Question

Horizontal and Vertical tangent lines. Parametric Eq.

Answer 1

\[x=\cos(\theta),y=\cos(3\theta)\]
So horizontal tangent line is when dy/dx=0, correct?

Answer 2

whoops, when dy/dtheta = 0

Answer 3

Hope I'm not second-guessing you by offering advice, but yes: If your derivative dy/dx is zero, you have a horiz. tan. line. Likewise if your deriv. dy/dx is undefined, you have a vert. tan. line.

I understand theta is your parameter here, but would still suggest that you go all the way to find dy/dx, not dy/ d(theta).

Answer 4

how do I do it with dy/dx? I know how to find dy/dx but not how to find what is vertical or horizontal then @mathmale

Answer 5

\[\frac{ dy }{ dx }=\frac{ -3\sin 3 \theta }{ -\sin \theta }\]

Answer 6

First, I'll assume your result is correct, and that our purpose right now is to determine where we have one or more horiz. tan. lines and/or where we have one or more vert. tan. lines. I'm not checking your work up until now.

When working with parametric equations for x and y,

Answer 7

\[\frac{ dy }{ dx }=\frac{ \frac{ dy }{ d \theta } }{ \frac{ dx }{ d \theta } }\]

Answer 8

Horiz. asy.: Let the numerator = 0 (ignore the den.) and solve for theta..

Vert. asy.: Let the denom = 0 (ignore the numerator) and solve for theta.

Answer 9

Okay :) let me see

Answer 10

Once you've found the relevant theta values, stick those into the original equations for x and y to find points (x,y) at which you have horiz. tan. line(s) and at which you have vert. tan lines.

Answer 11

\[-3\sin3\theta=0\]\[\sin3\theta=0\] now what do I do xD

Answer 12

@jim_thompson5910 :)

Answer 13

@Kainui

Answer 14

Well if you're solving for when \(\sin( 3 \theta)=0\) you can use the fact that you know:
\[\sin(0)=0\]\[\sin(\pi)=0\]
(I stopped before going to \(2 \pi \) since it's one complete revolution which is the same as 0.)

So now you can see that by comparison of the inside parts, this must be true:

\[3\theta = 0\]\[3\theta=\pi\]

Answer 15

Just like a quadratic has two solutions, so does \(\theta\), just wanted to clarify that each of those equations are different conditions now.

Answer 16

@Kainui, sorry! I left the pc for a while. I am back now.

Answer 17

So now that I have those two values, do I need to find a corresponding point? I guess there would be two points.

Answer 18

\[\theta=0, \pi/3\]