Fortunately I found this:
http://en.wikipedia.org/wiki/Polar_coor … l_calculus
The slope of the line would then be dy/dx, and the angle said line makes with the x-axis would be atan( dy/dx ). Right?
The angle the tangent makes with the horizontal is tan[sup]−1[/sup](dy/dx).
]]>thats how i work it out.
]]>Anyway, I'll cut to the chase. A spiral is easy to describe in polar coordinates. It has the form r = aθ / 2π , where a is the amount by which r increases in each complete revolution. It's also not hard to map this equation to cartesian coordinates:
x = r*cos(θ)
y = r*sin(θ)
Here's the goofy problem: how do I find the (cartesian) slope of the line tangent to the curve at a given value of r and θ? I've tried implicit differentiation, like so:
dr/dθ = 1 / 2π
dx/dθ = (dr/dθ)*(-sin(θ))
dx/dθ = -sin(θ)/2π
dy/dθ = (dr/dθ)*(cos(θ))
= cos(θ)/2π
The slope of the line would then be dy/dx, and the angle said line makes with the x-axis would be atan( dy/dx ). Right? Am I getting this right? Because the angles produced from these expressions seem a bit off.
Thanks!
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