Problem 101 - Thomas Calculus

Question

Finding displacement from an antiderivative of velocity a. Suppose that the velocity of a body moving along the s-axis is $$ \frac{d s}{d t}=v=9.8 t-3 $$ \begin{equation} \begin{array}{l}{\text { i) Find the body's displacement over the time interval from }} \\ {t=1 \text { to } t=3 \text { given that } s=5 \text { when } t=0 \text { . }} \\ {\text { ii) Find the body's displacement from } t=1 \text { to } t=3 \text { given }} \\ {\text { that } s=-2 \text { when } t=0 \text { . }} \\ {\text { iii) Now find the body's displacement from } t=1 \text { to } t=3} \\ {\text { given that } s=s_{0} \text { when } t=0 .}\end{array} \end{equation} b. Suppose that the position \(s\) of a body moving along a coordinate line is a differentiable function of time \(t .\) Is it true that once you know an antiderivative of the velocity function \(d s / d t\) you can find the body's displacement from \(t=a\) to \(t=b\) even if you do not know the body's exact position at either of those times? Give reasons for your answer.

Step-by-Step Solution

Answer

Displacement from \( t=1 \) to \( t=3 \) is 27.4 for any initial position. Antiderivatives give displacement without known exact positions.

Step 1: Calculate the Antiderivative for Problem a

The given velocity function is \( v = 9.8t - 3 \). We need to find the position function \( s(t) \) by integrating the velocity function. The integration of \( v \) will give \( s(t) = \int (9.8t - 3) \, dt = 4.9t^2 - 3t + C \), where \( C \) is the integration constant representing the initial position.

Step 2: Solve for Part i using Initial Condition

Given \( s=5 \) when \( t=0 \), substitute into the position function to find \( C \): \[ 5 = 4.9(0)^2 - 3(0) + C \] Thus, \( C = 5 \). Therefore, the position function is \( s(t) = 4.9t^2 - 3t + 5 \). Now calculate the displacement from \( t=1 \) to \( t=3 \):\[ s(3) - s(1) = (4.9(3)^2 - 3(3) + 5) - (4.9(1)^2 - 3(1) + 5) = 34.3 - 6.9 = 27.4 \].

Step 3: Solve for Part ii using New Initial Condition

Now given \( s=-2 \) when \( t=0 \), substitute into the position function: \[ -2 = 4.9(0)^2 - 3(0) + C \] Thus, \( C = -2 \). Therefore, the position function is \( s(t) = 4.9t^2 - 3t - 2 \). Calculate the displacement from \( t=1 \) to \( t=3 \):\[ s(3) - s(1) = (4.9(3)^2 - 3(3) - 2) - (4.9(1)^2 - 3(1) - 2) = 27.4 \].

Step 4: Solve for Part iii with Generic Initial Position

For initial position \( s=s_0 \) when \( t=0 \), equate it to find \( C \): \[ s_0 = 4.9(0)^2 - 3(0) + C \] Thus, \( C = s_0 \). Therefore, \( s(t) = 4.9t^2 - 3t + s_0 \). Calculate the displacement from \( t=1 \) to \( t=3 \):\[ s(3) - s(1) = 27.4 \], independent of \( s_0 \).

Step 5: Answer Problem b Regarding Displacement

Even without knowing the exact position, knowing an antiderivative gives displacement between \( t=a \) and \( t=b \) because displacement depends on the difference in position: \( s(b) - s(a) = F(b) - F(a) \), where \( F \) is an antiderivative. Changes in position don't depend on absolute position but on position differences.

Key Concepts

Understanding Antiderivatives

An antiderivative is a concept from calculus, involving the reverse process of differentiation. When you know the rate of change or the derivative of a function, finding the original function requires determining its antiderivative. For example, if you have a velocity function, integrating it gives you the position function. This is because the velocity is the derivative of the position function with respect to time.

In mathematical terms, if the velocity, represented as \( v(t) \), is given by \( 9.8t - 3 \), then the antiderivative of this expression \( s(t) \) is found through integration. Thus, \( s(t) = \int (9.8t - 3) \, dt \). Solving it provides \( s(t) = 4.9t^2 - 3t + C \), where \( C \) is the constant representing the initial condition or starting position.

When calculating an antiderivative, it's crucial to consider the constant \( C \), as it can vary based on different initial conditions, which influence the position function derived from the velocity function.

The Role of the Velocity Function

The velocity function describes how quickly something changes its position over time. In physics and calculus, it is essential for understanding motion. For the problem at hand, the velocity function is given by \( v = 9.8t - 3 \).

A velocity function offers insight into both the speed and direction of motion in a single expression. The function is the first derivative of the position function. It shows how much the position changes with time and whether the movement is forwards or backwards. The formula \( \frac{d s}{d t} = v \) demonstrates this dynamic, with \( s \) representing the position at time \( t \).

Analyzing the velocity function can help you determine the nature of motion—the higher the velocity value, the faster the object moves. Integration of this function gives insight into how an object's position changes over a time interval.

Integration in Calculus

Integration is a fundamental concept in calculus. It allows you to determine an area's, volume's, or, in this context, position function from a given rate of change. You integrate to find the antiderivative or the original function from its derivative.

To find the position of a moving object, you integrate the velocity function. For example, given the velocity function \( v(t) = 9.8t - 3 \), integrating it helps obtain the position function: \( s(t) = 4.9t^2 - 3t + C \). This antiderivative includes an arbitrary constant \( C \), influenced by initial conditions.

Integration also aids in computing displacement over a time interval \([a, b] \) by evaluating the definite integral \( \int_{a}^{b} v(t) \, dt \). This approach helps identify the total change in position without requiring the exact position at each time point, solely depending on differences in the values of the antiderivative.

The Importance of Initial Conditions

In calculus, initial conditions play a crucial role when integrating to find antiderivatives. They help pinpoint the constant \( C \) in the antiderivative function. Without known initial conditions, the antiderivative includes \( C \), which remains undetermined.

For instance, if you are informed that \( s = 5 \) when \( t = 0 \), you can input these values into the antiderivative to find \( C \). With \( s(t) = 4.9t^2 - 3t + C \), substituting \( t = 0 \) and \( s = 5 \) results in \( C = 5 \). Thus, your position function becomes \( s(t) = 4.9t^2 - 3t + 5 \).

Initial conditions are more than just arbitrary starting points—they are key to refining mathematical models and ensuring accurate calculations in terms of physical positioning and movement over time.