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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
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

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inkyvoydBest ResponseYou've already chosen the best response.0
For some reason I don't think I'm doing this right...
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
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?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
I tried to turn sin(theta)=1/2 into secant theta, but the answer I'm getting is cos(theta)=pm 2/(sqrt 3)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
@amistre64 , got any ideas what I"m donig wrong?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the puals website is a great resource indeed
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
Oh I c, I'm supposed to convert it into a cartesian equation?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
The question is, why can't I work with it while it's in polar form?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
no need to convert  look at notes in the link I gave and all will become apparent :)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
i could be wrong, but i think it is \[\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)r\sin(\theta)}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
^^  that is basically the summary result
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
so you just need to calculate dr/d(theta)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
oh look! paul saves the day again. whew.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
plug the result in the equation above, set it to zero and solve
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
there is one slight /twist/ in the solution  hopefully you will be able to see it :)
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1353710144960:dw
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
@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?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
corrction, th=pi*(any integer)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
I calculated the derivative as:\[\frac{dy}{dx}=\frac{\cos(\theta)(2\sin(\theta)+1)}{1\sin(\theta)2\sin^2(\theta)}=\frac{\cos(\theta)(2\sin(\theta)+1)}{(1+\sin(\theta))(12\sin(\theta))}\]
 one year ago
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