Problem 28 - 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. $$a(t)=e^{-t}, v(0)=60, s(0)=40$$

Step-by-Step Solution

Answer

The position function for the object is: s(t) = e^(-t) + 61t + 39 The velocity function for the object is: v(t) = -e^(-t) + 61

Step 1: Integrate the acceleration function to find the velocity function

To find the velocity function, v(t), we need to integrate the acceleration function, a(t), with respect to time: $$v(t) = \int a(t) dt = \int e^{-t} dt$$ The integral of e^{-t} is -e^{-t} + C_1, where C_1 is the integration constant: $$v(t) = -e^{-t} + C_1$$

Step 2: Apply the initial condition for velocity

We're given that the initial velocity at time t=0 is 60, or v(0) = 60. Using this information, we can find the integration constant C_1: $$v(0) = -e^{-0} + C_1 = 60$$ $$C_1 = 61$$ So the velocity function is given by: $$v(t) = -e^{-t} + 61$$

Step 3: Integrate the velocity function to find the position function

Now, we need to find the position function, s(t), by integrating the velocity function, v(t), with respect to time: $$s(t) = \int v(t) dt = \int (-e^{-t} + 61) dt$$ We will integrate term by term: $$\int (-e^{-t} + 61) dt = -\int e^{-t} dt + 61 \int dt$$ Now, the integral of -e^{-t} is e^{-t} and the integral of 61 is 61t. Adding our integration constant C_2: $$s(t) = e^{-t} + 61t + C_2$$

Step 4: Apply the initial condition for position

We're given that the initial position at time t=0 is 40, or s(0) = 40. Using this information, we can find the integration constant C_2: $$s(0) = e^{-0} + 61(0) + C_2 = 40$$ $$C_2 = 39$$ So the position function is given by: $$s(t) = e^{-t} + 61t + 39$$ In conclusion, the position function for the object is: $$s(t) = e^{-t} + 61t + 39$$ And the velocity function for the object is: $$v(t) = -e^{-t} + 61$$

Key Concepts

Integration

When we talk about integration in calculus, we refer to the process of finding the "anti-derivative" of a function. This essentially means reversing differentiation. If you know the rate at which something is changing, like acceleration (rate of change of velocity), integration helps you find the original function, such as velocity or position.

In the exercise, we start with an acceleration function, represented as \(a(t) = e^{-t}\). To find the velocity, we integrate this function with respect to time. Integration here helps us transition from the rate of change (acceleration) back to velocity, giving us the function \(v(t) = -e^{-t} + C_1\). The integration constant \(C_1\) arises because integration is a cumulative process, and without initial conditions, there are infinite possibilities for the starting point.

• Integration can be understood as finding the area under a curve.
• Every integrated result includes a constant of integration, represented as \(C\).

Once you have this first integration done, the next step (still using integration) is to calculate the position, \(s(t)\), by integrating the velocity function \(v(t)\). This gives you the broader picture of the object's motion.

Differential Equations

Differential equations are equations that relate a function with its derivatives. In the world of physics and engineering, they are extremely useful because they often describe dynamic systems, such as the movement of an object along a line, as in our problem.

The exercise deals with a straightforward differential equation for acceleration, \(a(t)\), a known function of time. Our task is to integrate this to find velocity \(v(t)\), and then integrate again to find position \(s(t)\).

• Differential equations help relate the rate of change of quantities with the quantities themselves.
• Solving them often involves integration and applying initial conditions.

In our example, the differential equation starts from \(a(t)\) and results in a velocity equation after integration. This process of solving until we find solution functions (like velocity or position) relies on the rules of calculus, specifically integration, to work back from derivatives to original functions.

Initial Conditions

Initial conditions are essential to solve differential equations completely, ensuring that we derive specific and accurate functions for velocity and position in this context. They provide a starting point or a specific value at a particular instance in time, which helps determine the constants of integration that pop up during integration.

In the exercise, you were given initial conditions: \(v(0) = 60\) for velocity and \(s(0) = 40\) for position. These values are critical because after integrating our equations, we acquire general solutions with constants \(C_1\) and \(C_2\). By plugging initial conditions into these equations, we solve for these constants.

• Initial conditions remove ambiguity from integration results by solving the constants.
• They make the solutions applicable to real-world scenarios by tailoring the general solutions to specific cases.

Without initial conditions, we would have multiple (or infinite) possible values for our solution functions, meaning we couldn't accurately predict the system's behavior.