Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Show that for \(s(t)=\frac{g}{2} t^{2}+v_{0} t+\gamma, s^{\prime}(t)=g t+v_{0}\).

Step-by-Step Solution

Answer

\( s'(t) = g t + v_0 \).

Step 1: Identify the Function

We are given the function \( s(t) = \frac{g}{2} t^2 + v_0 t + \gamma \). This is a position function of time \( t \).

Step 2: Differentiate the Function

To find \( s'(t) \), differentiate \( s(t) \) with respect to \( t \). Use the power rule for differentiation: \( \frac{d}{dt} \left( \frac{g}{2} t^2 \right) = g t \), \( \frac{d}{dt} \left( v_0 t \right) = v_0 \), and \( \frac{d}{dt} (\gamma) = 0 \).

Step 3: Calculate the Derivative

Apply the differentiation rules: \( s'(t) = \frac{d}{dt} \left( \frac{g}{2} t^2 + v_0 t + \gamma \right) = g t + v_0 + 0 = g t + v_0 \).

Step 4: Interpret the Result

The derivative \( s'(t) = g t + v_0 \) represents the velocity function derived from the original position function \( s(t) \). The calculation confirms that \( s'(t) \) is indeed \( g t + v_0 \).

Key Concepts

Position Function

The position function is a fundamental concept in calculus that relates to how an object's position changes over time. In mathematical terms, it is designated as the function \( s(t) \), where \( s \) represents the position and \( t \) represents time. This function can include several terms, each with a different rate of contribution to the position.

In our exercise, the position function \( s(t) = \frac{g}{2} t^2 + v_0 t + \gamma \), involves:

• \( \frac{g}{2} t^2 \): This term represents the influence of acceleration due to gravity over time. It's a quadratic component that shows how position changes in relation to time squared.
• \( v_0 t \): This linear term accounts for initial velocity. It shows how the object's position changes over time at a constant speed.
• \( \gamma \): A constant term, often used to signify an initial position at time \( t=0 \).

Understanding the position function allows us to predict where an object will be at any given time, making it an indispensable tool in physics and engineering.

Velocity Function

The velocity function is derived from the position function and provides information about the speed and direction of an object at any point in time. It is denoted as \( s'(t) \), which signifies the derivative of the position function \( s(t) \).

In simple terms, differentiating the position function gives you the velocity function. From the exercise, by differentiating \( s(t) = \frac{g}{2} t^2 + v_0 t + \gamma \), we obtain the velocity function \( s'(t) = gt + v_0 \). Here,

• \( gt \): Represents the acceleration as it contributes linearly to velocity over time.
• \( v_0 \): Shows the constant component of the velocity, which is the initial speed at the start.

By understanding an object's velocity, we can understand how quickly it's moving and in what direction, allowing for important calculations in motion analysis.

Power Rule of Differentiation

The power rule of differentiation is a key technique in calculus for finding the derivative of polynomials. This simplifies the process of differentiation by providing a straightforward formula to apply. For a function of the form \( x^n \), the power rule states that the derivative is \( nx^{n-1} \).

In the context of our exercise, the power rule is applied to each term of the position function \( s(t) = \frac{g}{2} t^2 + v_0 t + \gamma \). Here's how:

• For \( \frac{g}{2} t^2 \), the derivative is \( 2 \cdot \frac{g}{2} t^{2-1} = gt \).
• For \( v_0 t \), the derivative is \( v_0 \times 1 \cdot t^{1-1} = v_0 \).
• For \( \gamma \), a constant, the derivative is \( 0 \).

Using the power rule simplifies finding the velocity function \( s'(t) = gt + v_0 \). Understanding and applying the power rule is crucial for solving calculus problems efficiently.