use the desmos graph linked here, where you can graph parametric equations to match the graph with its…

Explanation:

Step1: Analyze option a

For $\begin{cases}x(t)=t^{2}\y(t)=t^{3}\end{cases}$, when $t = 0$, $x = 0,y = 0$. As $t$ varies, the curve has a self - intersecting and non - circular shape.

Step2: Analyze option b

For $\begin{cases}x(t)=t\cos(t)\y(t)=t\sin(t)\end{cases}$, this represents a spiral - like curve since the distance from the origin $r=\sqrt{x^{2}+y^{2}}=\sqrt{t^{2}\cos^{2}(t)+t^{2}\sin^{2}(t)} = |t|$ which increases as $|t|$ increases.

Step3: Analyze option c

For $\begin{cases}x(t)=2+\cos(t)\y(t)=\sin(t)\end{cases}$, we know that $(x - 2)^{2}+y^{2}=\cos^{2}(t)+\sin^{2}(t)=1$. This is a circle centered at $(2,0)$ with radius $r = 1$.

Step4: Analyze option d

For $\begin{cases}x(t)=t+\cos(t)\y(t)=t+\sin(t)\end{cases}$, as $t$ increases, the curve moves in a non - circular, non - spiral way with a kind of linear - like growth with some wiggles due to the trigonometric terms.

Step5: Analyze option e

For $\begin{cases}x(t)=e^{t}\y(t)=t^{2}\end{cases}$, as $t$ ranges from $- 5$ to $5$, when $t$ is negative, $x=e^{t}\in(0,e^{5})$ and $y = t^{2}\geq0$. When $t$ is positive, $x$ grows exponentially and $y$ is a quadratic function of $t$.

We need to match the given graphs with the parametric equations. Without seeing the actual graphs clearly, assume the circle - like graph matches with option c.

Answer:

c. $\begin{cases}x(t)=2+\cos(t)\y(t)=\sin(t)\end{cases}$