Calculus II - Calculus in Polar Coordinates

[GreenPrint]

Homework Statement

Find the points at which the following polar curves have a horizontal and vertical tangent line.

(a) r = 3 + 6 cos(theta)

Homework Equations

The Attempt at a Solution

x = r cos(theta) = (3 + 6 cos(theta)))cos(theta) = 3cos(theta) + 6 cos(theta)^2
y = rsin(theta) = (3 + 6cos(theta) )sin(theta) = 3 sin(theta) + 6 cos(theta)sin(theta)

dy/dtheta = 3 cos(theta) + 6[cos(theta)^2 - sin(theta)^2]
dx/dtheta = -3 sin(theta) - 12 cos(theta)sin(theta)

dy/dx = - [cos(theta) + 2[cos(theta)^2 - sin(theta)^2]]/[sin(theta) - 4cos(theta)sin(theta)]

for vertical tangent
sin(theta) = 4 cos(theta)sin(theta)
1 = 4 cos(theta)
theta = arccos(1/4)

for horizontal tangent
cos(theta) + 2cos(theta)^2 = 2sin(theta)^2 = 2(1-cos(theta)^2) = 2 - 2cos(theta)^2
cos(theta) + 2cos(theta)^2 + 2cos(theta)^2 - 2 = 0
4cos(theta)^2 + cos(theta) - 2 = 0
cos(theta) = (-1 +/- sqrt(1 - 4(-2)(4) ))/(2(4)) = (-1 +/- sqrt(33) )/8
theta = arccos( (-1 +/- sqrt(33) )/8 )

I don't see what I'm doing wrong, the answer key says that the vertical tangent occurs at theta = arcsin(1/4) and the horizontal tangents occur at arcsin( (-1 +/- sqrt(33) )/8 )

Thanks for any help!

 

[dynamicsolo]

Are you sure that is the answer that goes with this problem? This is the equation of a limacon with an interior loop, so there should be four vertical tangents (the solver appears to have discarded the same factor you did). I agree with your calculations otherwise as far as I've checked. I wonder if they did something like r = 3 + 6 \sin \theta...

 

[GreenPrint]

Interesting... I don't see were I discarded a factor. Were did I do so. Ya, that's what the answer key says, it was made by someone so it's possible it's wrong. huh... I just checked and for some reason like you have suggested the person who made the answer key was doing r = 3 + 6 sin(theta) despite the question being r = 3 + sin(theta)

 

[dynamicsolo]

dx/dtheta = -3 sin(theta) - 12 cos(theta)sin(theta)

...

for vertical tangent
sin(theta) = 4 cos(theta)sin(theta)
1 = 4 cos(theta)
theta = arccos(1/4)

To find the vertical and horizontal tangents, you only need to set dx/dt or dy/dt , respectively, individually to zero. But when working with an equation involving a sum or difference like this,

3 \sin \theta + 12 \cos \theta \sin \theta = 0 ,

it is better to factor it and find where the individual factors equal zero, hence,

\sin \theta ( 1 + 4 \cos \theta) = 0 ,

implying \sin \theta = 0 \Rightarrow \theta = 0 or \pi and

4 \cos \theta = -1 for the other two angles .

(Ah, I see the other problem: you dropped a minus sign when you went to

"sin(theta) = 4 cos(theta)sin(theta)
1 = 4 cos(theta)" .)

In any case, when you divide through by a factor in an equation, it is important to make sure that factor cannot be zero (which it can be for this curve). In removing that factor, you are throwing away other possible solutions. That is why there are four vertical tangents to be found, rather than just two. (I'm a bit surprised the solver didn't consider this, but you can inspect the polar curve for yourself...)

 

[GreenPrint]

So for vertical tangents that still gives me only 3 points
pi/2, (3pi)/4, arccos(1/4)
the fourth point is from the fact that arccos(1/4) has two solutions? I thought we always ignored the other one were arccos(theta) was defined from 0 to pi and we ignored the values when solving equations like
cos(theta) = 1/4
but we don't ignore the other values if it was just arccos(1/4) or something rather

 

[dynamicsolo]

In dealing with a result like cos(theta) = -1/4 for polar curves, you are asking what angles have -1/4 as their cosine, rather than being limited to the range of the inverse cosine function. So there are two points there.

The other two are from sin(theta) = 0 , for which we get theta = 0 or pi . Thus, there are four vertical tangents. (But you don't have to take my word for it: plot the curve!)

 

[GreenPrint]

Hm interesting, thank you very much, that makes sense.