Problem 6 - Algebra and Trigonometry

Question

\begin{array}{l}{1-22 \text { a pair of parametric equations is given. }} \\\ {\text { (a) Sketch the curve represented by the parametric equations. }} \\\ {\text { (b) Find a rectangular-coordinate equation for the curve by }} \\\ {\text { eliminating the parameter. }}\end{array} $$ x=t^{2}, \quad y=t^{4}+1 $$

Step-by-Step Solution

Answer

The curve is a parabola, \(y = x^2 + 1\).

Step 1: Set up the Parameter

We are given the parametric equations \(x = t^2\) and \(y = t^4 + 1\). Our goal is to sketch the curve and find the rectangular coordinates by eliminating the parameter \(t\).

Step 2: Find Expressions for t

To eliminate the parameter, we need to express \(t\) in terms of \(x\) or \(y\). From \(x = t^2\), we solve for \(t\), giving us \(t = \pm \sqrt{x}\).

Step 3: Substitute into y Equation

Next, substitute \(t = \pm \sqrt{x}\) into the equation for \(y\). Thus, \(y = (\pm \sqrt{x})^4 + 1\). Simplifying, \(y = x^2 + 1\).

Step 4: Sketch the Curve

The rectangular equation \(y = x^2 + 1\) represents a parabola shifted one unit up from the standard position. Sketch a parabola opening upwards with its vertex at (0, 1).

Step 5: Verify the Rectangular Equation

Double-check by substituting back: if \(y = x^2 + 1\), then \(t^4 + 1 = (t^2)^2 + 1\), confirming our elimination process was correct.

Key Concepts

Rectangular Coordinate Equation

To convert from parametric equations to a rectangular coordinate equation, we are essentially trying to express the relationship between \(x\) and \(y\) without the intermediary parameter \(t\). This can make equations easier to analyze and graph. Here, we had the parametric equations \(x = t^2\) and \(y = t^4 + 1\). Our goal was to eliminate \(t\) and express \(y\) purely in terms of \(x\).
First, we solved the equation \(x = t^2\) for \(t\). Solving gives \(t = \pm \sqrt{x}\). This result allows us to replace \(t\) in the equation for \(y\).
By substituting \(t = \pm \sqrt{x}\) into \(y = t^4 + 1\), we get \(y = (\pm \sqrt{x})^4 + 1\). This simplifies straightforwardly to \(y = x^2 + 1\). We've successfully eliminated the parameter, leaving us with a rectangular coordinate equation: \(y = x^2 + 1\), which is much simpler to work with when plotting the graph of the curve.

Eliminating the Parameter

Eliminating the parameter in parametric equations involves removing the variable \(t\) to find a direct relationship between \(x\) and \(y\). This process is crucial for converting parametric equations into a form that can be easily graphed and interpreted.
To eliminate the parameter \(t\), we take the equation that relates \(x\) to \(t\) (here, \(x = t^2\)) and solve for \(t\). The solution was \(t = \pm \sqrt{x}\).
Following this, we substitute this expression for \(t\) into the second equation (\(y = t^4 + 1\)). This substitution simplifies to \(y = x^2 + 1\).

• Start with one of the parametric equations to solve for \(t\).
• Substitute the expression for \(t\) into the other parametric equation.
• Simplify the results to arrive at the final rectangular coordinate equation.

This technique is a powerful tool in calculus and algebra for simplifying complex systems by reducing the number of variables.

Curve Sketching

Sketching the curve deduced from parametric equations helps understand the shape and nature of the function. In this exercise, we converted the parametric form to the rectangular coordinate equation \(y = x^2 + 1\).
The rectangular coordinate equation \(y = x^2 + 1\) describes a parabola. Understanding the parabola's properties allows us to accurately visualize it:

• The graph is a parabola opening upwards.
• The vertex of this parabola is located at \((0, 1)\) because the term \(+1\) represents a vertical shift.
• The curve is symmetric about the y-axis due to the absence of linear \(x\)-terms.

With these characteristics, sketching becomes straightforward. We plot the vertex point, recognize the symmetry, and trace the curve opening upwards. Curve sketching through these techniques is not only essential for visualizing mathematical concepts but also vital for applications in physics and engineering, where understanding curve behavior is critical.