{8}\right)^2 - \left(\frac{1}{8}\right)^2 - \frac{1}{2} \\ y &= \pm \sqrt{ 1 - \cos^2{t} } \left( 1 + 4\cos{t} \right) \\ \\ \frac{x}{4} &= \left( \cos{t} + \frac{1}{8} \right)^2 - \frac{1}{64} - \frac{32}{64} \\ y &= \pm \sqrt{ 1 - \cos^2{t} }\left( 1 + 4\cos{t} \right) \\ \\ \frac{16x}{64} &= \left( \cos{t} + \frac{1}{8} \right)^2 - \frac{33}{64} \\ y &= \pm \sqrt{ 1 - \cos^2{t} } \left( 1 + 4\cos{t} \right) \end{align*}$

$\displaystyle \begin{align*} \frac{16x + 33}{64} &= \left( \cos{t} + \frac{1}{8} \right)^2 \\ y &= \pm \sqrt{ 1 - \cos^2{t} }\left( 1 + 4\cos{t} \right) \\ \\ \frac{\pm \sqrt{ 16x + 33 }}{8} &= \cos{t} + \frac{1}{8} \\ y &= \pm \sqrt{ 1 - \cos^2{t} }\left( 1 + 4\cos{t} \right) \\ \\ \frac{-1 \pm \sqrt{ 16x + 33 }}{8} &= \cos{t} \\ y &= \pm \sqrt{ 1 - \cos^2{t} } \left(1 + 4\cos{t}\right) \end{align*}$

And now you can substitute $\displaystyle \begin{align*} \cos{t} \end{align*}$ into the second equation.

**Soroban** · August 28th 2012, 10:05 PM

Hello, TenaciousE!

I too tried to eliminate the parameter
. . and got into an awful mess.

$\begin{Bmatrix}x &=& \cos t + 2\cos2t & [1] \\ y &=& \sin t + 2\sin2t & [2] \end{Bmatrix}$

$\text{Graph the curve to discover where it crosses itself}$
. . $\text{and find both tangent lines at that point.}$

$\begin{array}{cccccccc}\text{Square [1]:} & x^2 &=& \cos^2\!t + 4\cos2t\cos t + 4\cos^2\!2t \\ \text{Square [2]:} & y^2 &=& \sin^2\!t + 4\sin2t\sin t + 4\sin^2\!2t \end{array}$

Add . . . and recall that $\sin^2\!\theta + \cos^2\!\theta \:=\:1$

We have: . $x^2 + y^2 \:=\:1 + 4(\cos2t\cos t + \sin2t\sin t) + 4$

n . . . . . . . $x^2 + y^2 \:=\:5 + 4\cos t$

Hence: . $\cos t \:=\:\frac{x^2+y^2 - 5}{4}\;\;[3]$

Then: . $\cos2t \:=\:2\cos^2t - 1 \:=\:2\left(\frac{x^2+y^2-5}{4}\right)^2 - 1$

n . . . . $\cos2t \:=\:\frac{(x^2+y^2-5)^2-8}{8}\;\;[4]$

Substitute [3] and [4] into [1]: . $x \;=\;\frac{x^2+y^2-5}{4} + 2\left(\frac{(x^2+y^2-5)^2 - 8}{8}\right)$

And you can try to simplify that . . .

**Bingk** · August 29th 2012, 03:49 AM

I'm not sure about this, but I think the problem was designed for you to practice graphing functions in parametric form (that's why it's not easy to convert it, and that's why the problem asks you to graph it to "discover" where it crosses itself). I think you're meant to sketch the graph to get an idea of where it crosses itself (i.e. find distinct values t1 and t2 so that x(t1)=x(t2) and y(t1)=y(t2)), and get the tangents at those points.

Just some ideas that might help ...

x(t) = cos(t) + 2cos(2t) is basically a wave with a certain length (i.e. if it's wavelength is L, then x(t) = x(t+L))
y(t) will also have a wavelength, which may be different from that of x(t).

If they're different, their common wavelength would be the lcm.

**skeeter** · August 29th 2012, 10:33 AM

Originally Posted by TenaciousE

Problem:
x = cos(t) + 2cos(2t)
y = sin(t) + 2sin(2t)

graph the curve to discover where it crosses itself and find both tangent lines at that point.

as per the directions ...

**TenaciousE** · August 29th 2012, 01:51 PM

I finally found the tangent lines to be y= -(15)^(1/2)(x+2) and y=15^(1/2)(x+2)
I wasn't able to successfully eliminate the parameter, so I eventually used a calculator to graph the curve and I found that the graph crossed itself where y =0, so I was able to find the slopes by looking at the graph.
I feel like a bit of a cheater because I used a graphing calculator, but I simply could not follow the math required to solve it with algebra. Thanks for all your help though. It got me thinking in the right direction.

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