Problem 39 - Applied Calculus

Question

Let \(f(t)=t^{2}-4 t+5\) (a) Find \(f^{\prime}(t)\). (b) Find \(f^{\prime}(1)\) and \(f^{\prime}(2)\). (c) Use a graph of \(f(t)\) to check that your answers to part (b) are reasonable. Explain.

Step-by-Step Solution

Answer

(a) \( f'(t) = 2t - 4 \). (b) \( f'(1) = -2 \), \( f'(2) = 0 \). (c) Graph shows these as decreasing and vertex points.

Step 1: Find the Derivative

To find the derivative of the function \( f(t) = t^2 - 4t + 5 \), apply the power rule for differentiation. The power rule states that the derivative of \( t^n \) is \( nt^{n-1} \). Therefore, differentiate each term:- The derivative of \( t^2 \) is \( 2t \).- The derivative of \( -4t \) is \( -4 \).- The derivative of the constant \( 5 \) is \( 0 \).Thus, \( f'(t) = 2t - 4 \).

Step 2: Evaluate f'(t) at Specific Points

To find \( f'(1) \) and \( f'(2) \), substitute \( t = 1 \) and \( t = 2 \) into the derivative function \( f'(t) = 2t - 4 \).- For \( f'(1) \): \[ f'(1) = 2(1) - 4 = 2 - 4 = -2 \]- For \( f'(2) \): \[ f'(2) = 2(2) - 4 = 4 - 4 = 0 \]

Step 3: Graphical Verification

To verify the results graphically, consider the graph of \( f(t) = t^2 - 4t + 5 \). This is a parabola opening upwards, with its vertex occurring where the first derivative equals zero.- At \( t = 2 \), \( f'(2) = 0 \), which corresponds to the vertex, indicating a minimum point on the graph.- At \( t = 1 \), \( f'(1) = -2 \), which indicates a negative slope, showing the function is decreasing at \( t = 1 \).Both these insights are consistent with the behavior of a standard upward-opening parabola.

Key Concepts

Power Rule

The power rule is an essential technique in calculus for finding derivatives of polynomial functions. When you have a function of the form \( t^n \), the power rule lets you find its derivative easily by multiplying the exponent \( n \) by the base \( t \) raised to the power of \( n-1 \). So, for any term \( t^n \), the derivative \( \frac{d}{dt}(t^n) = nt^{n-1} \).

Consider the function \( f(t) = t^2 - 4t + 5 \). Here's how the power rule is applied to differentiate each term:

• The term \( t^2 \) becomes \( 2t \), as you multiply 2 (the exponent) by \( t \) raised to the power of \( 2-1 \).
• The term \( -4t \) simply becomes \( -4 \), since \( t^1 \) follows the rule \( 1\cdot t^{0}=1 \).
• Constant terms like \( 5 \) have a derivative of zero, because their slope doesn’t change.

Combining these, the derivative is \( f'(t) = 2t - 4 \). This concise rule speeds up the differentiation process.

Graphical Verification

Graphical verification is a powerful tool to confirm the derivative calculations. By analyzing the graph of a function, we can visually interpret changes in slope.

For our function \( f(t) = t^2 - 4t + 5 \), we notice it represents a parabola that opens upwards. The vertex, where the slope changes and \( f'(t) = 0 \), is a key feature. This vertex represents the minimum point on the graph.

When we calculate \( f'(2) = 0 \), it suggests \( t = 2 \) is where the slope is zero, confirming the minimum at the vertex of the parabola. Similarly, at \( t = 1 \), \( f'(1) = -2 \), this negative slope indicates the graph is descending at that point. Through this, graphical verification solidifies our understanding of the function's behavior.

Bear in mind:

• A positive derivative indicates an increasing function.
• A negative derivative indicates a decreasing function.
• A zero derivative indicates stationary points — potential minima or maxima.

Derivative Evaluation

Evaluating derivatives involves substituting specific points into the derivative function to assess the function's rate of change at those points.

For the derivative \( f'(t) = 2t - 4 \):

• To find \( f'(1) \), substitute \( t = 1 \) into the equation: \[ f'(1) = 2(1) - 4 = -2 \]
• Similarly, to find \( f'(2) \), substitute \( t = 2 \): \[ f'(2) = 2(2) - 4 = 0 \]

These evaluations tell us that at \( t = 1 \), the function is decreasing with a slope of \(-2\), showing a downward trend. At \( t = 2 \), the slope is zero, indicating a change in direction from decreasing to increasing — a critical point.

Derivative evaluation provides precise points to reflect the function’s behavior and verify the actual shape of the graph.