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? $$x=\sin (2 \pi t), \quad y=\cos (2 \pi t), \quad 0 \leq t \leq 1$$

Short answer

The parametric equations \(x=\sin (2 \pi t)\), \(y=\cos (2 \pi t)\) trace a circle with center at the origin and radius 1 in counterclockwise direction starting and ending at (0,1). The Cartesian equation equivalent is \(x^2 + y^2 = 1\). The entire graph of the Cartesian equation is traced.

Step-by-step solution

- Graph the curve

We have the parametrized equations for the curve as \(x=\sin (2 \pi t)\), \(y=\cos (2 \pi t)\) where \(0 \leq t \leq 1\). This traces out a circle once counter-clockwise about the origin with radius 1. Because we’re moving counter-clockwise around the circle we know that at \(t = 0\) the point on the circle will be at (0, 1) and at \(t = 1\) we will be at the same point.

- Determine the direction of the curve

Since \(t\) goes from 0 to 1, we start from point (0, 1). As \(t\) increases, \(\sin(2 \pi t)\) and \(\cos(2 \pi t)\) traces out the circle once in a counterclockwise direction and then return to point (0, 1). Thus, the curve is traced in a counterclockwise direction from (0,1) back to itself.

- Find a Cartesian equation

Since \(x = \sin (2 \pi t)\) and \(y = \cos (2 \pi t)\), then \(x^2 + y^2 = \sin^2(2 \pi t) + \cos^2(2 \pi t) = 1\). So, the Cartesian equation for the curve is \(x^2 + y^2 = 1\). This is simply a circle with radius 1, centered at the origin. The graph of the parametric curve is exactly the same as the graph of this circle. Being more precise, it starts at (0,1) and traces out the circle once, ending at (0,1).

- Identify the traced portion

Since the parametrized equations trace out the whole circle from \(t = 0\) to \(t = 1\), so the portion of the graph of the Cartesian equation that is traced by the parametric curve is the whole circle.