Problem 34 - Calculus Early Transcendentals

Question

Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position. $$a(t)=\frac{2 t}{\left(t^{2}+1\right)^{2}}, v(0)=0, s(0)=0$$

Step-by-Step Solution

Answer

Given an object with acceleration function \(a(t) = \frac{2 t}{\left(t^{2}+1\right)^{2}}\), and initial conditions \(s(0) = 0\) and \(v(0) = 0\), we found its position and velocity functions: Position function: \(s(t) = -\arctan(t) + t\) Velocity function: \(v(t) = -\frac{1}{t^2 +1} + 1\)

Step 1: Integrate acceleration to find velocity function

We will integrate the acceleration function, \(a(t)\), with respect to time to find the velocity function, \(v(t)\). Remember to add a constant, \(C_1\), which we will determine using the initial velocity condition. $$v(t) = \int a(t) dt = \int \frac{2 t}{\left(t^{2}+1\right)^{2}} dt$$ To integrate this function, we can use substitution. Let: $$u = t^2 + 1$$ Then: $$\frac{du}{dt} = 2t,~~ dt = \frac{du}{2t}$$ Now we can rewrite our integral: $$v(t) = \int \frac{2t}{u^2} \cdot \frac{du}{2t} = \int \frac{1}{u^2} du$$ Now, we can integrate the function with respect to \(u\): $$v(t) = -\frac{1}{u} + C_1 = -\frac{1}{t^2 +1} + C_1$$

Step 2: Use initial velocity condition to find integration constant \(C_1\)

To find the value of \(C_1\), we'll use the initial condition \(v(0) = 0\). $$0 = -\frac{1}{0^2 +1} + C_1$$ Solving for the constant \(C_1\), we get: $$C_1 = 1$$ This gives us our velocity function as: $$v(t) = -\frac{1}{t^2 +1} + 1$$

Step 3: Integrate velocity to find position function

Now, we'll integrate the velocity function, \(v(t)\), with respect to time to find the position function, \(s(t)\). Remember to add another constant, \(C_2\), which we will determine using the initial position condition. $$s(t) = \int v(t) dt = \int \left(-\frac{1}{t^2 +1} + 1\right) dt$$ We can split the integral: $$s(t) = \int -\frac{1}{t^2 +1} dt + \int 1 dt$$ The first term is an arctangent integral, and the second term is a simple integral. Hence: $$s(t) = -\arctan(t) + t + C_2$$

Step 4: Use initial position condition to find integration constant \(C_2\)

To find the value of \(C_2\), we'll use the initial condition \(s(0) = 0\). $$0 = -\arctan(0) + 0 + C_2$$ Solving for \(C_2\), we get: $$C_2 = 0$$ This gives us our position function as: $$s(t) = -\arctan(t) + t$$

Step 5: Conclusion

Thus, we have found the position and velocity functions for the given problem: $$s(t) = -\arctan(t) + t$$ $$v(t) = -\frac{1}{t^2 +1} + 1$$

Key Concepts

Integration

Integration is a fundamental concept in calculus, crucial for finding quantities like area under a curve or, as in this problem, determining velocity and position from acceleration. When we integrate a function, we essentially accumulate the quantity represented by that function. For example, integrating acceleration gives us velocity, while integrating velocity gives us position.

In this particular problem, the acceleration function is given by:

• \( a(t) = \frac{2t}{(t^2 + 1)^2} \)

To find the velocity, we integrate this function over time. The key technique here is substitution, a method used to simplify complex integrals by changing variables. We set:\( u = t^2 + 1 \), which simplifies our integral to:

• \( v(t) = \int \frac{1}{u^2} \, du \)

By performing the integration, we find a general solution, adding an integration constant to capture all possible values that satisfy initial conditions. This constant is crucial as it allows us to tailor solutions to specific real-world scenarios.

After integrating and applying initial conditions, the velocity function becomes:

• \( v(t) = -\frac{1}{t^2 +1} + 1 \)

This process demonstrates how integration turns abstract mathematical operations into practical tools for mechanics.

Differential Equations

Differential equations connect functions and their derivatives and are prevalent in modeling how quantities change. In kinematics, they help relate acceleration, velocity, and position.

In this exercise, the problem of finding velocity from acceleration translates into solving a simple ordinary differential equation (ODE). The acceleration given is:

• \( a(t) = \frac{2t}{(t^2 + 1)^2} \)

This function represents the rate of change of the velocity, such that integrating over time yields the velocity function:

• \( v(t) = \int a(t) \, dt \)

The differential equation approach involves identifying both the derivative (acceleration here) and antiderivative (integrated velocity), allowing us to solve for specific behaviors of objects in motion.

Moreover, every step, from setting up the integral to solving it, requires checking initial conditions to determine constants of integration accurately. This illustrates the iterative nature of solving differential equations.

Kinematics

Kinematics is the branch of mechanics dealing with motion without considering its causes. It involves quantities such as position, velocity, and acceleration.

In this example, the task is to determine an object's velocity and position as it moves along a straight path, given acceleration. Let's summarize:

• Acceleration \( a(t) \): Tells how quickly velocity changes.
• Velocity \( v(t) \): Found by integrating acceleration, revealing how the position changes.
• Position function \( s(t) \): Obtained by integrating velocity, giving the exact location of the object at any time.

The idea is to "build up" from acceleration to position. Once you have the object's velocity as \( v(t) = -\frac{1}{t^2 + 1} + 1 \), its position can be determined by another integration:

• \( s(t) = \int v(t) \, dt \)

This leads to:

• \( s(t) = -\arctan(t) + t \)

Kinematics gives us a clear understanding of motion dynamics, essential for solving complex motion problems mechanics present. It is a practical application of calculus that translates theoretical mathematics into tangible real-world analysis.