# Match the parametric equations with the graph (labeled I- VI). Give reasons for

Match the parametric equations with the graph (labeled I- VI). Give reasons for your choices.
$$\displaystyle{x}={\cos{{8}}}{t}$$ , $$\displaystyle{y}={\sin{{8}}}{t}$$ , $$\displaystyle{z}={e}{0.8}{t}$$, $$\displaystyle{t}\geq{0}$$

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

Cherry McCormick
Step 1
Elimination. Previously, we learned that the parametric equation $$\displaystyle{x}={\cos{{8}}}{t}$$ and $$\displaystyle{y}={\sin{{8}}}{t}$$ lie on a circle on the xy-plane in $$\displaystyle{R}^{{{2}}}$$. In $$\displaystyle{R}^{{{3}}}$$, these parametric equations lie on a cylinder pointing to the z-axis. Only IV and VI fits this criterion as they are circular and the circles are oriented along the xy-plane. We have answered VI earlier so this must be IV. Let's check!
Step 2
Further elimination and conclusion.
Between the two equations, observe that the 2-component:
$$\displaystyle{z}={e}^{{{2}{t}}}$$ is oscillating at an increasing amount. z is defined for all t and is not bounded by any values of t. Graph VI seems to be asymptotic to 0 whereas graph IV has circles that have an increasing period between them. Therefore, this equation must be describing graph IV.
###### Have a similar question?
giskacu
Step 1
Since $$\displaystyle{x}{\left({t}\right)}={\cos{{8}}}{t}$$ and $$\displaystyle{y}{\left({t}\right)}={\sin{{8}}}{t}$$ we have that the curve lies on the cylinder $$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={1}$$ (this is because $$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={1}$$ implies that the projection of the curve on the xy plane is the unit circle).
From $$\displaystyle{z}={e}^{{{0.8}{t}}}$$ we see that z will increase exponentially as t increases, The only graph that has these properties is
GRAPH IV
GRAPH IV