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

yogi55hr 2021-11-19 Answered
Match the parametric equations with the graph (labeled I- VI). Give reasons for your choices.
\(\displaystyle{x}={\cos{{8}}}{t}\) , \(\displaystyle{y}={\sin{{8}}}{t}\) , \(\displaystyle{z}={e}{0.8}{t}\), \(\displaystyle{t}\geq{0}\)

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Expert Answer

Cherry McCormick
Answered 2021-11-20 Author has 1319 answers
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.
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giskacu
Answered 2021-11-21 Author has 517 answers
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
Answer
GRAPH IV
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