Problem 47 - University Calculus: Early Transcendentals

Question

Give the acceleration \(a=d^{2} s / d t^{2},\) initial velocity. and initial position of an object moving on a coordinate line. Find the object's position at time \(t\). $$a=e^{t}, \quad v(0)=20, \quad s(0)=5$$

Step-by-Step Solution

Answer

The position function is \( s(t) = e^t + 19t + 4 \).

Step 1: Integrate the Acceleration to Find Velocity

The given acceleration function is \( a(t) = e^t \). To find the velocity function \( v(t) \), integrate the acceleration with respect to time:\[ v(t) = \int e^t \, dt = e^t + C_1, \]where \( C_1 \) is a constant of integration.

Step 2: Apply Initial Condition for Velocity

We know the initial velocity \( v(0) = 20 \). Substitute \( t = 0 \) into the velocity equation:\[ v(0) = e^0 + C_1 = 1 + C_1 = 20. \]Solve for \( C_1 \):\[ C_1 = 19. \]Thus, the velocity function is:\[ v(t) = e^t + 19. \]

Step 3: Integrate the Velocity to Find Position

Now, integrate the velocity function \( v(t) = e^t + 19 \) to find the position function \( s(t) \):\[ s(t) = \int (e^t + 19) \, dt = e^t + 19t + C_2, \]where \( C_2 \) is another constant of integration.

Step 4: Apply Initial Condition for Position

We know the initial position \( s(0) = 5 \). Substitute \( t = 0 \) into the position equation:\[ s(0) = e^0 + 19(0) + C_2 = 1 + C_2 = 5. \]Solve for \( C_2 \):\[ C_2 = 4. \]Thus, the position function is:\[ s(t) = e^t + 19t + 4. \]

Key Concepts

Differential equations

Differential equations are mathematical equations that involve functions and their derivatives. They are crucial for modeling various real-world phenomena, especially those involving rates of change, like velocity and acceleration. In our problem, we deal with a second-order differential equation, which is indicated by the acceleration function \(a = \frac{d^2 s}{dt^2}\).

To understand differential equations, let’s break them down:

• Derivative: Indicates the rate of change of a function concerning one of its variables. The first derivative of a position function \(s(t)\) is velocity \(v(t)\), and the second derivative is acceleration \(a(t)\).
• Order: The order of a differential equation is determined by the highest derivative present. Here, \(a(t) = e^t\) is a second-order differential equation since it involves a second derivative.

Differential equations use calculus concepts to help solve problems related to change and motion, making them powerful tools in physics and engineering.

Initial conditions

Initial conditions are values given at the start of a problem that help ensure a unique solution to a differential equation. Think of them as clues given to solve a mystery. In our example, we have initial conditions for both velocity and position given as \(v(0) = 20\) and \(s(0) = 5\).

Let’s look at how initial conditions are used:

• They allow us to determine the constants of integration that appear when solving differential equations by integration.
• They provide specific values for the unknown functions at a particular point in time, which narrows down the solution from infinitely many possibilities to a single unique solution.

In our case, these initial conditions were used to find specific values for \(C_1\) and \(C_2\), ensuring that the calculated velocity and position functions are unique and accurate.

Constant of integration

The constant of integration represents an unknown constant added when we integrate a function. Whenever we perform indefinite integration on a function, this constant appears, since integrating an indefinite integral gives a family of functions.

Here's why the constant is essential:

• It accounts for all possible vertical shifts of the original function. This means without the constant, our integrated function might not align properly with the problem's initial conditions or constraints.
• The integration process results in a family of functions differing by some constant \(C\), and initial conditions help us find the exact value of \(C\) that matches the problem.

In this problem, after integrating acceleration to find velocity, we get \(v(t) = e^t + C_1\). The initial velocity condition \(v(0) = 20\) helps find \(C_1 = 19\). Similarly, while finding the position, \(C_2 = 4\) is determined using the initial position condition \(s(0) = 5\). These steps ensure the final functions accurately match the problem scenario.