Problem 16 - Differential Equations and Linear Algebra

Question

Determine \(f(t)\). $$f(t-3)=t e^{-(t-3)}$$.

Step-by-Step Solution

Answer

The function \(f(t)\) is given by \(f(t) = (t+3)e^{-t}\).

Step 1: Substitute t' + 3 for t

We will substitute \(t' + 3\) in the place of \(t\), so the function becomes: \(f((t'+3)-3) = (t'+3)e^{-((t'+3)-3)}\)

Step 2: Simplify the function

Now let's simplify the function. Notice that the term \((t'+3)-3\) in the argument of \(f()\) is equal to \(t'\). We can now rewrite the function as: \(f(t') = (t'+3)e^{-(t')}\)

Step 3: Rename the variable

Now that we have found the function \(f(t')\), we can rename the variable \(t'\) back to \(t\), giving us the final function: \(f(t) = (t+3)e^{-t}\) Therefore, the function \(f(t)\) is given by \(f(t) = (t+3)e^{-t}\).

Key Concepts

Function Transformation

When dealing with functions, transformations are an essential concept that helps us manipulate and understand the graphical representation of equations. In this context, **function transformation** applies to altering functions to simplify or solve them. If you encounter a problem involving function transformation, often it comes down to shifting, stretching, or compressing the function in some way.

• A simple example is shifting a function horizontally or vertically.
• For our specific case with the function \( f(t-3) = t e^{-(t-3)} \), we apply a horizontal shift.
• This shift happens because of the "\(+3\)" inside the \( f(t) \), which means we move the entire function graph 3 units to the right.

This concept is useful when trying to identify the underlying original form of a given function. In practice, you recalibrate the input, as was done by substituting \( t' + 3 \). This results in obtaining \( f(t') = (t'+3)e^{-t'} \), which when back-substituted and simplified, allows for identifying \( f(t) \).

Exponential Functions

Exponential functions are pivotal in both mathematics and real-world applications, characterized by a constant base raised to a variable exponent. An example is \( e^{-t} \), which exponentially decreases as \( t \) increases. Understanding how these functions behave is crucial:

• Exponential functions grow or decay rapidly; this behavior is determined by the sign of the exponent.
• For \( e^{-t} \), notice how it represents exponential decay, meaning the function's values taper off as \( t \) becomes large.
• In our problem, this exponential decay is coupled with a linear term \( (t+3) \), modifying the standard exponential curve.

When paired with linear functions, as seen in \( (t+3)e^{-t} \), exponential functions yield a curve that starts slow, peaks, and then declines, revealing a transformation of both rate and amplitude associated with the curve itself. This dual nature gives us unique and powerful tools to model phenomena like radioactive decay or cooling processes.

Variable Substitution

Variable substitution serves as a bridge to simplify complex expressions and functions, particularly beneficial in solving equations or evaluating integrals. This method involves accurately replacing a variable to make an equation easier to grasp or solve.

• The substitution of \( t' + 3 = t \) in the problem helps eliminate confusion and directly leads to simplifying the function.
• By focusing on simplifying the expression with \( t' \), the equation becomes more transparent as we progress.
• Once simplified, the renaming of \( t' \) back to \( t \) is essential for making the function \( f(t) \) consistent with commonly used notation.

This is particularly insightful in dynamic systems where understanding the basic form after substitution lays the groundwork for further interpretations or integrations. In essence, variable substitution is akin to peeling layers of an onion, where each step reduces complexity for clearer insight into the original problem statement.