What is the tangential angle for a point at some distance down

>>8468759
is it even defined?

what is it on a circle?

>>8468759
wat

>>8468759
Tangential angle is always zero. That's what "tangent" means.

I suspect OP means what is the slope of the tangential line at a point some distance down the Archimedes spiral...

>>8468796
The slope of the curve at some distance

>>8468800
the slope of polar r = theta:
dy/dx = [theta cos(theta) + sin(theta)]/[-theta sin(theta) + cos(theta)]

All you ausitsic fags cant understand a basic question.

OP is asking that if you were to travel a specific distance along an archemedes spiral from the center starting point, what would the slope of the line you followed be at the moment you stop?

AKA What is the slope per distance traveled.

[math]r=b \theta[/math]
[math]x^2+y^2 =b^2 \theta ^2[/math]
[math] x^2+y^2 =b^2 \arctan(\frac{y}{x}) ^2 [/math]

Now just implicite differetiate respect to [math]x[/math].

>>8468820
Yes this is what im trying to ask
>>8468819
I don't want it in terms of arc length

>>8468831
>I don't want it in terms of arc length
Good thing I gave it to you in terms of theta, not arc length.

>>8468838
Am I retarded? I want to know the slope from the arc length

>>8468844

if you have theta you can easily find the tangent.

>>8468851
except I don't have theta

>>8468856
Apparently you are retarded.
Every arc length determines a unique theta. Just solve for it

You can use the the formula for length of a polar graph to find the arc length in terms of a change in angles, and perhaps invert this to solve for a corresponding angle. In the case of the spiral it would be the integral from angle intial to angle final of b*root(theta^2 + 1). I dont know if theres a closed solution to this integral, and it is even less likely that it is invertible if it is a closed solution, but if somehow you could get theta from that you could subtract it by 2pi until you got its branch on the 0 to 2pi interval and then do the usual stuff.

>>8468861
There's two components to this problem: Arc length and slope at some point on the cuve

there's no theta and there's no polar coordinates

I don't give a fuck i can solve for theta

Since I'm too dumb to ask a basic question:

I have some arc length x
what is the slope at that point

>>8468861
Actually you are the one who is retarded, because you cant invert arc length for theta in a spiral.
http://mathworld.wolfram.com/ArchimedesSpiral.html

Parametrize the spiral as a function of theta in cartesian coordinates.
r(theta) = b*theta*(cos(theta), sin(theta))
take the derivative with respect to theta, then take the norm. You are left with b*theta.
Integrate b*theta with respect from 0 to some arbitary theta. You are left with the formula
s(theta) = b*(theta)^2 / 2
So, if s is arc length, then theta = root(2s/b)
Now you can find the the tangent line.

>>8468896
Dont listen to this I just realized my mistake I didn't use the product rule when I differentiated r. I have no idea if theres a solution to this but I think this way is barking up the wrong tree, sorry bub.

File: torus_top_view.gif (13KB, 450x450px) Image search: [Google]

13KB, 450x450px

I've been quite interested in it for some time, can you guys offer me any books to start with?

>>8468759
Much like how area is only defined in our cartesian system of graphing, we must define slope in terms of dy and dx. Unfortunately, the polar coordinates equation r = a, where r is the radius, and a is the angle, is multiple valued for a function f(x,y). To compensate for this, we will need to add branches to arctangent. x = r*cos(a), y = r*sin(a), so x = a*cos(a) and y = a*sin(a), and arctan(y/x) = a s.t. -pi/2<a<pi/2. To compensate for this, we will need to define g(x,y) = arctan(y/x) + n*pi, where n = floor([sqrt(x^2 + y^2)+pi/2]/pi]). So, by implicit differentiation, we get d/dx [g(x,y)]^2 = d/dx[x^2 + y^2], 2g(x,y)*[-y/(y^2 + x^2) + d/dx n*pi] = 2x + 2ydy/dx. But, as long as a is not of the form (2k-1)*pi/2, where k is a positive integer, then n*pi, is constant in a sufficiently small neighborhood, and d/dx n*pi = 0, so dy/dx = -g(x,y)*/(x^2 + y^2) - x/y = dy/dx. Clearly, x = 0 and y = 0 cases are undefined, so we will need to treat them as special cases. As long as y is not 0, then dy/dx is the slope m, and we can use the point slope formula to find the tangent line.

>>8468960
*multiple valued, as in choosing x does not give a unique y, and choosing y does not give a unique x.

>>8468960
doesn't answer the question
see: >>8468880

>>8468976
>x = a*cos(a), y = a*sin(a)
> -g(x,y)*/(x^2 + y^2) - x/y = dy/dx
you can do it

>>8468759
that picture is fucked an angle is referenced from two straight lines, not one curvature and a line

>>8468984
No, because you cant solve the cartesian coordinates from the arc length, you buffoon.

>>8468829
It might be easier to take the derivative in polar and then convert to cartesian

it's zero

I think it is theta=cos^-1((cos(a2)-a2*sin(a2))/(a2+1))-cos^-1((cos(a1)-a1*sin(a1))/(a1+1)) where a1 and a2 are the two points in question on the curve

if you want d(theta)/ds=(-1/(a+1))-(cos(a)/((1+a)(sin(a)+a*cos(a))) for the curvature at a