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inkyvoyd
 4 years ago
Polar functions and derivatives:
1. show that the cardioid r=1+sin(theta) has a horizontal tangent line when r=1/2
inkyvoyd
 4 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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inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0For some reason I don't think I'm doing this right...

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0I 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?

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0@amistre64 , got any ideas what I"m donig wrong?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2you 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0the puals website is a great resource indeed

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0Oh I c, I'm supposed to convert it into a cartesian equation?

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

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

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

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2^^  that is basically the summary result

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2so you just need to calculate dr/d(theta)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh look! paul saves the day again. whew.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2plug the result in the equation above, set it to zero and solve

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353710144960:dw

inkyvoyd
 4 years ago
Best 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?

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0corrction, th=pi*(any integer)

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