Calculus Question: Tangents on Polar Coordinate of Polar Curve r=cos(theta)+sin(theta)

Question

How do I find the polar coordinates of the points on the polar curve r=cos(theta)+sin(theta), 0(greater than or equal to)(theta)(less than or equal to)(pi), where the tangent line is horizontal or vertical?

 

I know that I need to convert the coordinates to x & y and then take the derivative of the two (dy/dx) but can't seem to figure it out.

 

Using the chain rule I was able to figure out dx/d(theta)=2(-sin(theta))+cos^2(theta)-sin^2(theta)  and  dy/d(theta)=2cos(theta)+cos^2(theta)-sin^2(theta) but have no idea if I am on the correct path or what I need to do next. I thought about using the double angle trig identity but am not sure if it applies in this situation. It has been awhile since I had to utilize trig.

 

Any help would be awesome. Thank you.

Answer 1

i would start here:

 

r = cos(theta) + sin(theta) ; multiply both sides by r

r^2 = r[cos(theta) + sin(theta)] = r*cos(theta) + r*sin(theta)

 

we know r^2 = x^2 + y^2; r*cos(theta) = x; r*sin(theta) = y; substitute

 

x^2 + y^2 = x+y      from here we derivitize

D[x^2 + y^2] = D[x+y]

2x + 2y*y' = 1 + y'    solve for y'

2y*y' - y' = 1-2x

y'*(2y-1) = 1-2x

y' = (1-2x)/(2y-1)

 

let y' = 0 for horizontal tangent

1-2x = 0

x = 1/2

 

let y' = undefined at vertical tangent

2y-1 = 0

y = 1/2

 

using these values, you should be able to get back into polar coordinates (keep in mind the question is only looking for solutions where theta is between 0 and 2pi)