Vector Functions — Question 6

Question 6 – Solution

Strategy. Find a parameter substitution connecting the two formulas, then use the first two components to test repeated points.

Step 1: Same trace For $0<u<2\pi$, put $t=2\pi-u$; at $u=0$, use $t=0$. For the interior parameters, $$\cos(2\pi-u)=\cos u,\qquad \sin(2\pi-u)=-\sin u,$$ and $$\cos(2t)=\cos(4\pi-2u)=\cos 2u.$$ Hence $\mathbf r(2\pi-u)=\mathbf q(u)$, so every point of one trace occurs on the other. Thus their geometric traces are identical.

Step 2: Orientation Both begin at $(1,0,1)$. For small positive $t$, the second component of $\mathbf r(t)$ is positive. For small positive $u$, the second component of $\mathbf q(u)$ is negative. Therefore they traverse the trace in opposite orientations.

Step 3: Repeated-point test Suppose $\mathbf r(t_1)=\mathbf r(t_2)$. Its first two coordinates give $$(\cos t_1,\sin t_1)=(\cos t_2,\sin t_2),$$ so $t_1-t_2$ is an integer multiple of $2\pi$. Since both parameters lie in $[0,2\pi)$, this forces $t_1=t_2$. The same argument applies to $\mathbf q$.

Thus