Question

Make a sketch of \(f(t)\) for \(-25 s \leq t \leq 25\) s when \(f(t)\) is given by the following expression: $$\begin{aligned}f(t)=&-(20 t+400) u(t+20)+(40 t+400) u(t+10) \\\&+(400-40 t) u(t-10) \\ &+(20 t-400) u(t-20)\end{aligned}$$

Short answer

Piecewise linear segments with transitions at \( t = -20, -10, 10, \) and \( 20 \).

Step-by-step solution

Understanding the Unit Step Function

The function given, \( f(t) \), includes several terms multiplied by the unit step function, \( u(t) \). The unit step function \( u(t) \) is equal to 0 for \( t < 0 \) and equal to 1 for \( t \geq 0 \). Each term in the function will activate or deactivate based on the value of \( t \) relative to the step position.

Breaking Down Each Term of the Function

The function \( f(t) \) can be broken into four terms:1. \(-(20t + 400) u(t + 20)\), active for \( t \geq -20 \).2. \( (40t + 400) u(t + 10) \), active for \( t \geq -10 \).3. \( (400 - 40t) u(t - 10) \), active for \( t \geq 10 \).4. \( (20t - 400) u(t - 20) \), active for \( t \geq 20 \).Notice that each term adds or subtracts based on the timing and condition of the unit step function it is multiplied with.

Evaluating Intervals of Activation

Consider the intervals determined by the unit step functions:- For \( t < -20 \): All terms are off, so \( f(t) = 0 \).- For \( -20 \leq t < -10 \): Only the first term is active, so \( f(t) = -(20t + 400) \).- For \( -10 \leq t < 10 \): The first and second terms are active. Thus, \( f(t) = -(20t + 400) + (40t + 400) = 20t \).- For \( 10 \leq t < 20 \): The first, second, and third terms are active, so \( f(t) = 20t + (400 - 40t) = 400 - 20t \).- For \( t \geq 20 \): All terms are active, therefore \( f(t) = 400 - 20t + (20t - 400) = 0 \).

Sketching the Function

Using the evaluated intervals:- From \( t = -25 \) to \( -20 \), the function is 0.- From \( t = -20 \) to \(-10 \), the function is \( -(20t + 400) \), a negatively sloped line.- From \( t = -10 \) to \( 10 \), the function is \( 20t \), a positively sloped line through the origin.- From \( t = 10 \) to \( 20 \), the function is \( 400 - 20t \), a negatively sloped line reaching 0 at \( t = 20 \).- For \( t > 20 \), the function is again 0.Sketch each of these line segments accordingly on the interval \(-25 \leq t \leq 25\).

Confirm the Sketch Consistency

Review the endpoints and verify that the slopes and values match each transition point. Ensure continuity and accuracy in transitions across different sections of \( f(t) \). Specifically, confirm matching function values at \( t = -20, -10, 10, \) and \( 20 \) with the calculated piecewise segments. Revisiting each piecewise component for accuracy reinforces understanding.

Key concepts

Piecewise Functions

Piecewise functions, like the one given in the problem for the function \( f(t) \), are comprised of multiple sub-functions, each of which applies to a certain interval of the domain. In the provided function, each piece is defined by a linear equation active over specific time intervals, controlled by the unit step function.

These pieces activate at different points \( t \) and contribute to the overall shape and value of the function. For \( f(t) \), the function is broken down into four distinct segments, each determined by the time intervals dictated by combinations of linear terms and unit step functions.

• -20t + 400 is active for \( t \geq -20 \)
• 40t + 400 is active for \( t \geq -10 \)
• 400 - 40t is active for \( t \geq 10 \)
• 20t - 400 is active for \( t \geq 20 \)

By understanding how each piece contributes over its respective time range, students can gain insights into how piecewise functions behave and switch between different rules and equations depending on the value of the variable.

Graphing Functions

Graphing piecewise functions involves plotting each segment of the function over its respective interval and ensuring proper transition between pieces. For \( f(t) \), the graph consists of different linear segments according to the dictated formulas within certain ranges.

Start by plotting the function over the interval from \(-25 \leq t \leq 25\):
- Between \( t = -25 \) and \( t = -20 \), the function value is 0 because no piece is active.
- From \( t = -20 \) to \( t = -10 \), the piece \(-20t - 400 \) is active, creating a line with a negative slope.
- For \( t = -10 \) to \( t = 10 \), the equation becomes \(20t\), indicating a positively sloped line through the origin.
- Between \( t = 10 \) and \( t = 20 \), the line \(400 - 40t\) emerges, decreasing towards zero.
- From \( t = 20 \) to \( t = 25 \), the function returns to 0 because both terms cancel out.

It is important to carefully sketch these segments to ensure they connect correctly at their transition points at \( t = -20, -10, 10, \) and \( 20 \). Consistency and accuracy across these transitions help sketch a reliable graph demonstrating the behavior of \( f(t) \).

Function Evaluation

Evaluating piecewise functions such as \( f(t) \) involves determining the function's value point-by-point across its domain, mindful of the intervals where different pieces are active. The function's behavior changes as \( t \) moves from one interval to another. Here’s how to evaluate \( f(t) \) given specific values:

- For any \( t < -20 \), \( f(t) \) is 0 because no terms are activated by the unit steps.
- At \( t = -15 \), for instance, only the first term \(-20t-400\) contributes, resulting in \( f(-15) = -100\).
- When \( t = 0 \), both of the first two terms are operational, giving \( f(0) = 400 \).
- Moving to \( t = 15 \), active terms change to \(20t - 400\), yielding \( f(15) = 100 \).
- Once \( t \ge 20 \), all terms contribute equally inversely, neutralizing \( f(t) \) back to 0 again.

Knowing which segment contributes at these different points allows accurate calculation of the function's output. This methodical evaluation enhances understanding of when and how each piece affects the overall function.