we know x^2 + y^2 = r^2
and cosØ = x/r
so, ...
r = 2-cos theta
r = 2 - x/r
r^2 = 2 - x
x^2 + y^2 = 2-x
2x + 2y d/dx = -1
2y dy/dx = -1 - 2x
dy/dx = (-1-2x)/(2y
= 0 for a horizontal tangent
-1 - 2x = 0
x = -1/2
and subbing back into x^2 + y^2 = 2-x
1 + y^2 = 2 + 1
y^2 = 2
y = ± √2
there are 2 horizontal tangents
at ((-1/2) , √2) and (-1/2 , - √2)
for a vertical tangents, dy/dx must be undefined,
or 2y = 0
y = 0
then x^2 + 0 = 2 - x
x^2 + x - 2 = 0
(x+2)(x-1) = 0
x = -2 or x = 1
two vertical tangents, at
(-2,0) and (1,0)
Convert the polar equation to rectangular coordinates: r = 2-cos theta
...and find all points of vertical and horizontal tangency.
2 answers
Ahem. I think there's an unfortunate typo in there.
r = 2-cosθ
r^2 = 2r-rcosθ
x^2+y^2 = 2√(x^2+y^2) - x
This complicates things a bit, as there are more vertical tangents. The curve is, after all, (almost) a limaçon, not a circle.
r = 2-cosθ
r^2 = 2r-rcosθ
x^2+y^2 = 2√(x^2+y^2) - x
This complicates things a bit, as there are more vertical tangents. The curve is, after all, (almost) a limaçon, not a circle.
1 answer