Line integral approximation through a Riemann sum.

248 Views Asked by At

"A curve's length is the length of polygonal segments: $\sum_{i=1}^{n} |r(t)_{i}-r(t)_{i-1}| = \sum_{i=1}^{n} \dfrac{|r(t)_{i}-r(t)_{i-1}|}{\Delta t} \Delta t = \sum_{i=1}^{n} |r'(t_{i})| \Delta t$."

Why would $|r(t)_{i}-r(t)_{i-1}| \neq \dfrac{|r(t)_{i}-r(t)_{i-1}|}{\Delta t} \Delta t = r'(t) \Delta t$?

For instance on the curve $r(t)=\langle t,t^{2} \rangle$ I calculate the derivative is $r'(t) = \langle 1,2t \rangle$; so on the interval $[0,2]$ when I pick $n = 2, \Delta t = 1$ the value $|(1,1)-(0,0)|= |\langle 1,1 \rangle| = \sqrt{2}$ is not equal to $|r'(1)|\Delta t = \sqrt{5}$ or $|r'(0)|\Delta t = 1$.

Wikipedia 6/20/2021

1

There are 1 best solutions below

0
On BEST ANSWER

I noticed that choosing $n=2$ doesn't make the segment into the derivative.