91Ó°ÊÓ

When discussing parametric equations like those given in the exercise, continuity is a key concept. A function is continuous if there are no breaks or jumps in its graph. For our exercise, both parametric definitions, like the one for \( x(t) = 2t - |t-1| \), need to be continuous over real values of \( t \).
For different values of \( t \), the absolute value impacts the expression: when \( t \geq 1 \), the expression becomes \( x = t + 1 \), and when \( t < 1 \), it becomes \( x = 3t - 1 \). Since these are linear expressions, they're continuous over their respective intervals.
Likewise, \( y(t) = 2t^2 + t|t| \) is combined into simple polynomial expressions for different \( t \)'s: \( y = 3t^2 \) for \( t \geq 0 \) and \( y = t^2 \) for \( t < 0 \). Polynomials are inherently continuous, so \( y(t) \) is continuous as well.

• Thus, the function described by these parametric equations provides a continuous mapping for all real numbers \( t \).
• The continuity translates to the function \( y = f(x) \) being continuous over all real numbers \( x \).

Differentiability indicates whether a function has a defined slope at every point, meaning it must be smooth without any sharp turns or corners.
For parametric equations, checking differentiability often involves calculating derivatives. We take \( dx/dt \) and \( dy/dt \), and use them to find \( dy/dx \).

For \( x(t) \), depending on the value of \( t \) and due to the presence of an absolute value, the derivative changes:

• For \( t \geq 1 \), \( dx/dt = 1 \).
• For \(t < 1\), \( dx/dt = 3 \).

These reflect that there's a potential point of non-differentiability at \( t = 1 \).

For \( y(t) \),

• When \( t \geq 0 \), \( dy/dt = 6t \).
• When \( t < 0 \), \( dy/dt = 2t \).

It shows that \( y'(t) \) changes form as \( t \) switches between positive and negative.
Assessing \( dy/dx \), we notice undefined behavior at \( t = 0 \) and \( t = 1 \) (due to division in \( dy/dx \)), implying the function is not differentiable at these \( x \) points.

• This results in exclusions from the differentiable domain for points in \( x \) corresponding to \( t = -1, 2 \).

Absolute value functions are crucial in parametric equations since they affect how expressions behave at certain points. An absolute value, like in \( |t-1| \) from our exercise, modifies the expression based on the condition of its argument.
If \( t \geq 1 \), then \( |t-1| = t-1 \), otherwise, it becomes \( 1-t \) when \( t < 1 \). This leads to different linear expressions for different intervals of \( t \).

• This behavior creates potential points of non-differentiability or non-smoothness in the function as slope may suddenly change.
• It's essential to analyze these critical points (where the behavior changes) to understand the function's behavior entirely.

Using absolute values in functions contributes to piecewise definitions. Each piece is continuous on its interval, but transitions between pieces must be carefully checked, especially for continuity and differentiability.