Finding the vertical and horizontal tangents to the polar curve r=θ?

[tsp216]

So there's this curve in polar coordinates, with:

. . . . .\(\displaystyle r\, =\, \theta\, \mbox{ for }\, 0\, \leq\, \theta\, \leq\, 2\pi\)

I need to find the \(\displaystyle \,\theta\,\) values for all of its vertical and horizontal tangents. So I've got:

. . . . .\(\displaystyle x\, =\, r\cos(\theta)\, =\, \theta\, \cos(\theta)\)

. . . . .\(\displaystyle y\, =\, \theta\, \sin(\theta)y\, =\, \theta\, \sin(\theta)\)

which gives:

. . . . .\(\displaystyle \dfrac{dx}{d\theta}\, =\, \cos(\theta)\, -\, \theta\, \sin(\theta)\)

. . . . .\(\displaystyle \dfrac{dy}{d\theta}\, =\, \sin(\theta)\, +\, \theta\, \cos(\theta)\)

So the slope should be:

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \sin(\theta)\, +\, \theta\, \cos(\theta)\, \cos(\theta)\, -\, \theta\, \sin(\theta)\)

So for the tangent to be horizontal I need:

. . . . .\(\displaystyle \sin(\theta)\, +\, \theta\, \cos(\theta)\, =\, 0\, \sin(\theta)\, +\, \theta\, \cos(\theta)\, =\, 0\)

But I can't see how to solve this, except by just seeing that it works when \(\displaystyle \, \theta\, =\, 0,\,\) but I can't see how to solve this, except just by seeing that it works when \(\displaystyle \, \theta\, =\, 0.\,\) As for the vertical tangent, I'm not sure how to find \(\displaystyle \, \theta\, \) so that \(\displaystyle \, \cos(\theta)\, -\, \theta\, \sin(\theta)\, =\, 0\)

I've tried asking this question on StackExchange - Mathematics (here), but haven't really received any helpful replies.

 

[stapel]

So there's this curve in polar coordinates, with:

. . . . .\(\displaystyle r\, =\, \theta\, \mbox{ for }\, 0\, \leq\, \theta\, \leq\, 2\pi\)

I need to find the \(\displaystyle \,\theta\,\) values for all of its vertical and horizontal tangents. So I've got:

. . . . .\(\displaystyle x\, =\, r\cos(\theta)\, =\, \theta\, \cos(\theta)\)

. . . . .\(\displaystyle y\, =\, \theta\, \sin(\theta)y\, =\, \theta\, \sin(\theta)\)

Is the last line above perhaps meant to be as follows?

. . . . .\(\displaystyle y\, =\, r\, \sin(\theta)\, =\, \theta\, \sin(\theta)\)

which gives:

. . . . .\(\displaystyle \dfrac{dx}{d\theta}\, =\, \cos(\theta)\, -\, \theta\, \sin(\theta)\)

. . . . .\(\displaystyle \dfrac{dy}{d\theta}\, =\, \sin(\theta)\, +\, \theta\, \cos(\theta)\)

Assuming the correction mentioned above, I agree with your derivatives.

So the slope should be:

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \sin(\theta)\, +\, \theta\, \cos(\theta)\, \cos(\theta)\, -\, \theta\, \sin(\theta)\)

The derivative, dy/dx, should have been a fraction, as is explained and illustrated here. How did you arrive at your expression? What were your steps?

So for the tangent to be horizontal I need:

. . . . .\(\displaystyle \sin(\theta)\, +\, \theta\, \cos(\theta)\, =\, 0\, \sin(\theta)\, +\, \theta\, \cos(\theta)\, =\, 0\)

How did you arrive at this expression? Why is the sine multiplied by zero? How did you get this equation from your derivative expression?

Please reply showing all of your steps. Thank you!

 

[Deleted member 4993]

So there's this curve in polar coordinates, with:

. . . . .\(\displaystyle r\, =\, \theta\, \mbox{ for }\, 0\, \leq\, \theta\, \leq\, 2\pi\)

I need to find the \(\displaystyle \,\theta\,\) values for all of its vertical and horizontal tangents. So I've got:

. . . . .\(\displaystyle x\, =\, r\cos(\theta)\, =\, \theta\, \cos(\theta)\)

. . . . .\(\displaystyle y\, =\, \theta\, \sin(\theta)y\, =\, \theta\, \sin(\theta)\)

which gives:

. . . . .\(\displaystyle \dfrac{dx}{d\theta}\, =\, \cos(\theta)\, -\, \theta\, \sin(\theta)\)

. . . . .\(\displaystyle \dfrac{dy}{d\theta}\, =\, \sin(\theta)\, +\, \theta\, \cos(\theta)\)

So the slope should be:

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \sin(\theta)\, +\, \theta\, \cos(\theta)\, \cos(\theta)\, -\, \theta\, \sin(\theta)\)

So for the tangent to be horizontal I need:

. . . . .\(\displaystyle \sin(\theta)\, +\, \theta\, \cos(\theta)\, =\, 0\, \sin(\theta)\, +\, \theta\, \cos(\theta)\, =\, 0\)

But I can't see how to solve this, except by just seeing that it works when \(\displaystyle \, \theta\, =\, 0,\,\) but I can't see how to solve this, except just by seeing that it works when \(\displaystyle \, \theta\, =\, 0.\,\) As for the vertical tangent, I'm not sure how to find \(\displaystyle \, \theta\, \) so that \(\displaystyle \, \cos(\theta)\, -\, \theta\, \sin(\theta)\, =\, 0\)

I've tried asking this question on StackExchange - Mathematics (here), but haven't really received any helpful replies.

Sin(Θ) + Θ*cos(Θ) = 0

cannot be solved (for Θ) by known algebraic manipulations.

Using numerical method the (approximate) solutions are:

Θ = 0, 2.028758, 4.91318 ​radians

 

[tsp216]

@stapel Yeah, sorry about the obscure resulting equations; I copied and pasted them from stackexchange.com so the formatting was messed up. Indeed my initial results were exactly the same as yours. Sorry for the delayed reply as well. I had to use Wolfram Alpha to find the solutions for both the numerator and denominator when they're = 0, and so am still unsure if there is actually a way to find an exact solution. As Subhotosh Khan's reply above indicates, maybe there isn't a way to find an exact derivative for each of these equations. Thnx again for the reply.