Please help with vector calculus?

Question

If f is a continuously differentiable function, show how

L(x) = integral(from a to b) : ||x'(t)|| dt

may be used to establish the formula

L = integral(from a to b) : sqrt(1 + [f '(x)]^2) dx

for the length of the curve y = f(x) between (a , f(a)) and (b , f(b))

Any help would be extremely helpful.

supastremph · Accepted Answer

So, x is a vector in this context, let's call it r for obvious reasons.

x' = dr/dt = (dx/dt,dy/dt)

|dr/dt| = sqrt((dx/dt)^2 + (dy/dt)^2)

|dr/dt|dt = sqrt((dx/dt)^2 + (dy/dt)^2)dt = sqrt(dx^2 + dy^2) = sqrt( (1+(dy/dx)^2)dx^2) 
= sqrt(1+(f')^2)dx

? · Answer

If you have y=f(x) you can write this as a vector,

R= x i + j f(x)    if x= x(t)  (Parametric)  then your vector X is now,
X= x(t) i  + j f(x(t))     or
X= x(t) i + y(t)  j        

Now , X(t+dt) - X(t)   is tangent vector at t  , as well as dt goes to 0
  dR =X(t+dt) - X(t) = DX/dt dt   (Similar if dt goes to 0 )

Now Remember that a vector A dot B = IAI IBI cos T  if A=B  T=0 ,cos T =1  and
A dot A = IAI^2  , we are looking for IdRI
sqrt (dR dot dR) =IdRI = IDx/dtI dt  then 
Lengh = INT IDx/dt I  dt 

Other form is  dy/dx = (dy/dt)/(dx/dt) 
1+(dy/dx)^2 = ((dx/dt)^2 +(dy/dt)^2 )/(dx/dt)^2
L = INT ( sqrt ( ((dx/dt)^2 +(dy/dt)^2 ))/(dx/dt) dx  , but dx = (dx/dt) dt   (Remember x=x(t))
L=INT ( sqrt ( ((dx/dt)^2 +(dy/dt)^2 )) dt      The same .-
I hope i have help you, from CHILE.-

 
  
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Please help with vector calculus?

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