Question

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=t^{2}+t, \quad y=t^{2}-t $$

Short answer

The corresponding rectangular equation is: \( x^2 - 2xy + y^2 - 6x + 2y = 0 \). It represents a circle with centre at (3, -1) and radius \(\sqrt{10} \). The orientation of the curve is counterclockwise.

Step-by-step solution

Rewrite the equations

Start by considering the given parametric equations. \(x=t^{2}+t\), \(y=t^{2}-t\).

Eliminate the parameter t

In order to eliminate the parameter t from these equations, subtract the second equation from the first, which will cancel out the \(t^{2}\) term: \(x - y = 2t\). Now, solve for \(t\) and get \(t= \frac{x - y}{2}\). Substitute this into one of the original parametric equations, for example \(x\), to get the rectangular equation: \(x = (\frac{x - y}{2})^2 + (\frac{x - y}{2})\).

Simplify the equation

Simplify the equation from step 2 to obtain: \(x = \frac{(x^2 - 2xy +y^2) + 2(x - y)}{4}\), which simplifies further to \(4x = x^2 - 2xy + y^2 + 2x -2y\) and by reordering, we get \(x^2 - 2xy + y^2 -2x + 2y - 4x = 0\). Hence, the corresponding rectangular equation is: \( x^2 - 2xy + y^2 - 6x + 2y = 0\).

Sketch the curve

It is a second-degree equation with both \(x^2\) and \(y^2\) appearing and thus should be a conic section. To decide which type of conic section it is and get further information about the location of the curve, we complete the square regarding x and regarding y, which results in \((x-3)^2 + (y+1)^2 = 10\). This is a circle with centre at (3, -1) and radius \(\sqrt{10}\).

Determine the orientation of the curve

The orientation of the curve is given by how \(t\) changes from its smallest to its largest value. For \(t<-0.5\), \(x\) and \(y\) both decrease, for \(-0.560_1\), \(x\) and \(y\) both increase. Consider these to draw the curve's orientation which is counterclockwise.