Parametric Equations and Curves — Question 6

Question 6 – Solution

Solution

1. A line segment from $A$ to $B$ can be parametrized by $$\mathbf r=A+u(B-A), \qquad 0\le u\le1.$$

2. The parameter interval must be changed from $6\le t\le8$ to $0\le u\le1$. The linear change of variables is $$u=\frac{t-6}{8-6}=\frac{t-6}{2}.$$

3. Compute the displacement vector: $$B-A=(4,-1)-(-2,5)=(6,-6).$$ Therefore, $$(x,y)=(-2,5)+\frac{t-6}{2}(6,-6).$$

4. Simplify each coordinate: $$\begin{aligned} x&=-2+3(t-6)=3t-20,\\ y&=5-3(t-6)=23-3t. \end{aligned}$$ Hence a suitable parametrization is $$\boxed{x=3t-20,\qquad y=23-3t,\qquad 6\le t\le8}.$$

5. At $t=6$ it gives $A=(-2,5)$, and at $t=8$ it gives $B=(4,-1)$. At $t=7$ it gives $$(x,y)=(1,2)=\frac{A+B}{2},$$ so the midpoint condition is satisfied.