Problem 9 - Calculus Concepts

Question

a. use the limit definition of the derivative (algebraic method) to write an expression for the rate-of-change function of the given function. b. evaluate the rate of change as indicated. \(\quad g(t)=4 t^{2}-3 ;\left.\frac{d g}{d t}\right|_{t=4}\)

Step-by-Step Solution

Answer

The rate of change at \( t = 4 \) is 32.

Step 1: Apply the Limit Definition

The limit definition of the derivative of a function \( g(t) \) is given by \( \lim_{{h \to 0}} \frac{{g(t+h) - g(t)}}{h} \). Substitute \( g(t) = 4t^2 - 3 \) into this formula.

Step 2: Substitute the Function into the Definition

Substitute \( g(t+h) = 4(t+h)^2 - 3 \) and \( g(t) = 4t^2 - 3 \) into the limit definition: \[ \lim_{{h \to 0}} \frac{{4(t+h)^2 - 3 - (4t^2 - 3)}}{h}. \] Simplify the expression to: \[ \lim_{{h \to 0}} \frac{{4(t^2 + 2th + h^2) - 4t^2}}{h}. \]

Step 3: Simplify the Expression

Expand \( 4(t+h)^2 \) to \( 4t^2 + 8th + 4h^2 \). Simplify the numerator: \( 4t^2 + 8th + 4h^2 - 4t^2 \). It simplifies to \( 8th + 4h^2 \). The expression becomes: \[ \lim_{{h \to 0}} \frac{{8th + 4h^2}}{h}. \]

Step 4: Factor and Simplify Further

Factor out \( h \) in the numerator: \( h(8t + 4h) \). The expression simplifies to \( \lim_{{h \to 0}} (8t + 4h) \).

Step 5: Evaluate the Limit as h Approaches 0

As \( h \) approaches 0, the limit \( 8t + 4h \) becomes \( 8t \). Thus, the rate-of-change function (derivative) is \( \frac{dg}{dt} = 8t \).

Step 6: Evaluate the Derivative at \( t = 4 \)

Plug \( t = 4 \) into the derivative \( \frac{dg}{dt} = 8t \). Calculate \( 8(4) = 32 \). Thus, \( \left.\frac{dg}{dt}\right|_{t=4} = 32 \).

Key Concepts

Rate of Change

The concept of rate of change is essential in understanding how a function behaves as its input variable changes. In calculus, the rate of change of a function is described by its derivative. The derivative provides a way to measure how a small change in the input of a function affects the output. For instance, if we have a function like \( g(t) = 4t^2 - 3 \), we can understand how this function changes as \( t \) changes by using the derivative.The derivative, in this context, represents the instantaneous rate of change of the function \( g(t) \). So, when we evaluate the derivative, we are essentially calculating how steep the curve of the function is at any specific point. This is particularly helpful in practical scenarios like physics, where you might calculate how quickly a car's position changes over time, indicating its velocity.

Algebraic Methods

Algebraic methods play a crucial role in finding the derivative of a function using the limit definition. This process involves a series of steps that simplify the expression to eventually find the derivative.

1. First, we use the limit definition of the derivative \( \lim_{{h \to 0}} \frac{{g(t+h) - g(t)}}{h} \). For our function \( g(t) = 4t^2 - 3 \), we need to substitute each \( t \) with \( t + h \) to find \( g(t+h) \).
2. Next, we substitute \( g(t+h) = 4(t+h)^2 - 3 \) and \( g(t) = 4t^2 - 3 \) into the limit formula, which then involves some algebra to simplify \( 4(t^2 + 2th + h^2) - 4t^2 \).
3. Continuing with the algebra, the expression simplifies to \( 8th + 4h^2 \).
4. Factor out \( h \) from the numerator which allows us to further reduce the expression to \( 8t + 4h \).

With these methods, students can approach derivatives in a systematic way that reduces errors and clarifies each step.

Derivative Evaluation

Evaluating the derivative involves finding the derivative at a specific point. Once the algebraic simplification is complete, the next step is calculating the actual value of the derivative.We found the derivative of \( g(t) = 4t^2 - 3 \) to be \( \frac{dg}{dt} = 8t \). To evaluate the derivative at a given point, such as \( t = 4 \), simply substitute this value into the derivative formula:

• Plug \( t = 4 \) into the derivative function \( 8t \).
• Calculate \( 8 \times 4 = 32 \).

Hence, \( \left.\frac{dg}{dt}\right|_{t=4} = 32 \), which means at \( t = 4 \), the rate of change of the function \( g(t) \) is 32. This specific evaluation offers insight into how rapidly the function is changing at that particular moment.