Question

Polar functions and derivatives:
1. show that the cardioid r=1+sin(theta) has a horizontal tangent line when r=1/2

Answer 1

For some reason I don't think I'm doing this right...

Answer 2

I calculated d(theta)/dr implicitly, getting
1=0+cos(theta)d(theta)/dr
sec(theta)=d(theta)/dr
Then I tried to express theta in terms of r, so I used the equation and substituted r=1/2
1/2=1+sin(theta)
sin(theta)=-1/2
Now what?

Answer 3

I tried to turn sin(theta)=-1/2 into secant theta, but the answer I'm getting is cos(theta)=pm 2/(sqrt 3)

Answer 4

@amistre64 , got any ideas what I"m donig wrong?

Answer 5

you need to work out dy/dx and set it to zero.
see here for a good guide on how to do this: http://tutorial.math.lamar.edu/Classes/CalcII/PolarTangents.aspx

Answer 6

the puals website is a great resource indeed

Answer 7

Oh I c, I'm supposed to convert it into a cartesian equation?

Answer 8

The question is, why can't I work with it while it's in polar form?

Answer 9

no

Answer 10

no need to convert - look at notes in the link I gave and all will become apparent :)

Answer 11

i could be wrong, but i think it is
\[\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}\]

Answer 12

^^ - that is basically the summary result

Answer 13

so you just need to calculate dr/d(theta)

Answer 14

oh look! paul saves the day again. whew.

Answer 15

:)

Answer 16

Oh wow - thanks!

Answer 17

plug the result in the equation above, set it to zero and solve

Answer 18

there is one slight /twist/ in the solution - hopefully you will be able to see it :)