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inkyvoyd
Polar functions and derivatives: 1. show that the cardioid r=1+sin(theta) has a horizontal tangent line when r=1/2
For some reason I don't think I'm doing this right...
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?
I tried to turn sin(theta)=-1/2 into secant theta, but the answer I'm getting is cos(theta)=pm 2/(sqrt 3)
@amistre64 , got any ideas what I"m donig wrong?
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
the puals website is a great resource indeed
Oh I c, I'm supposed to convert it into a cartesian equation?
The question is, why can't I work with it while it's in polar form?
no need to convert - look at notes in the link I gave and all will become apparent :)
i could be wrong, but i think it is \[\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}\]
^^ - that is basically the summary result
so you just need to calculate dr/d(theta)
oh look! paul saves the day again. whew.
plug the result in the equation above, set it to zero and solve
there is one slight /twist/ in the solution - hopefully you will be able to see it :)
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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?
corrction, th=pi*(any integer)
I calculated the derivative as:\[\frac{dy}{dx}=\frac{\cos(\theta)(2\sin(\theta)+1)}{1-\sin(\theta)-2\sin^2(\theta)}=\frac{\cos(\theta)(2\sin(\theta)+1)}{(1+\sin(\theta))(1-2\sin(\theta))}\]