Tangent lines on the Golden Spiral

[Jdo300]

Hello all, I'm working on a diagramming program that uses the golden spiral to make templates for something I'm working on. I'm wondering if someone can tell me how to get the equation of the tangent line at point (x,y) on the spiral. My program graphs the spiral parametrically and I use these equations:

X = Cos (T) * Phi ^ ((2 / π) * T)
Y = Sin (T) * Phi ^ ((2 / π) * T)

I've uploaded a sample picture to illustrate the spiral graph my program creates. I added a picture of a tangent and a point to show what I'm trying to find. So far, the graphing part of my program works just fine but I'm not quite sure how to do the tangents. I heard that you can use derivatives to do it but I haven't taken Calculus yet so I have no clue how to do them yet. Any help would be greatly appreciated.

Thanks,
Jason O

 

[Gokul43201]

This will be more meaningful if you had 1/2Pi instead of 2/Pi. Then, 'T' would represent the angle in the x-y plane.

Anyways, you'll need to use calculus (or have someone do it for you) to find the tangent at any point. This is what you get :

The slope of the tangent at some point, P(T) is = {cos(T) + (2/Pi)sin(T)}/{(2/Pi)cos(T) - sin(T)}
You can use this slope and the co-ordinates of the point (x,y) given by your parametric equations to get the equation of the tangent line at that point.

If you are not allowed to use a calculus result, there is a reasonable approximation that may be used. Pick 2 nearby points on the curve, on either side of the given point, and find the slope of the line joining these points. That will be pretty close to the slope of the tangent.

NOTE : (1) Pick points that are equidistant to the required point,
(2) The distance between them should be small compared to the value of a^(2T/Pi)
(3) The distance between them should be sufficiently large compared to the resolution of the display or the underlying XY coordinate frame.
(4) Obeying (2) and (3) simultaneously requires that you avoid points with very small values of T

 

[Jdo300]

Thank you very much for your help! As it turns out, I was able to get the derivatives for the X and Y equations using a TI-89 calculator that my Calculus teacher had. I have one more question that I'm not sure about. If I have a point (point X) and it is on the Golden spiral, how would I calculate the next point on the spiral that is a certain distance away from the first point? Since I am using parametric equations, I thought about just incrementing T but I see that when I plot the points, they increase in distance from each other. Is there a formula I could use to do this if I know the X, Y coordinates of the first point on the spiral and the distance I know I want the second point to be from the first?