Problem 21 - Calculus

Question

Draw the graph of \(f(t)=\left|1-t^{2}\right|\) for \(0 \leqslant t \leqslant 2\). Find a three-part formula for \(v(t)\).

Step-by-Step Solution

Answer

The three-part formula is \(v(t) = \begin{cases} 1-t^2, & 0 \leq t < 1 \\ 0, & t = 1 \\ t^2-1, & 1 < t \leq 2 \end{cases}\).

Step 1: Analyze the Function

The given function is \(f(t) = |1 - t^2|\). The absolute value bars tell us that the output of the function is always non-negative. We need to determine when \(1 - t^2\) is positive, zero, or negative to remove the absolute value.

Step 2: Find Critical Points

To find when \(1 - t^2 = 0\), solve: \(1 - t^2 = 0\) which gives \(t^2 = 1\) leading to \(t = \pm 1\). Since we are considering \(0 \leq t \leq 2\), the critical point is \(t = 1\).

Step 3: Determine Piecewise Intervals

Based on the critical point found, the function can be divided into intervals: \(0 \leq t < 1\), \(t = 1\), and \(1 < t \leq 2\).

Step 4: Define Piecewise Function for Each Interval

For \(0 \leq t < 1\), \(1 - t^2\) is positive, so \(f(t) = 1 - t^2\). At \(t = 1\), \(f(t) = |1 - 1^2| = 0\). For \(1 < t \leq 2\), \(1 - t^2\) is negative, therefore \(f(t) = -(1 - t^2) = t^2 - 1\).

Step 5: Find the Three-Part Formula for \(v(t)\)

The three-part formula is derived from the piecewise expression of \(f(t)\):* For \(0 \leq t < 1\), \(v(t) = 1 - t^2\),* For \(t = 1\), \(v(t) = 0\),* For \(1 < t \leq 2\), \(v(t) = t^2 - 1\).

Key Concepts

Understanding Absolute Value Functions

Absolute value functions transform any input into a non-negative output by taking the absolute value of the expression. In the given exercise, the function is defined as \(f(t) = |1-t^2|\). This means no matter what value \(t\) takes, the function outputs \(1-t^2\) or its opposite if \(1-t^2\) is negative. The expression inside the absolute value, \(1-t^2\), is crucial as it determines the behavior of the function.
Analyzing \(1-t^2\), we see that when \(1-t^2\) is positive or zero, the absolute value doesn't change its sign. However, if \(1-t^2\) is negative, the absolute value will reverse the sign to ensure positivity. This flipping point, where \(1-t^2\) changes from positive to negative or vice versa, helps identify where to adjust the function for graphing using piecewise sections.

Graphing Piecewise Functions

Graphing functions like \(f(t) = |1-t^2|\) involves understanding the intervals where the underlying quadratic expression changes sign. This specific function is defined over the interval \(0 \leq t \leq 2\). Let's break it into defined segments:

• For \(0 \leq t < 1\), \(1-t^2\) remains positive, thus the graph will take on the form \(f(t) = 1-t^2\).
• At \(t = 1\), the function becomes zero because \(1-1^2 = 0\), a critical point where the graph touches the horizontal axis.
• For \(1 < t \leq 2\), \(1-t^2\) turns negative, so the function adjusts to \(f(t) = t^2 - 1\), flipping the sign to keep values positive.
  

The resulting piecewise function segments can be graphed to visually represent \(f(t)\) by sketching each segment on the same set of axes. This will show a combination of a downward-opening parabola, a single point at \(t = 1\), and then an upward-opening parabola.

Finding and Using Critical Points

Critical points are values where a function's behavior changes, often where derivatives equal zero or do not exist. In the scenario for \(f(t) = |1-t^2|\), the critical point is when \(1-t^2 = 0\), resulting in \(t = 1\). Why is this point critical? Because it's where the expression inside the absolute value changes sign, affecting the definition of the function.
At \(t = 1\), the quadratic switches from producing positive to negative values, which necessitates adjusting the function into two behaviors split by this point. For piecewise definitions like \(f(t)\), it's important to locate these critical thresholds. They signal shifts in the mathematical rules governing the function. Keeping track of these will aid in accurately graphing and understanding piecewise functions like those involving absolute values.