Question

For the polar curve r=3cos(2theta) find the locations of all horizontal and vertical tangents

Answer 1

The issue was that this one really blew up in my face when I found dy/dx, and am having a hard time finding where it equals zero/is undefined

Answer 2

@amistre64

Answer 3

r=3cos(2t)

rr=3 rcos(2t)

x^2 + y^2 =3 r(cos^2(t)-sin^2(t))

x^2 + y^2 =3 (rcos^2(t)-rsin^2(t))

x^2 + y^2 =3 (x cos(t)-y sin(t))

2x x' + 2y y' =3 (x' cos(t)-xt' sin(t)-y' sin(t)-yt' cos(t))
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dy/dx..... let x=1, y' = dy/dx

2x + 2y y' =3cos(t) -3xt' sin(t) -3y' sin(t) -3yt' cos(t))

3y' sin(t) + 2y y' =3cos(t) -3xt' sin(t) -3yt' cos(t)) - 2x

y'(3sin(t) + 2y) =3cos(t) -3xt' sin(t) -3yt' cos(t)) - 2x

y' =(3cos(t) -3xt' sin(t) -3yt' cos(t)) - 2x)/(3sin(t) + 2y)
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when the top is zero we are tangent horizontal

when the bottom is zero we are tangent vertical
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bottom seems simplest

3sin(t) + 2y = 0

3sin(t) + 2r sin(t) = 0

3sin(t) + 6cos(2t)sin(t) = 0

3sin(t) (1+ 2cos(2t)) = 0

sin(t) = 0 or cos(2t) = -1/2
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top part, lets give it a whirl

3cos(t) -3xt' sin(t) -3yt' cos(t) - 2x = 0

3cos(t) -3rt' cos(t)sin(t) -3rt' sin(t) cos(t) - 2r cos(t) = 0

cos(t) (1 -6t' cos(2t) sin(t) - 2cos(2t)) = 0

cos(t) = 0 is one spot

1 -6t' cos(2t) sin(t) - 2cos(2t) = 0

this one might be trickiest :)

Answer 4

You got a different dy/dx than I did

Answer 5

http://www.wolframalpha.com/input/?i=polar+plot+r%3D3cos%282t%29

the pictures nice :)

Answer 6

I found dy/dtheta and dx/dtheta by using x=rcostheta and y=rsintheta. Then I substituted r into each equation and got x=(3cos(2theta))costheta and y=(3cos(2theta))sintheta

Answer 7

Then I said dy/dx=(dy/dtheta)(dx/dtheta), and I derived my x and y equations. Which exploded...

Answer 8

dy = dy dt
dx dt dx

Answer 9

Does t=theta?

Answer 10

of course

Answer 11

Oh yeah, I meant (dy/dtheta)OVER(dx/dtheta) :P Sorry

Answer 12

But yeah, it got really complicated

Answer 13

we can evaluate it from 0 to pi/4 and the rest of the graph is copies

Answer 14

Evaluate what?

Answer 15

the graph .... i posted what it looks like if you didnt take the time to sketch it

Answer 16

Yeah, I know what it looks like. I have to prove the tangent lines though.
For my derivative, I got dy/dx=\[\frac{ 3\cos(2\theta)\cos(\theta)+\sin(\theta)(-6\sin(2\theta)) }{ -3\cos(2\theta)\sin(\theta)-6\sin(2\theta)\cos(\theta)}\]

Answer 17

Which looks different than what you got

Answer 18

yeah i just went for the straight forward implicit derivative
\[\frac{dy}{dx} =\frac{3cos(t) -3xt' sin(t) -3yt' cos(t) - 2x}{3sin(t) + 2y}\]

Answer 19

Did we get the same thing though?

Answer 20

dunno :) what do you get for your verticals?

Answer 21

Thats the problem. I was trying to find the tangents and I didn't really know how to do that because of all of the trig functions

Answer 22

when the slope is undefined, its vertical

a fraction is undefined when its denom is zero

3sin(t) + 2y = 0, but y = r cos(t) = 3cos(2t) sin(t)

3sin(t) (1 + 2cos(2t)) = 0 its a product, so its zero when either of its factors or 0

sin(t) = 0, or cos(2t) = -1/2

Answer 23

i cant verify that your derivative is good ....

Answer 24

Can we use my derivative? That is the way that my teacher would rather see

Answer 25

post all of your work to get there, and i can follow it

Answer 26

Ok

Answer 27

r=3cos(2theta)

Answer 28

x=rcos(theta)=(3cos(2theta))cos(theta)
y=rsin(theta)=(3cos(2theta))sin(theta)

Answer 29

\[\frac{ dy }{ dx }=\frac{ 3\cos(2\theta)\cos(\theta)+\sin(\theta)(-3\sin(2\theta))2 }{ (3\cos(2\theta))(-\sin(\theta))+\cos(\theta)(-3\sin(2\theta))2 }\]

Answer 30

And I simplified that a little bit.

Answer 31

dy/dt = -6sin(2t) sin(t) + 3cos(2t) cos(t) = 0 give horizontal

dx/dt = -6sin(2t)cos(t) - 3cos(2t)sin(t) = 0 give vertical

agreed?

Answer 32

Yes

Answer 33

Exactly

Answer 34

And I wasn't really sure how to get the zeros of those

Answer 35

well dunno if itll simplfy it but

cos(2t) = cos^2(t) - sin^2(t)
=1 -2sin^2(t)
=2 cos^2(t) - 1

sin(2t) = 2 sin(t)cos(t)

Answer 36

I tried doing some stuff with those identities, but it just messed it up. I tried simplifying to get cot(2theta)=2tan(theta)

Answer 37

But that doesn't really seem to work either

Answer 38

-6sin(2t)cos(t) - 3cos(2t)sin(t)

-12sin(t)cos(t)cos(t) - 3(2cos^2(2) - 1)sin(t)

-12sin(t)cos^2(t) - 6sin(t)cos^2(t) +3sin(t)

-18sin(t)cos^2(t) +3sin(t)

3sin(t)(-6cos^2(t) +1) = 0

sin(t) = 0, and 1/6 = cos^2(t); cos(t) = sqrt(6)/6

might be some errors in that along the way ....

Answer 39

Well its not cos(theta)=root6/6

Answer 40

oh wait. I forgot that the sides of the petals have tangents too

Answer 41

I see! Thank you!!

Answer 42

youre welcome

Answer 43

and yes, cos(t) = rt6/6 is a zero on the bottom