Help me with my HW

[HAULN-SS]

I need to prove that the segment PT is the same length as the arc PR..stumped

the spiral has equation r = a * theta

[Ryan T]

Keith feel free to chime in here anytime, mr math major...

[Mike]

rofl for bullshit questions!

[HAULN-SS]

oops..correction..OT is equal to the lenght of the arc PR...now it's supposed to be easier

[Dave]

whut the debil is that!?!?

[HAULN-SS]

It's a spiral, part of a circle, a tangent line to the point P at the spiral, a line that is drawn from the origin to P, and a line perpindicular to that line

[damnit458]

Keith feel free to chime in here anytime, mr math major...

LOL I have no idea. What class is this for?

Come to me when you have questions on Calculus, Differential Equations, Linear Algebra, or Numerical Analysis

[BLINGMW]

oops..correction..OT is equal to the lenght of the arc PR...now it's supposed to be easier

THAT'S easier? At least there IS a line segment PT. There isn't even an "O" on the damn thing! In conclusion, I think the answer is 5.

[.RJ]

look at the funny lines :lol:

[Dragon]

oops..correction..OT is equal to the lenght of the arc PR...now it's supposed to be easier

THAT'S easier? At least there IS a line segment PT. There isn't even an "O" on the damn thing! In conclusion, I think the answer is 5.

5?!?!?! are you kidding, the answer is always C

[Feersty]

Yeah man there ain't no O, so where do you get OT?

[JohnC]

42, bitches.

[damnit458]

Yeah man there ain't no O, so where do you get OT?

He probably means the Origin for O...

Edit: Nvm the arc thing. Does she give you any more information?

[HAULN-SS]

O for origini...i must've left that off the picture inadvertently...i finally got a solution..i'll post it later today when I have time to write it up

[HAULN-SS]

Ok..I know you all have been holding your breaths wondering..is that really a true statement? Well, I present a proof. Any one who wants to read proof and/or proof read..notify me if I need corrections.


I will use "o" to mean theta and "a" to mean alpha, seeing as I dont have a greek keyboard.


let P have polar coordinates (r, theta) = (ao, o). Since x = r*cos(o) and y = r * sin(o) then we can substitute in with r = ao.

P = (ao*cos(o), ao*sin(o))

So the vector OP = (aocos(o), aosin(o)). But OP is perpindicular to the vector OT. So OT = (-aosin(o), aocos(o))
The Slope of OP is x/y = aosin(o)/aocos(o) = sin(o)/cos(o) cancelling out the ao's on top and bottom. Thus OT must have slope -cos(o)/sin(o)

(this is the negative reciprocal)

also, we now know the slop of the line k, so writing it in point slope form we can get y = mx = -cos(o)/sin(o) * x => sin(o)y + cos(o)x = 0

Set this up in parametric form with k = { x = -aosin(o)t, y = aocos(o)t}

Since the line i is tangent at the point P, where r = ao, the slope of i is the derivative at the point theta.

so take the derivative of x and y to get x' = acos(o) - aosin(o) and y' = asin(o) + aocos(o)

Put these in parametric form for the line i

i = { x = x_0 + x'(o)s, y = y_0 + y'(o)s} where x_0, y_0 are the points at P.

Set these equations for x(s) and x(t) equal AND y(s) and y(t) equal since were are looking for an intersection of two lines.

-aosin(o)t = aocos(o) + (acos(o)-ao(sin(o))s
AND
aocos(o)t = aosin(o) + (asin(o) + aocos(o))s


By inspection (or solving a system of 2 equations with 2 unknowns s and t, which archimedes wouldnt have known how to do), T = (ao^2sin(o), -ao^2cos(o)) so s = o and t = -o

Now to find the length of OT. This is the Square root of the squares of the components of T.

After much simplification this = a(o^2)

To find the arc length of PR, we take circumference times the ratio of the arc angle to 2pi.

PR = 2pi*r x o/2pi. Notice the 2pi's cancel so we get PR = ao * o = a(o^2) = OT. We have precisely proven that if T is the intersection of i and k, then arcPR = length of OT. Q.E.D.

[ScottyB]

well, i made it through the first few sentences, but then i blacked out and fell out of my office chair.

[HAULN-SS]

THe vey last sentence is the only one that counts anyway. Anyone remember what Q.E.D means? It's latin for "it is shown" i think..but i dont know what the actual words are

[HAULN-SS]

Quad Erat Demonstratum - "what was to be shown"

found it