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? $$x=t^{2}, \quad y=\sqrt{4-t^{2}}, \quad 0 \leq t \leq 2$$

Short answer

The initial point of the curve is (0, 2) and the terminal point is (4, 0). The curve is traced in a descending direction. The Cartesian equation of the curve is \(x + y^{2} = 4\). The parametrized curve traces the upper half of the circle defined by this equation.

Step-by-step solution

Graph the Curve and Identify Points

To do this, plug in the boundary values of \(t\) into the parametric equations. For \(t = 0\), the point is \((x, y) = (0^{2}, \sqrt{4-0^{2}}) = (0, 2)\). This is the initial point. For \(t = 2\), the point is \((x, y) = (2^{2}, \sqrt{4 - 2^{2}}) = (4, 0)\). This is the terminal point. To sketch the curve, plot these points and note that the \(y = \sqrt{4 - t^{2}}\) part restricts to only positive \(y\) values.

Identify the Direction of the Curve

As \(t\) increases from 0 to 2, \(x\) increases from 0 to 4 and \(y\) decreases from 2 to 0. Therefore, the curve is traced from the initial point to the terminal point in a descending direction.

Convert to Cartesian Equations

Square both sides of the equation for \(y\) to eliminate the square root: \(y^{2}=4-t^{2}\). Substitute \(x=t^{2}\) from the given \(x\) parametrization: \(y^{2}=4-x\). Rearranging gives: \(x + y^{2} = 4\). This is the Cartesian equation of the curve.

Identify the Traced Portion

The parametrization traces out a semi-circle, because of the square root in the \(y\) equation which restricts \(y\) to non-negative values. While the Cartesian equation represents an entire circle of radius 2 centered at (2,0). The parametrized curve traces the upper half of this circle.