• May 20th 2005, 08:02 AM
zee
Question) The polar curve C is given by the polar equation r=1+cos(theta)
Find the length of C.

Answer) This is what I have so far:
r = 1 + cos(theta)

Since L = INT(sqrt[1 + (dr/d(theta))^2])d(theta)

Therefore,
L = int(sqrt[1+sin^2(theta)])d(theta) (with integral from 0 to 2)

zee
• May 20th 2005, 11:32 PM
ticbol
No.
In polar coordinates,
L = sqrt[r^2 +(dr/dT)^2]dT -------***

It is not
L = sqrt[1 +(dr/dT)^2]dT
You are confused with
L = sqrt[1 +(dy/dx)^2]dx
in cartesian coordinates.

------
r = 1 +cosT
dr/dT = -sinT

If the limits are from T=0 to T=2 radians,
L = INT.(0->2)[sqrt{(1+cosT)^2 +(-sinT)^2}]dT
L = INT.(0->2)[sqrt{(1 +2cosT +cos^2(T)) +sin^2(T)}]dT
L = INT.(0->2)[sqrt{(1 +2cosT +{cos^2(T) +sin^2(T)}}]dT
L = INT.(0->2)[sqrt{1 +2cosT +1}]dT
L = INT.(0->2)[sqrt{2 +2cosT}]dT ---(1)

The "dT" outside of the bracket is not the derivative of the sqrt{2 +2cosT} inside the bracket.

We need to express the sqrt{2 +2cosT} into a one-degree function.
See if we can convert the (2 +2cosT) into a square of cos(nT) or sin(nT).

cosT
= cos(T/2 +T/2)
= cos^2(T/2) -sin^2(T/2)
= cos^2(T/2) -[1 -cos^2(T/2)]
= 2cos^2(T/2) -1 ---***

Substitute that into (1),
L = INT.(0->2)[sqrt{2 +2cosT}]dT ---(1)
L = INT.(0->2)[sqrt{2 +2(2cos^2(T/2) -1)}]dT
L = INT.(0->2)[sqrt{2 +4cos^2(T/2) -2)}]dT
L = INT.(0->2)[sqrt{4cos^2(T/2)}]dT
L = INT.(0->2)[2cos(T/2)]dT

Now we manipulate the "dT" outside.
L = INT.(0->2)[2cos(T/2)](2dT/2)
L = INT.(0->2)[2cos(T/2)](dT/2)(2)
L = 2*2*INT.(0->2)[cos(T/2)](dT/2)
L = 4*INT.(0->2)[cos(T/2)](dT/2)

Now we can integrate,
L = 4[sin(T/2)](0->2)
L = 4[sin(2/2) -sin(0/2)]
L = 4[sin(1) -sin(0)]
L = 4[0.84147 -0]