Question

clarification,

Find the points on the curve where the tangent is horizontal or vertical.
x = cos theta
y = cos 3 theta

so i know it's horizontal when dy/dtheta = 0
and i've found the points to be (1/2, -1) and (-1/2, 1)
how would i determine when its vertical? dx/d theta = 0?
the solutions manual says there is no vertical so i'm not sure..

Answer 1

You're working with "parametric equations."

If I remember correctly, the slope of a tangent line in such a situation is given by \[\frac{ dy }{ dx }=\frac{ \frac{ dy }{ d \theta } }{ \frac{ dx }{ d \theta } }\]

Answer 2

where theta is the parameter in question.

Please check this out in your textbook or online.

This is not to say that your (dy / d theta) is wrong.

Answer 3

Your tangent line would be vertical wherever the denominator (dx / d theta) is zero, since division by zero is undefined.

Answer 4

I'd suggest that you find dy/dx as shown above in an equation involving the parameter theta. Reduce your result as much as possible. If it has a denominator, can that denom. every be zero?

Answer 5

it would come to be the same thing would it not?
dy/d theta = -3 sin 3 theta
dx/d theta = -sin theta

-3 sin 3 theta/- sin theta = 0
sin 3 theta = 0
at theta = pi/3, 2pi/3, -pi/3, -2pi/3

and for the vertical part
dx/d theta = - sin theta
theta = pi,2pi
but my solutions manual says there is no vertical tangent

Answer 6

I see y our point.

Would it make a positive difference were you to expand sin (3 theta) into terms that include sin theta and then reduce the ensuing quotient?

Answer 7

\[\sin 3 \theta = \sin (\theta + 2 \theta)=?\] Might work, might not, but perhaps worth a try.

Answer 8

ah i hadn't thought of that, would make more sense if it were reduced that way and therefore there were no verticals at all..

Answer 9

i came up with 3sin theta - 4sin^2 theta
and yeah taking out a sin theta reduces the quotient to one :)
my book went all the way on this one lol