Let c be a curve w/parameterization r(t). show that the line integral of T dr equals the length of the curve where T is the unit tangent.
am not even conversant with those .....i don,t know
Hmmm. Let me dig out my text on differential geometry. Meantime, try googling the topic in vector analysis.
Actually, just consider the definition of T.
Take a peek here, and it should be clear, since the arc length is just
∫ ds = ∫ |dr/dt| dt
oops: here is
http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/arc/arc.html#alfunction
hmmm i got to write these down wow
To show that the line integral of T dr equals the length of the curve, we can use the definition of the line integral and properties of the unit tangent vector.
First, let's recall the definition of the line integral of a vector field F along a curve C with parameterization r(t):
∫ F · dr = ∫ (F • T) |dr/dt| dt,
where T is the unit tangent vector of the curve and |dr/dt| is the magnitude of the derivative of r(t) with respect to t.
We want to compute the line integral of T dr, so we have:
∫ T · dr = ∫ (T • T) |dr/dt| dt,
Since T is the unit tangent vector, its magnitude is 1, so T • T = 1. This simplifies the line integral to:
∫ T · dr = ∫ |dr/dt| dt.
Now, let's consider the length of the curve.
The length of the curve C is given by:
L = ∫ |dr/dt| dt,
which is exactly the same as the integral we obtained for the line integral of T dr.
Therefore, we can conclude that the line integral of T dr equals the length of the curve:
∫ T · dr = L.