1. ## Parametric tangent lines

$x = cos(t) +2cos(2t)$
$y = sin(t) +2sin(2t)$

$dx = -4sin(2t)-sin(t)$
$dy = 4cos(2t)-cos(t)$

I have found the crossing point to be at (-2,0) by looking at my graphing calc, is there a more precise way?

How do i find the tangent lines to this point now?

Thanks

2. You're question doesn't really makes sense to me as the 2 equations

are one curve -- that's the idea of a parametrically defined curve.

Could you send the original question?

What is actually crossing Here?

A general answer to slope for a parametrically defined curve is:

dy/dx = (dy/dt)/(dx/dt)

3. Originally Posted by Calculus26
You're question doesn't really makes sense to me as the 2 equations

are one curve -- that's the idea of a parametrically defined curve.

Could you send the original question?

What is actually crossing Here?

A general answer to slope for a parametrically defined curve is:

dy/dx = (dy/dt)/(dx/dt)

ImageShack - Image Hosting :: 53923477.jpg

4. Ok now it makes sense the curve crosses itself.

to get the tangent lines

you need to find the times when it crosses itself

t = arccos(-1/4) which is the same as pi -arcos(1/4) and

also the 3d quadrant angle pi +arcos(1/4)

use dy/dx = dy/dt/dx/dt as mentioned before

To solve this I got the pt(-2,0) as you did graphically

Set cos(t) + 2cos(2*t) =-2

cos(t) +4cos^2(t) -2 = -2

cos(t)(1+4cos(t)) = 0

cos(t) = 0 yields pi/2,3pi/2

the second term gives

cos(t) = 1/4 t = arccos(-1/4) whiwhich yields the same results I mentioned above

plug in the values in y to see which ones yield 0 0r even easier

Now I'll let you do the same thing with y(t) = sin(t) +2sin(2t) = 0

5. Hello, silencecloak!

This is a long a tedious problem . . .

$x \:= \cos t +2\cos2t$
$y \:=\: \sin t +2\sin2t$

$\frac{dx}{dt} \:=\: \text{-}\sin t -4\sin2t$
$\frac{dy}{dt} \:=\: \cos t + 4\cos2t$

I have found the crossing point to be at (-2,0) by looking at my graphing calc.
Is there a more precise way? . . . . I don't know

How do i find the tangent lines to this point now?
Assuming that (-2,0) is indeed the intersection, we can work backwards.

Since $x = -2$, we have: . $\cos t + 2\cos2t \:=\:-2 \quad\Rightarrow\quad \cos t + 2(2\cos^2t-1) \:=\:-2$

. . $\cos t + 4\cos^2\!t - 2 \:=\:-2 \quad\Rightarrow\quad \cos t + 4\cos^2\!t \:=\:0 \quad\Rightarrow\quad \cos t(1 + 4\cos t) \:=\:0$

And we have: . $\begin{array}{cccc}\cos t \:=\:0 & \Rightarrow & t \:=\:\pm\frac{\pi}{2} \\
1+4\cos t \:=\:0 & \Rightarrow & t \:=\:\arccos\left(\text{-}\frac{1}{4}\right) \end{array}$

Since $y = 0$, we have: . $\sin t + 2\sin2t \:=\:0 \quad\Rightarrow\quad\sin t + 2(2\sin t\cos t) \:=\:0$

. . $\sin t + 4\sin t\cos t \:=\:0 \quad\Rightarrow\quad \sin t(1 + 4\cos t) \:=\:0$

And we have: . $\begin{array}{cccc}\sin t \:=\:0 & \Rightarrow & t \:=\:0,\pi \\
1 + 4\cos t \:=\:0 & \Rightarrow & t \:=\:\arccos\left(\text{-}\frac{1}{4}\right) \end{array}$

The only root common to both sets is: . $t \:=\:\arccos\left(\text{-}\tfrac{1}{4}\right)$

We can work out the trig values for $t$ and $2t$.

. . We have: . $t \:=\:\arccos\left(\text{-}\tfrac{1}{4}\right) \quad\Rightarrow\quad \boxed{\cos t \:=\:-\frac{1}{4}}$

. . $\sin t \:=\:\pm\sqrt{1-\cos^2\!t} \:=\:\pm\sqrt{1-\left(\text{-}\tfrac{1}{4}\right)^2} \:=\:\pm\sqrt{\tfrac{15}{16}} \quad\Rightarrow\quad\boxed{ \sin t\:=\:\pm\frac{\sqrt{15}}{4}}$

. . $\sin2t \:=\:2\sin t\cos t \:=\:2\left(\pm\tfrac{\sqrt{15}}{4}\right)\left(\t ext{-}\tfrac{1}{4}\right) \quad\Rightarrow\quad \boxed{\sin2t \:=\:\mp\frac{\sqrt{15}}{8}}$

. . $\cos2t \:=\:2\cos^2\!t - 1\:=\:2\left(\text{-}\tfrac{1}{4}\right)^2-1 \quad\Rightarrow\quad \boxed{\cos2t \:=\:-\frac{7}{8}}$

$\frac{dy}{dx} \:=\:\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \;=\;\frac{\cos t + 4\cos2t} {\text{-}\sin t - 4\sin 2t}$

Substitute the values we found above and get the two slopes of the tangents.

Then you can write the equations of the two tangents through (-2, 0).