Question

Explain why the parametrization $x=\sin t+1,y={\sin }^{2}t-4$for $t\in (-\infty ,\infty )$repeatedly traces the same small portion of the graph of the function $y=x^2-2x-3$.

Short answer

The parameterization of the curves $\sin t=x-1,y={\sin }^{2}t-4$ traces exactly the same curve $y=x^2-2x-3$as that of the equations $x=2t+1,y=4t^2-4.$

Step-by-step solution

Step: 1 Given information

Consider the parametric curves$x=\sin t+1,y={\sin }^{2}t-4,t\in (-\infty ,\infty ).$

Calculation

The goal is to figure out how to parametrize the curves.

Remove the parameter $t$from the parametric equations to parameterize the curves.

Take $x=\sin t+1$.

Add $-1$to both sides of the equation.

$$\begin{gathered} x-1=\sin t+1-1 \\ x-1=\sin t+1-1 \\ x-1=\sin t \end{gathered}$$

Now substitute $\sin t=x-1$in the equation $y={\sin }^{2}t-4$.

Then,

$$\begin{gathered} y=(x-1{)}^2-4 \\ y=x^2-2x+1-4 \\ y=x^2-2x-3 \end{gathered}$$

Further simplification

Here the $\sin t$function is periodic and lies in between $-1,1$. And x, y are also bounded.

Here the values that satisfy the equation$y=x^2-2x-3$ are

Thus, the parameterization of the curves $\sin t=x-1,y={\sin }^{2}t-4$ traces exactly the same curve $y=x^2-2x-3$ as that of the equations $x=2t+1,y=4t^2-4$.

Hence the explanation.