Question

In Exercises \(5-22,\) a parametrization is given for a curve. (a) Graph the curve. What are the initial and terminal points, if any? Indicate the direction in which the curve is traced. (b) Find a Cartesian equation for a curve that contains the parametrized curve. What portion of the graph of the Cartesian equation is traced by the parametrized curve? $$x=3 t, \quad y=9 t^{2}, \quad-\infty

Short answer

The parametric equations \(x = 3t\) and \(y = 9t^2\) represents the entire graph of the Cartesian equation \(y = x^2\), with no initial and terminal points.

Step-by-step solution

Sketch the Parametric Plot and Find Initial and Terminal Points

The given parametric equations are \(x = 3t\) and \(y = 9t^2\). As the domain of \(t\) extends from \(-\infty\) to \(+\infty\), the initial and terminal points don't exist. The graph can be plotted by substituting various values of \(t\) into \(x\) and \(y\). The direction of the curve is determined by the increasing values of \(t\).

Convert the Parametric Equation to Cartesian Form

To find the Cartesian form, one can first solve one of the parametric equations for \(t\) and then substitute that expression into the other parametric equation. Solving the \(x\)-equation for \(t\) gives \(t = x/3\). Substituting \(t = x/3\) into the \(y\)-equation, gives \(y = 9(x/3)^2 = x^2\).

Determine the Traced Portion of the Cartesian Graph

As the domain of \(t\) is \(-\infty < t < +\infty\), the graph of the Cartesian equation \(y = x^2\) is entirely traced by the parametric curve.