Question

Starting from the point (5,-3,-2) reparametrize the curve r(t)=(5+3t)i + (-3-3t)j + (-2-2t)k in terms of arc length.

Answer 1

Question 

Starting from the point (5,-3,-2) reparametrize the curve

r(s) = (5 + 3t)i + ( - 3 - 3t)j + ( - 2 - 2t)kr(s)=(5+3t)i+(−3−3t)j+(−2−2t)k
In term of arc length

<====✌️FIRST KNOW THE meaning of arc of reparmentrization......⏬⏬⏬

_____⭐Arc length reparametrization⭐__

☛ To reparametrize a curve r(t) by arc length, we need to modify the curve so that the path is the same but increasing the argument by 1 results in increasing the arc length by 1. 

☛ This way, inputting a value of sfor the curve will result in the curve having arc length s.

Answer 

☛ This is a linear curve, and each unit increase in t corresponds to an increase

===>> ||<3,-3,-2>|| 
= = > > \sqrt{9 + 9 + 4 =} \sqrt{22}==>>9+9+4=​22​ 
in arc length.

Thanks Thus, to reparametrize the curve in terms of arc length, we need to divide t by √22

r(s) = < 5 + \frac{3}{ \sqrt{22} } s - - - - - 1r(s)=<5+22​3​s−−−−−1 
( - 3 - \frac{3 }{ \sqrt{22} } s) - - - - - - - 2(−3−22​3​s)−−−−−−−2 

( - 2 - \frac{2}{ \sqrt{22} } s) - - - - - 3(−2−22​2​s)−−−−−3