Vector Functions — Question 5

Question 5 – Solution

Strategy Use $\vec r=A+s(t)(B-A)$ with a scalar progress function satisfying $s(0)=0$, $s(1)=1$, and $s'(0)=s'(1)=0$.

Progress function The cubic $s(t)=3t^2-2t^3$ has the required endpoint values and zero endpoint derivatives. Since $0\le s(t)\le 1$ on $[0,1]$, the path stays on the segment.

Vector function Because $B-A=\left\langle 5,-2,-4\right\rangle$, $$\boxed{\vec r(t)=\left\langle-2,1,4\right\rangle+(3t^2-2t^3)\left\langle 5,-2,-4\right\rangle},\quad 0\le t\le 1.$$

Verification Substitution gives the correct endpoints, and $\vec r'(0)=\vec r'(1)=\vec 0$.