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=\tan t, \quad y=-2 \sec t, \quad-\pi / 2

Short answer

The graph of the parametric curve is a semi-circle with a negative center, radius 2, initial and terminal points at \( (-1,0) \) and \( (1,0) \) respectively. The curve is traced from left to right. The Cartesian equation of the curve is \( x^2 + y^2/4 = 1 \) and the parametric curve traces the upper half of the circle.

Step-by-step solution

Graph the Function and Identify Key Points

Firstly, plot out the given parametric function over the range \( -\pi/2 < t < \pi/2 \). This can be done by calculating corresponding \( x \) and \( y \) values for a series of \( t \) values within the given range, then plotting these points on a graph. As \( t \) increases from \( -\pi/2 \) towards \( \pi/2 \), the curve will be traced out. Mark the initial point (the point corresponding to \( t = -\pi/2 \)) and the terminal point (the point corresponding to \( t = \pi/2 \)).

Find the Cartesian Form

Now, convert the parametric equations into a Cartesian equation. To do this, eliminate the parameter \( t \) using the trigonometric identity \( \tan^2 t + 1 = \sec^2 t \). Therefore, we rewrite the function as \( x = \tan t \) and \( y = -2 \sec t \) and rearrange the equations to eliminate \( t \). This gives us the Cartesian equation \( x^2 + y^2/4 = 1 \).

Determine the Traced Portion

Finally, the portion of the graph traced by the parametric curve is determined by the original range of \( t \). For this case, as \( -\pi/2 < t < \pi/2 \), this corresponds approximately to the top semi-circle of the graph of the equation \( x^2 + y^2/4 = 1 \). This means the parametric function traces the upper half of the circle.