Convert the polar equation to rectangular coordinates: r = 2-cos theta
...and find all points of vertical and horizontal tangency.
To convert the polar equation \(r = 2 - \cos(\theta)\) to rectangular coordinates, we can use the following conversions:
\(x = r \cdot \cos(\theta)\)
\(y = r \cdot \sin(\theta)\)
Let's substitute these expressions into the given equation:
\(x = (2 - \cos(\theta)) \cdot \cos(\theta)\)
\(y = (2 - \cos(\theta)) \cdot \sin(\theta)\)
Now we can find the points of vertical and horizontal tangency by determining when the derivative of \(y\) with respect to \(x\) equals zero.
Differentiating both equations with respect to \(\theta\), we get:
\(\frac{dx}{d\theta} = \frac{d}{d\theta}((2 - \cos(\theta)) \cdot \cos(\theta))\)
\(\frac{dy}{d\theta} = \frac{d}{d\theta}((2 - \cos(\theta)) \cdot \sin(\theta))\)
Using the product rule and chain rule, we can simplify the derivatives:
\(\frac{dx}{d\theta} = \sin(\theta) - (\sin(\theta) \cdot \cos(\theta))\)
\(\frac{dy}{d\theta} = \cos(\theta) - (\sin(\theta) \cdot \cos(\theta))\)
To find the points of tangency, we need to set \(\frac{dy}{dx}\) equal to zero and solve for \(\theta\):
\(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\cos(\theta) - (\sin(\theta) \cdot \cos(\theta))}{\sin(\theta) - (\sin(\theta) \cdot \cos(\theta))} = 0\)
Simplifying the equation above, we can obtain:
\(\cos(\theta) - (\sin(\theta) \cdot \cos(\theta)) = 0\)
Now we solve this equation to find the values of \(\theta\) that correspond to points of vertical and horizontal tangency.
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)
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.