Question

Use a graphing utility to graph the curve represented by the parametric equations (indicate the orientation of the curve). Eliminate the parameter and write the corresponding rectangular equation. $$ x=\cos ^{3} \theta, \quad y=\sin ^{3} \theta $$

Short answer

The graph is a loop with an anti-clockwise orientation. The rectangular equation for the given pair of parametric equations is \( (x+y)^3 = 1 \).

Step-by-step solution

Graphing the Parametric Equations

Use a graphing utility to graph the parametric equations. Here, the parameter is \( \theta \). For each value of \( \theta \), the point on the graph is (\( \cos ^{3} \theta \), \( \sin ^{3} \theta \)). This gives a general shape of the graph. For instance, for \( \theta = 0 \), point is (1, 0), for \( \theta = \frac{ \pi }{4} \), the point is ( \( \frac { \sqrt {2} }{2} \), \( \frac { \sqrt {2} }{2} \) ). Plot more points to get a clear picture of the graph.

Indicating the Orientation of the Curve

The orientation of the curve begins from where \( \theta = 0 \) and follows the anticlockwise direction as \( \theta \) increases. For \( \theta = 2\pi \), we round back to the starting point, indicating the graph is a loop.

Eliminating the Parameter

To eliminate the parameter (\( \theta \) ), make use of the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Cube both sides to get \( \sin^6 \theta + 3\sin^4 \theta \cos^2 \theta + 3\sin^2 \theta \cos^4 \theta + \cos^6 \theta = 1 \). Now replace \( \sin^3 \theta \) with y and \( \cos^3 \theta \) with x in the equation above and arrange terms to match with the identity \( a^3 + 3a^2b + 3ab^2 + b^3 = (a+b)^3 \). This gives us \( (x+y)^3 = 1^3 \). So the rectangular equation for the given pair of parametric equations is \( (x+y)=1 \).