9.6 - Projections of a curve

1. Consider the curve described by $\overrightarrow{\mathbf{r}}(t)=⟨t\cos t,t\sin t,t⟩$ Sketch three planar projections: $x(z)$ (ignoring $y$), $y(z)$ (ignoring $x$), and $y(x)$ (ignoring $z$). Can you visualize the entire curve by looking at the projections?

   Can you visualize the entire curve by looking at the projections? [Check yourself by using GeoGebra's Curve function where enter a parametric equation for each coordinate. In this case:
   Curve(t*cos(t), t*sin(t), t, t, t_min, t_max).
   
   
   
   

   
   
   
   
   
   
   
   

2. Consider the curve given by $\overrightarrow{\mathbf{r}}(t)=t\hat{\mathbf{i}}-\frac{\sqrt{3}}{2}{t}^2\hat{\mathbf{j}}+\frac{1}{2}{t}^2\hat{\mathbf{k}}$.
   
   Sketch three planar projections, $y(x)$, $z(y)$.

   Hint: When you're sketching $z(x)$ vs $y(x)$, and $z(y)$. It may be useful to think of $t^2$ as the parameter. Call it $t^2=s$ and sketch $z(s)$ vs $y(s)$. But because $t^2$ can never be negative, $s$ can only take on positive values.

   • Projection in the $xy$ plane: ignoring $z$, we have $\overrightarrow{\mathbf{r}}(t)=t\hat{\mathbf{i}}-\frac{\sqrt{3}}{2}{t}^2\hat{\mathbf{j}}$. Since $x=t$, we can graph $y(x)=-\frac{\sqrt{3}}{2}{x}^2$. So, $y(x)$ is parabola, opening downwards, with its vertex at the origin.
   • In the $xz$ plane: $z(x)=\frac{1}{2}{x}^2$. So, $z(x)$ is a parabola, open upwards, with its vertex at the origin.
   • In the $yx$ plane: $\overrightarrow{\mathbf{r}}(t)=-\frac{\sqrt{3}}{2}{t}^2\hat{\mathbf{i}}+\frac{1}{2}{t}^2\hat{\mathbf{j}}$. We could parameterize this using $s=t^2$, as $$\begin{array}{}\text{(1)}& \overrightarrow{\mathbf{r}}(s)=-\frac{\sqrt{3}}{2}s\hat{\mathbf{j}}+\frac{1}{2}s\hat{\mathbf{k}}\text{for }0\le s.\end{array}$$ Though $t$ is allowed to be any real number, $t^2=s$ is only ever 0 or positive. This looks like a line that stops at the origin. When $s=2$, $\overrightarrow{\mathbf{r}}(2)=(-\sqrt{3},1)$: