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. $$ x=|t-1|, \quad y=t+2 $$

Short answer

The sketch of the curve represented by the parametric equations \( x = |t-1| \), \( y = t + 2 \) consists of two lines that intersect at the origin with orientation from the bottom left to the top right. The corresponding rectangular equation is \( y = x + 3 \) for \( x \geq 0 \), and \( y = -x + 3 \) for \( x < 0 \).

Step-by-step solution

Understand the Parameter

In the function \( x = |t-1| \) we need to understand the parameter \( t \) can be any real number. We can separate this equation into two cases: \( t < 1 \) and \( t \geq 1 \), because of the absolute value. For \( t < 1 \), \( x = 1 - t \), and for \( t \geq 1 \), \( x = t - 1 \). For the second equation \( y = t + 2 \), \( y \) increases as \( t \) increases.

Draw the Curve

Firstly, we can draw the curve when \( t \geq 1 \). Here, as \( t \) ranges from 1 to infinity, \( x = t - 1 \) ranges from 0 to infinity. At the same time, \( y = t + 2 \) ranges from 3 to infinity. So, the curve in this case is a part of the line \( y = x + 3 \) for which \( x \geq 0 \). Secondly, for the case \( t < 1 \), as \( t \) ranges from negative infinity to 1, \( x = 1 - t \) ranges from infinity to 0, meanwhile \( y = t + 2 \) ranges from negative infinity to 3. Therefore, the curve here is a part of the line \( y = -x + 3 \) for which \( x \geq 0 \).

Indicate the Orientation

Given that \( x \) and \( y \) are both increasing as \( t \) moves from negative infinity through to positive infinity, the orientation of the curve is from the bottom left to the top right.

Find the Rectangular Equation

By replacing \( t \) in the \( y \)-equation with the \( x \)-equation, we can find the rectangular equation. However, this method itself has two cases. In the first case when \( t \geq 1 \), replacing \( t \) by \( x + 1 \) in the equation \( y = t + 2 \), we have \( y = x + 3 \). For the second case when \( t < 1 \), replacing \( t \) by \( 1 - x \) in the \( y \)-equation \( y = t + 2 \), we get \( y = -x + 3 \). So, due to the nature of the absolute value breaking down this relation into two pieces, the rectangular equation of \( y \) in terms of \( x \) also has two parts: \( y = x + 3 \) for \( x \geq 0 \), \( y = -x + 3 \) for \( x < 0 \).