Problem 29 - Applied Calculus

Question

For \(f(t)=4-2 e^{t}\), find \(f^{\prime}(-1), f^{\prime}(0)\), and \(f^{\prime}(1)\). Graph \(f(t)\), and draw tangent lines at \(t=-1, t=0\), and \(t=1 .\) Do the slopes of the lines match the derivatives you found?

Step-by-Step Solution

Answer

Derivatives are -\frac{2}{e}, -2, -2e at t = -1, 0, 1 respectively; these match tangent line slopes.

Step 1: Find the Derivative of the Function

The first step is to find the derivative of the function \(f(t) = 4 - 2e^{t}\). The derivative \(f'(t)\) can be found by differentiating each term separately. The derivative of 4, a constant, is 0. The derivative of \(-2e^{t}\) is \(-2e^{t}\), since the derivative of \(e^{t}\) is \(e^{t}\) and we multiply by -2. Therefore, the derivative \(f'(t) = -2e^{t}\).

Step 2: Evaluate the Derivative at t = -1

To find \(f'(-1)\), substitute \(t = -1\) into the derivative \(f'(t) = -2e^{t}\). Thus, \(f'(-1) = -2e^{-1}\). Simplify to \(f'(-1) = -\frac{2}{e}\).

Step 3: Evaluate the Derivative at t = 0

To find \(f'(0)\), substitute \(t = 0\) into the derivative \(f'(t) = -2e^{t}\). Thus, \(f'(0) = -2e^{0}\). Since \(e^{0} = 1\), we have \(f'(0) = -2\).

Step 4: Evaluate the Derivative at t = 1

To find \(f'(1)\), substitute \(t = 1\) into the derivative \(f'(t) = -2e^{t}\). Thus, \(f'(1) = -2e^{1}\). Simplify to \(f'(1) = -2e\).

Step 5: Graph the Function and Tangent Lines

Graph the function \(f(t) = 4 - 2e^{t}\) and plot points at \(t = -1, 0,\) and \(1\). Then draw tangent lines at these points using the slopes found: \(-\frac{2}{e}\), \(-2\), and \(-2e\) respectively. Check that these slopes match the corresponding derivative values at these points.

Key Concepts

Derivative Evaluation

Derivatives are fundamental in calculus, particularly for understanding how functions change. Evaluating a derivative at a specific point gives us the slope of the tangent line to the function at that point.
In our example, the function is given by \( f(t) = 4 - 2e^{t} \). To find the derivative, we differentiate each term individually.
- The derivative of a constant, like 4, is zero. - The term \( -2e^{t} \) differentiates to itself multiplied by -2, becoming \( -2e^{t}. \) Thus, the derivative of the entire function is \(-2e^{t}.\)
Once the derivative is found, we can plug in different values of \(t\) to evaluate it at specific points:

• At \(t = -1\), the derivative is \( -\frac{2}{e}. \)
• At \(t = 0\), it simplifies to \(-2\).
• At \(t = 1\), it is \(-2e.\)

Each of these gives us important information about how the function is behaving at those particular \(t\) values.

Tangent Lines

Tangent lines play a crucial role in visualizing how a function behaves at a specific point. The slope of a tangent line at any point on a graph indicates how steep the graph is at that point.
When we calculate the derivative at a certain point, we are essentially finding the slope of the tangent line at that point. This is why evaluating the derivative is so useful: it tells us about the function's instantaneous rate of change.
For the function \( f(t) = 4 - 2e^{t} \), tangent lines at specific points (like \(t= -1, 0,\) and \(1\)) help us visualize this rate of change.

• At \(t = -1\), the tangent slope is \( -\frac{2}{e}.\)
• At \(t = 0\), the slope is \(-2\), a straight line slope across.
• At \(t = 1\), the slope becomes steeper, at \(-2e.\)

These lines illustrate how rapidly the \(e^{t}\) component causes the overall function to change, highlighting the dramatic impacts of exponential growth or decay.

Exponential Functions

Exponential functions, like \(e^{t}\), are powerful mathematical tools used in many fields such as finance, physics, and biology. They have the distinct feature of constant proportional change, meaning they grow or decay at rates dependent on their current value.
In the example function \(f(t)=4-2e^{t}\), the exponential term \(e^{t}\) shows how one's simple expression can lead to a broad range of behaviors. The \'exponential\' character becomes clear as the function moves further along the \(t\)-axis.

• When \(t\) is negative, \(e^{t}\) brings down the value towards zero more gradually.
• At \(t=0\), \(e^{t}=1\), causing the shift in \(f(t)\) to be more moderate.
• As \(t\) becomes positive, the value of \(e^{t}\) increases swiftly, leading to rapid changes in the function's behavior.

Understanding exponential functions in calculus contexts is essential, as they demonstrate significant real-world phenomena and guide us in predicting future outcomes based on current trends.