Focuses on the algebraic process of converting a set of parametric equations into a single rectangular (Cartesian) equation using methods like substitution.
What does it mean to "eliminate the parameter" from a set of parametric equations?
It means to convert the pair of equations \(x = f(t)\) and \(y = g(t)\) into a single equation that directly relates \(x\) and \(y\), without involving the parameter \(t\). The result is a rectangular equation, like \(y = h(x)\) or \(G(x, y) = 0\).
Describe the general substitution method for eliminating the parameter.
1. Solve one of the parametric equations for the parameter \(t\) (e.g., solve \(x = f(t)\) to get \(t = g(x)\)).
2. Substitute the resulting expression for \(t\) into the other parametric equation.
Eliminate the parameter from the equations \(x = t + 4\) and \(y = 3t - 5\).
1. Solve for \(t\) in the first equation: \(t = x - 4\).
2. Substitute this into the second equation: \(y = 3(x - 4) - 5\), which simplifies to \(y = 3x - 12 - 5\), so \(y = 3x - 17\). This is a line.
Eliminate the parameter from \(x = t^2\) and \(y = t + 1\).
1. Solve for \(t\) in the second equation: \(t = y - 1\).
2. Substitute into the first equation: \(x = (y - 1)^2\). This is a parabola that opens to the right.
Eliminate the parameter from \(x = \sqrt{t}\) and \(y = t - 3\), for \(t \ge 0\).
From \(x = \sqrt{t}\), we know \(x \ge 0\) and \(t = x^2\). Substituting into the second equation gives \(y = x^2 - 3\). Because \(x\) must be non-negative, the graph is only the right half of the parabola.
When you eliminate a parameter, why is it crucial to consider the domain and range implied by the original parametric equations?
The original parametric equations may restrict the values of \(x\) and \(y\). The resulting rectangular equation might describe a larger curve, so you must state the restrictions on \(x\) and \(y\) to accurately represent the original parametric curve segment.
Eliminate the parameter from \(x = e^t\) and \(y = e^{2t}\).
Notice that \(y = e^{2t} = (e^t)^2\). Since \(x = e^t\), we can substitute to get \(y = x^2\). Since \(x=e^t\), \(x\) must be positive (\(x>0\)), so the graph is the portion of the parabola in the first quadrant.
Eliminate the parameter from \(x = 2 + \frac{1}{t}\) and \(y = t - 1\).
1. Solve for \(t\) in the second equation: \(t = y + 1\).
2. Substitute into the first equation: \(x = 2 + \frac{1}{y+1}\). This is a rational function.
Eliminate the parameter from the equations \(x=t^3\) and \(y=t^6+1\).
Recognize that \(t^6 = (t^3)^2\). Since \(x = t^3\), we can substitute directly: \(y = x^2 + 1\).
What type of rectangular curve results from the parametric equations \(x = at + x_0\) and \(y = bt + y_0\)?
These are the parametric equations for a line. Eliminating the parameter \(t\) will always result in a linear equation of the form \(y = mx + c\) or \(Ax + By = C\).
Eliminate the parameter from \(x=t^2-3\) and \(y=t^2+1\).
Solve for \(t^2\) in the first equation: \(t^2 = x+3\). Substitute this into the second equation: \(y = (x+3)+1\), which simplifies to \(y = x+4\). Since \(t^2 \ge 0\), then \(x+3 \ge 0\), so \(x \ge -3\). The graph is a ray.
