Problem 40 - Essential Calculus Early Transcendentals

Question

A particle is moving with the given data. Find the position of the particle. $$v(t)=1.5 \sqrt{t}, \quad s(4)=10$$

Step-by-Step Solution

Answer

The position function is \(s(t) = t^{3/2} + 2\).

Step 1: Understand the Problem

We need to find the position function, \(s(t)\), of a particle given its velocity function \(v(t) = 1.5 \sqrt{t}\) and the information that \(s(4) = 10\). The velocity function is the derivative of the position function.

Step 2: Set Up the Integral

The velocity \(v(t)\) is the derivative of position \(s(t)\), i.e., \(v(t) = \frac{ds}{dt}\). To find \(s(t)\), we will integrate \(v(t)\). The indefinite integral is given by\[ s(t) = \int v(t) \, dt = \int 1.5 \sqrt{t} \, dt.\]

Step 3: Perform the Integration

Calculate the indefinite integral.\[ s(t) = \int 1.5 \sqrt{t} \, dt = 1.5 \int t^{1/2} \, dt = 1.5 \left( \frac{t^{3/2}}{3/2} \right) + C = t^{3/2} + C.\]So, \(s(t) = t^{3/2} + C\).

Step 4: Use Initial Condition to Solve for C

We use the initial condition \(s(4) = 10\) to find the constant \(C\). Substitute \(t = 4\) and \(s(4) = 10\) into the position function:\[10 = 4^{3/2} + C.\]Calculate \(4^{3/2} = 8\):\[10 = 8 + C \, \Rightarrow \, C = 2.\]

Step 5: Write the Position Function

With \(C = 2\), the position function is:\[s(t) = t^{3/2} + 2.\]

Key Concepts

Position Function

The position function is a mathematical expression that describes the position of a particle at any given time, often denoted as \( s(t) \). In calculus, the position function is an important tool for understanding the path or location of a moving object over time. It gives us a way to determine where the particle is at any point in time \( t \), based on the particle's velocity and initial starting conditions.

To find the position function, we often start with the velocity function, which represents how fast and in what direction the particle is moving. By integrating the velocity function, we can determine the position function. This is because the derivative of the position function is the velocity function. Therefore, integration essentially "reverses" the process of finding the derivative.

• Identifying the position over time is crucial for understanding motion or planning further analysis.
• The position function helps in predicting future locations or in analyzing past movements of an object.

Whenever you need to determine the position function, look for the given velocity function and any initial conditions, such as the particle's position at a specific time.

Velocity Function

The velocity function is a fundamental component in understanding motion, specifically how fast and in which direction an object is moving. It's denoted usually by \( v(t) \), where \( t \) is time. The velocity function provides the rate of change of position with respect to time. Hence, it's the derivative of the position function \( s(t) \).

For the given exercise, the velocity function \( v(t) = 1.5 \sqrt{t} \) showcases that the speed of the particle changes with time in a specific way. As time progresses, the velocity changes depending on the square root of the time multiplied by a constant.

• Velocity indicates both speed and direction of an object.
• Understanding the velocity allows us to compute how the position of an object is evolving over time.

By integrating the velocity function, we will be able to find the position function which gives us the actual position at any given time point.

Definite Integral

A definite integral is a fundamental concept in calculus used to compute the accumulation of quantity, like area under a curve, or in this context, the net change in position. When we perform a definite integral, we evaluate an integral over a specific interval, providing a precise value.

For instance, if you want to determine the exact position of a particle between two time points, a definite integral of the velocity function over that time period will offer the exact change in position. This process helps to calculate the interval of interest, thereby converting velocity into position with the bounds of integration based on the initial and final time points.

• It's essential for finding the total displacement of an object over a set period.
• Definite integrals are important in not only understanding motion but also in physics and engineering for real-world problem solving.

Although the exercise focuses on finding an indefinite integral, definite integrals play a critical role when boundaries and specific intervals are involved.

Initial Condition

Initial conditions are constraints or known values at a particular time point that help in solving differential equations or integrals. In this case, the initial condition is given by \( s(4) = 10 \), indicating that at time \( t = 4 \), the position of the particle is 10 units.

Initial conditions are vital because they allow us to determine constants of integration when finding indefinite integrals. Without these known values, we would only be able to determine a family of possible functions rather than a specific one.

• Initial conditions refine our solutions, ensuring they fit the specific scenario.
• Knowing the precise position at a starting point helps in calculating constants and thereby the exact form of position or other functions.

In any problem involving integration, applying initial conditions helps tailor the function to accurately reflect the situation being analyzed.