Derivative of a spiral

ryos · 2007-03-02 21:26:52

Hey guys! Long time no post. I only come back when I've got a goofy problem. Sorry about that.

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!

luca-deltodesco · 2007-03-02 21:47:34

thats how i work it out.

JaneFairfax · 2007-03-03 03:41:24

ryos wrote:

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).

Last edited by JaneFairfax (2007-03-03 04:24:40)

John E. Franklin · 2007-03-03 03:59:13

luca-dd said: wrote:

Last edited by John E. Franklin (2007-03-03 04:00:47)

JaneFairfax · 2007-03-03 04:26:59

Good point, John! I completely missed that myself.

Fortunately I found this:
http://en.wikipedia.org/wiki/Polar_coor … l_calculus

ryos · 2007-03-03 07:20:18

D'oh! I forgot the chain rule. Thanks Jane.

Math Is Fun Forum

#1 2007-03-02 21:26:52

Derivative of a spiral

#2 2007-03-02 21:47:34

Re: Derivative of a spiral

#3 2007-03-03 03:41:24

Re: Derivative of a spiral

#4 2007-03-03 03:59:13

Re: Derivative of a spiral

#5 2007-03-03 04:26:59

Re: Derivative of a spiral

#6 2007-03-03 07:20:18

Re: Derivative of a spiral

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