Question

True or False. Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The graph of the parametric equations \(x=t^{2}\) and \(y=t^{2}\) is the line \(y=x\).

Short answer

The statement that the graph of the parametric equations \(x=t^{2}\) and \(y=t^{2}\) is the line \(y=x\) is false as it does not account for instances where 't' is negative.

Step-by-step solution

Understand parametric equations

Parametric equations are a set of equations that express a set of quantities as explicit functions of a number of independent variables, known as 'parameters.' Here, parameters are \(x=t^{2}\) and \(y=t^{2}\)

Analyze the given parametric equations

In the given parametric equations \(x=t^{2}\) and \(y=t^{2}\), for a given 't' value, both x and y will be the same. Therefore, it seems like the statement 'The graph of the parametric equations \(x=t^{2}\) and \(y=t^{2}\) is the line \(y=x\)' is true because each 't' gives the same value for both 'x' and 'y'.

Check if the relationship forms \(y=x\)

The line equation \(y=x\) entails that for every 'x' value there exists the same 'y' value, and vice versa. However, in our parametric equations \(x=t^{2}\) and \(y=t^{2}\), when 't' is negative, 'x' and 'y' become positive. For instance, at 't=-2', 'x' and 'y' become 4. This contradicts the function \(y=x\), where negative 'x' value would correspond to a negative 'y' value. Thus, the given statement is false.

Provide a contradicting example

For example, when 't'=-2, then 'x'=4 and 'y'=4, which is a point in the first quadrant. However, in the linear equation \(y=x\), this point would not exist. So the statement is false as it does not account for instances when 't' is negative.

Key concepts

Understanding Parametric Equations Graphs

Parametric equations play a crucial role in calculus by allowing us to graph curves that cannot be represented by a single function in the standard form 'y=f(x)'. Instead of using one equation to define 'y' as a function of 'x', parametric equations use a system where both 'x' and 'y' are defined in terms of an independent parameter, typically 't'.

For instance, the parametric equations given in the exercise, with both coordinates described as squares of the parameter 't' \( x=t^2 \text{ and } y=t^2 \text{ evoke the image of the line } y=x \text{ upon initial consideration. However, this superficial resemblance can be misleading.\)
In a true parametric graph of these equations, every value of 't' produces a coordinate pair (x, y), mapping out a curve on the two-dimensional plane. By plotting various values of 't' and their corresponding (x, y) pairs, we construct a visual representation of the relationship, often revealing a curve distinctly different from a straight line, especially when considering positive and negative values of 't'.

The Anatomy of a Line Equation

To further elaborate, a line equation like \( y = x \) represents a straight line where every point has an identical 'x' and 'y' value, creating a 45-degree diagonal line through the origin of the coordinate system. The nature of a line equation is linear, meaning that it extends infinitely in both directions, crossing all quadrants of the plane.

But why doesn't the parametric equation in the exercise match this simple line equation? It comes down to how each equation behaves. A linear equation is direct; it doesn't alter the sign of the input value—an input of -2 for 'x' will result in an output of -2 for 'y'. In contrast, the parametric equations given squaring 't' to find 'x' and 'y', only produce positive values, confining the graph to the first quadrant, which does not encompass the negative values that a line \( y = x \) would.

Calculus Contradiction Example

In calculus, contradictions such as the one described in the exercise are crucial for understanding the limitations and specific behaviors of certain equations. Through contradiction we realize that a graph of the parametric equations \( x=t^2 \) and \( y=t^2 \) does not represent the straight line \( y=x \) since it fails to account for negative 't' values.

For example, let's consider the scenario where 't' equals -2. According to the parametric equations we get 'x' = 4 and 'y' = 4, which places the point firmly in the first quadrant. However, on the line \( y=x \) if 'x' were -2, then 'y' would also be -2, placing the point in the third quadrant, not the first. This offers a clear contradiction, demonstrating that the parametric equations do not graph as the line \( y=x \) in all cases, thereby disproving the initial statement.