## inkyvoyd Group Title Polar functions and derivatives: 1. show that the cardioid r=1+sin(theta) has a horizontal tangent line when r=1/2 one year ago one year ago

1. inkyvoyd Group Title

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

2. inkyvoyd Group Title

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?

3. inkyvoyd Group Title

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

4. inkyvoyd Group Title

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

5. asnaseer Group Title

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

6. amistre64 Group Title

the puals website is a great resource indeed

7. inkyvoyd Group Title

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

8. inkyvoyd Group Title

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

9. asnaseer Group Title

no

10. asnaseer Group Title

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

11. satellite73 Group Title

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

12. asnaseer Group Title

^^ - that is basically the summary result

13. asnaseer Group Title

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

14. satellite73 Group Title

oh look! paul saves the day again. whew.

15. asnaseer Group Title

:)

16. inkyvoyd Group Title

Oh wow - thanks!

17. asnaseer Group Title

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

18. asnaseer Group Title

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

19. mahmit2012 Group Title

|dw:1353710144960:dw|

20. inkyvoyd Group Title

@asnaseer , for my discussion I also need to determine when the function has verticle tangent lines. The only problem is that I'm getting expression (2sin th+1)(cos th)/(-2sin^2 th) for dy/dx - meaning that I should have a vertical tangent line when -2sin^2 th=0, but solving for that equation gives me sin th=pi(any integer), which is obviously incorrect. What am I doing incorrectly?

21. inkyvoyd Group Title

corrction, th=pi*(any integer)

22. asnaseer Group Title

I calculated the derivative as:$\frac{dy}{dx}=\frac{\cos(\theta)(2\sin(\theta)+1)}{1-\sin(\theta)-2\sin^2(\theta)}=\frac{\cos(\theta)(2\sin(\theta)+1)}{(1+\sin(\theta))(1-2\sin(\theta))}$