Polar Tangent Lines: Finding Slopes at the Pole

[j9mom]

Homework Statement

r=2-3cosθ Find the tangent line at any point, and at the point (2,∏) Find the tangent line(s) at the pole

Homework Equations

Do I have to use x=rcosθ and y=rsinθ to convert it to rectangular to find slopes?

The Attempt at a Solution

Is the point 2∏ even a point in the graph. It is a limicon (sp?) graph?

 

[clamtrox]

You can find the tangent in any coordinate system. You can also convert into rectangular coordinates but it will probably require a lot more work.

 

[HallsofIvy]

"The point 2\pi" doesn't make any sense. Was that a typo for (2, \pi), the given point? When \theta= \pi, [/itex]r= 2- 3cos(\pi)= 5[/itex] so, no, that point is not on the graph. However, r= 2- 3cos(\pi/2)= 2 so perhaps that was what was meant. Alternatively, perhaps the problem intended r= 2- 3sin(\theta).

The slope of the tangent line at any point is, by definition, dy/dx. Since, in polar coordinates, x= r cos(\theta) and y= r sin(\theta), we have dx= dr cos(\theta)- r sin(\theta)d\theta and dy= dr sin(\theta)d\theta+ r cos(\theta) d\theta so that
\frac{dy}{dx}= \frac{cos(\theta) dr- r sin(\theta)d\theta}{sin(\theta)dr+ r cos(\theta)d\theta}= \frac{cos(\theta)\frac{dr}{d\theta}- r sin(\theta)}{sin(\theta)\frac{dr}{d\theta}+ r cos(\theta)}
That can be simplified.