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
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)
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
Hello, silencecloak!
This is a long a tedious problem . . .
Assuming that (-2,0) is indeed the intersection, we can work backwards.
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?
Since , we have: .
. .
And we have: .
Since , we have: .
. .
And we have: .
The only root common to both sets is: .
We can work out the trig values for and .
. . We have: .
. .
. .
. .
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).