Problem 45 - Applied Calculus

Question

Estimate the relative rate of change of \(f(t)=t^{2}\) at \(t=4\). Use \(\Delta t=0.01\).

Step-by-Step Solution

Answer

The relative rate of change at \( t = 4 \) is 0.5.

Step 1: Define Relative Rate of Change

The relative rate of change of a function \( f(t) \) at a point \( t \) is given by \( \frac{f'(t)}{f(t)} \), where \( f'(t) \) is the derivative of \( f(t) \). This measures how fast the function \( f(t) \) is changing at this point relative to its value.

Step 2: Compute the Derivative

Find the derivative of \( f(t) = t^2 \). Using the power rule, the derivative is \( f'(t) = 2t \).

Step 3: Calculate the Derivative at t=4

Substitute \( t = 4 \) into the derivative to get \( f'(4) = 2 \times 4 = 8 \). This represents the instantaneous rate of change of \( f(t) \) at \( t = 4 \).

Step 4: Evaluate the Function at t=4

Calculate \( f(4) = 4^2 = 16 \). This is the value of the function at \( t = 4 \).

Step 5: Compute the Relative Rate of Change

Now, calculate the relative rate of change using the formula \( \frac{f'(t)}{f(t)} \) at \( t = 4 \). This becomes \( \frac{8}{16} = 0.5 \).

Step 6: Short Conclusion

The relative rate of change of \( f(t) = t^2 \) at \( t = 4 \) is \( 0.5 \), meaning the function's rate of change is 50% of its value at this point.

Key Concepts

Derivative Calculation

When we talk about derivative calculation, we are diving into the heart of calculus. Derivatives help us understand how a function changes at any point. Consider a function, like \( f(t) = t^2 \). The derivative, denoted as \( f'(t) \), tells us how fast \( f(t) \) is changing with respect to \( t \).

The process involves taking the derivative of the given function. In our exercise, after applying calculus techniques, we find the derivative of \( f(t) = t^2 \) by using the power rule. The derivative is calculated as \( f'(t) = 2t \). This formula shows that for any value of \( t \), \( f'(t) \) gives the slope of the tangent line to the curve at that point.

Understanding derivatives and how to calculate them is crucial in many real-world applications. For instance, derivatives are vital in determining rates and predicting trends.

Instantaneous Rate of Change

The instantaneous rate of change of a function at a specific point gives us the slope of the curve at that point. It tells us how the function behaves at \'t\' right now. For \( f(t) = t^2 \) and \( t=4 \), the instantaneous rate of change is found by evaluating the derivative at \( t=4 \).

Substituting \( t = 4 \) into the derivative \( f'(t)=2t \), we calculate \( f'(4) = 2 \times 4 = 8 \).

• This means that at \( t = 4 \), the function \( f(t) \) is changing at a rate of 8 units per unit \( t \).
• This \'instantaneous\' measure helps us understand how the function \'f\' is increasing or decreasing precisely at that moment.

This concept is essential in physics, where it explains how fast something happens exactly at a specific point in time, like speed in an instant.

Power Rule

The power rule is one of the most simple and useful tools for finding derivatives of polynomial functions. It states: if you have a function \( f(t) = t^n \), then its derivative \( f'(t) = nt^{n-1} \). This rule simplifies the process of differentiation and makes it much faster.

In the exercise, we had \( f(t) = t^2 \). Using the power rule here, we bring down the exponent \( 2 \) and reduce it by one, leading to \( f'(t) = 2t^{2-1} = 2t \).

• This rule is fundamental because it applies to all polynomial functions and is often used in conjunction with other rules to solve more complex problems.
• Knowing when and how to apply it enables students to tackle broader calculus scenarios easily.

Applying the power rule is a crucial skill in calculus, especially for quickly finding the rates at which problems involving polynomials change.