Vector Functions — Question 2

Question 2 – Solution

Strategy. Solve the simplest component for the parameter and substitute it into the other two components.

Step 1: Recover the parameter Since $x=2t-1$, $$t=\frac{x+1}{2}.$$

Step 2: Eliminate $t$ Substitution into $y=t^2+1$ gives $$y=\frac{(x+1)^2}{4}+1.$$ Also, $$z=3\left(\frac{x+1}{2}\right)+4=\frac 32x+\frac{11}{2},$$ or $3x-2z+11=0$. Hence the trace is the part of the parabola $$\boxed{y=\frac{(x+1)^2}{4}+1,\qquad 3x-2z+11=0}$$ for which $-5\le x\le 3$. It lies in the stated plane.

Step 3: Endpoints and orientation $$\mathbf r(-2)=\left\langle -5,5,-2\right\rangle,\qquad \mathbf r(2)=\left\langle 3,5,10\right\rangle.$$ As $t$ increases, $x=2t-1$ increases, so the curve is traversed from the first point to the second.

Verification. Substituting either endpoint into both Cartesian equations confirms both constraints exactly.