Question

In Exercises 45 and \(46,\) 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=-\sec t, \quad y=\tan t, \quad-\pi / 2

Short answer

The parametric curve \(x = -\sec t\), \(y = \tan t\) where \(-\pi / 2<t<\pi / 2\) does not have initial and terminal points. It's graphed as a hyperbola traced from left to right, and corresponds to the portion of the graph of the Cartesian equation \(x^2 - y^2 = 1\) where \(x<0\).

Step-by-step solution

Find the Initial and Terminal Points

These points are obtained by substituting the limiting values of the parameter \(t\) into the parametric equations. The parametric equations for the curve are \(x=-\sec t\) and \(y=\tan t\). The parameter \(t\) ranges from \(-\pi / 2\) to \(\pi / 2\). \n\nSubstituting \(t = -\pi / 2\) into the parametric equations gives \(x = -\sec (-\pi / 2)\) and \(y= \tan (-\pi / 2)\). And, substituting \(t = \pi / 2\) gives \(x = -\sec (\pi / 2)\) and \(y = \tan (\pi / 2)\). Since \(sec(\pi / 2)\) and \(sec(-\pi / 2)\) are undefined and \(tan(\pi / 2)\) and \(tan(-\pi / 2)\) are also undefined, these equations have no valid points, so the curve does not have initial and terminal points.

Graphing the Curve

Due to the nature of the parameters \(x=-\sec t\) and \(y=\tan t\), where \(t\) ranges from \(-\pi / 2\) to \(\pi / 2\), the graph would take the form of a hyperbola. This is because the secant function can yield negative values, and tangent function increases without bound as \(t\) approaches \(\pm \pi / 2\). The curve is traced from left to right for the interval \(-\pi / 2<t<\pi / 2\).

Find a Cartesian Equation for the Curve

To find a Cartesian equation, eliminate the parameter \(t\). We can use trigonometric identity \(1 + \tan^2 t = \sec^2 t\) or \(1 = \sec^2 t - \tan^2 t\). Replacing \(\sec t\) with \(-x\) and \(\tan t\) with \(y\) gives the Cartesian equation \(x^2 - y^2 = 1\).

Determine the Portion of the Cartesian Curve

The graph of the Cartesian equation \(x^2 - y^2 = 1\) is a hyperbola with vertices at \((1, 0)\) and \((-1, 0)\), and asymptotes \(y = x\) and \(y = -x\). While the entire graph of the hyperbola is not traced by the parametric curve, the portion of the graph of \(x^2 - y^2 = 1\) that lies to the left of the y-axis, i.e., where \(x<0 \), matches the graph of the parametrized curve, because the range parameter \(t\), which is \(-\pi / 2<t<\pi / 2\), results in negative \(x\) values for \(x = -\sec t\).

Conclusion

From the above discussion, it's determined that the parametric curve \(x = -\sec t\), \(y = \tan t\) where \(-\pi / 2<t<\pi / 2\) ,does not have initial and terminal points, is graphed as a hyperbola traced from left to right in the Cartesian plane, and corresponds to the portion of the graph of the Cartesian equation \(x^2 - y^2 = 1\) where \(x<0\).