Vector Functions — Question 10

Question 10 – Solution

Strategy. Eliminate the parameter using the first two components. For uniqueness, retain the ordered pair $(\cos t,\sin t)$ rather than examining $z$ alone.

Step 1: Eliminate the parameter The first two components give $$x^2+y^2=4(\cos^2t+\sin^2t)=4.$$ Also, $$xy=(2\cos t)(2\sin t)=4\sin t\cos t=4z,$$ so $z=xy/4$. The trace is therefore contained in $$\boxed{x^2+y^2=4,\qquad z=\frac{xy}{4}}.$$ Conversely, every point of this intersection has $(x/2,y/2)$ on the unit circle and hence equals $(\cos t,\sin t)$ for some $t$; thus these equations describe the full trace.

Step 2: Intersections with $z=0$ Since $z=xy/4$, we need $x=0$ or $y=0$. On $x^2+y^2=4$, this yields $$\boxed{(2,0,0),\ (-2,0,0),\ (0,2,0),\ (0,-2,0)}.$$

Step 3: One-to-one traversal If $\mathbf r(t_1)=\mathbf r(t_2)$, then the first two coordinates imply $$(\cos t_1,\sin t_1)=(\cos t_2,\sin t_2).$$ Thus $t_1-t_2=2\pi k$ for an integer $k$. Inside $[0,2\pi)$, only $k=0$ is possible, so $t_1=t_2$. Hence no point is repeated.