Find the arc length: (2cost, 2sint, t) for 0

Question

Find the arc length: (2cost, 2sint, t) for 0 <= t <= 2pi

anonymous · Answer

if you're curious, the answer is 2sqrt(5pi)

anonymous · Answer

What is this acr lenght? because  the way I was thinking i got 4pi

anonymous · Answer

integrate

anonymous · Answer

this is a parametric equation intergrate from 0 to 2pi for the function given

anonymous · Answer

$$l=\int |r'(t)| \ \text{d}t $$

anonymous · Answer

c'(t) = ( -2sint, 2cost, 1) ?

anonymous · Answer

yeah

anonymous · Answer

ok, now do I use the dot product on itself?

anonymous · Answer

or sqaure each one and add then together?

anonymous · Answer

yes and finally take square root ...(thats a vector in 3D)

anonymous · Answer

all that in a square root to the integral 0 to 2pi

anonymous · Answer

i think i am messing up somewhere when i try that. can you help me with those steps?

anonymous · Answer

@mukushla  use the correct term, magnitude of the vector as we hav |r'(t)|

anonymous · Answer

first find r'(t) = (-2sint, 2cost,1)
magnitude of r'(t)= sqrt (4sin^2+4cos^t+1)= sqrt(5)

anonymous · Answer

then integrate  sqrt(5) from 0 to 2pi

anonymous · Answer

can I show you the formula i was given?

anonymous · Answer

sure

anonymous · Answer

L(c) = ||c'(t)||dt from integral a to b,  = Sqrt( c'(t) (dot prod) c'(t) ) from integral a to b

anonymous · Answer

||   ||  is the Norm of a vector

anonymous · Answer

which is same formula as given by @mukushla

anonymous · Answer

did not know

anonymous · Answer

look up pauls online notes calculus II or calculus III and I am sure it will be of great help to you : )