Vectors : Arc length of a space curve Use the integration capabilities of you graphing utility to approximate the arc length of the space curve given by: vector $$r(t) = t i + t^3 j + 3k$$ from the point $(1,1,3)$ to the point $(2,8,6)$. Round your answer to the nearest three decimal places.

QUESTION: Vectors : Arc length of a space curve
Use the integration capabilities of you graphing utility to approximate the arc length of the space curve given by: vector 
$$r(t) = t i + t^3 j + 3k$$
from the point $(1,1,3)$ to the point $(2,8,6)$. Round your answer to the nearest three decimal places. 

SOLUTION  =======================
Theory. 
If you have a curve $r(t) = x(t) i + y(t) j + z(t) k$  then the arc length between two points where t=a to t=b is given by 

$$s = \int^b_a \sqrt{ x'(t)^2 + y'(t)^2 + z'(t)^2 } dt $$
( this is the integral between $t=a$ and $t=b$ ) 
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It is clear that the two points given arise from $t=1$ and $t=2$. 

since when $t=1$, $r(1) = 1 i + 1 j + 3 k$ which corresponds to $(1,1,3)$ 
and when $t=2$, $ r(1) = 2 i + 8 j + 6 k$ which corresponds to $(2,8,6)$ 

For us, 
$x(t) = t$ 
$y(t) = t^3$ 
$z(t) = 3t $

so 

$x'(t) = 1 $
$y'(t) = 3t^2$ 
$z'(t) = 3 $

and $s = \int^b_a \sqrt{ 1 + 9t^4 + 9} dt$ 

      $s = \int^b_a \sqrt{ 10 + 9t^4} dt $

when I use the graphic program to approximate this integral I obtain, 

   $s = 7.763$ to 3 decimal places. 

cheers 
Jan Hansen