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inkyvoyd

  • 2 years ago

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

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  1. inkyvoyd
    • 2 years ago
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    For some reason I don't think I'm doing this right...

  2. inkyvoyd
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    @amistre64 , got any ideas what I"m donig wrong?

  5. asnaseer
    • 2 years ago
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    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
    • 2 years ago
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    the puals website is a great resource indeed

  7. inkyvoyd
    • 2 years ago
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    Oh I c, I'm supposed to convert it into a cartesian equation?

  8. inkyvoyd
    • 2 years ago
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    The question is, why can't I work with it while it's in polar form?

  9. asnaseer
    • 2 years ago
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    no

  10. asnaseer
    • 2 years ago
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    no need to convert - look at notes in the link I gave and all will become apparent :)

  11. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    ^^ - that is basically the summary result

  13. asnaseer
    • 2 years ago
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    so you just need to calculate dr/d(theta)

  14. satellite73
    • 2 years ago
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    oh look! paul saves the day again. whew.

  15. asnaseer
    • 2 years ago
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    :)

  16. inkyvoyd
    • 2 years ago
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    Oh wow - thanks!

  17. asnaseer
    • 2 years ago
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    plug the result in the equation above, set it to zero and solve

  18. asnaseer
    • 2 years ago
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    there is one slight /twist/ in the solution - hopefully you will be able to see it :)

  19. mahmit2012
    • 2 years ago
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    |dw:1353710144960:dw|

  20. inkyvoyd
    • 2 years ago
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    @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
    • 2 years ago
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    corrction, th=pi*(any integer)

  22. asnaseer
    • 2 years ago
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    I calculated the derivative as:\[\frac{dy}{dx}=\frac{\cos(\theta)(2\sin(\theta)+1)}{1-\sin(\theta)-2\sin^2(\theta)}=\frac{\cos(\theta)(2\sin(\theta)+1)}{(1+\sin(\theta))(1-2\sin(\theta))}\]

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