Question

Given the following velocity functions of an object moving along a line, find the position function with the given initial position. $$v(t)=e^{t}+4 ; s(0)=2$$

Short answer

Question: Find the position function for the given velocity function v(t) = e^t + 4 and initial position s(0) = 2. Answer: The position function for the given velocity function v(t) and initial position s(0) is s(t) = e^t + 4t + 1.

Step-by-step solution

Identify the given information

The velocity function v(t) and the initial position s(0) are given as follows: $$v(t)=e^{t}+4$$ $$s(0)=2$$

Integrate the velocity function

To find the position function s(t), we need to find the antiderivative of the velocity function v(t). $$s(t)=\int v(t) dt = \int (e^{t} + 4) dt$$

Calculate the antiderivative of the velocity function

Compute the antiderivative of each term: $$s(t)=\int e^{t} dt + \int 4 dt$$ The antiderivative of \(e^{t}\) is \(e^{t}\) and the antiderivative of 4 is 4t. So, we get: $$s(t)=e^{t}+4t+C$$ where C is the constant of integration.

Use the initial position to find the constant of integration

We know that s(0) = 2. Substitute t = 0 and s(t) = 2 into the equation we derived in Step 3: $$2=e^{0} + 4(0) + C$$ Solve for C: $$C=2-1$$ $$C=1$$

Write the final position function

Now that we have the constant of integration (C=1), we can write the final position function s(t) as: $$s(t)=e^{t}+4t+1$$ So, the position function for the given velocity function is: $$s(t)=e^{t}+4t+1$$

Key concepts

Integrating the Velocity Function

Understanding how to derive the position function from a given velocity function is crucial in calculus, especially in the context of physics and engineering. By integrating the velocity function, we're essentially calculating the area under the velocity-time curve, which represents the displacement, or change in position, of an object.

Imagine the velocity function as a graph with time on the x-axis and velocity on the y-axis. When we integrate this function, we're adding up all the tiny distances covered in tiny time intervals to find the total displacement over a period. This integration returns the position function, but with one missing piece, the constant of integration. In our exercise, after integrating the given velocity function, which is \(v(t)=e^{t}+4\), we found the indefinite integral to be \(s(t)=e^{t}+4t+C\), where \(C\) is the unknown that needs to be determined using the initial conditions.

Antiderivative

An antiderivative, in a broader sense, is the inverse of a derivative. It's a function that reverses the process of differentiation. When the derivative of a function gives us the rate of change, the antiderivative does the opposite by recovering a function from its known rate of change.

For example, if the rate of change of an object’s position is known to be \(e^{t}\), its position is an antiderivative of this rate function. Each term's antiderivative in the velocity function \(v(t)=e^{t}+4\) is straightforward; the exponential term remains \(e^{t}\) and the constant 4 becomes \(4t\), when integrated. Notice how each term of the given function transforms, and how this concept is key to building the solution to our example problem.

Constant of Integration

During the integration of a function, we encounter the 'constant of integration'. This mathematical concept is represented by the letter \(C\) and arises because integration is the inverse process of differentiation. Think of it as an 'adjustment' factor that helps us align the newly found antiderivative with the specific conditions of the problem.

After finding the antiderivative of a velocity function, the constant of integration is crucial to pin down the specific position function that adjusts to the real-world scenario being described. In our exercise, we used the initial condition \(s(0)=2\) to find the value of \(C\), giving us the specific solution that fits the given context.

Initial Position

The initial position is a term used to describe the position of an object at the starting point of an observation, often noted as \(s(0)\). This information is essential in determining the constant of integration when integrating a velocity function.

In our exercise, the initial position was given as \(s(0)=2\). This means that, at time \(t=0\), the object's position was 2 units from the reference point. By substituting \(t=0\) into the integrated position function and setting \(s(t)\) equal to 2, we were able to solve for the constant of integration. This allowed us to provide the exact position function that not only reflects the overall motion of the object but also its specific starting point, which is crucial for accurate predictions and analysis of the object's future positions.