Problem 18 - Calculus Early Transcendentals

Question

Consider an object moving along a line with the following velocities and initial positions. a. Graph the velocity function on the given interval and determine when the object is moving in the positive direction and when it is moving in the negative direction. b. Determine the position function, for \(t \geq 0,\) using both the antiderivative method and the Fundamental Theorem of Calculus (Theorem 6.1 ). Check for agreement between the two methods. c. Graph the position function on the given interval. $$v(t)=3 \sin \pi t \text { on }[0,4] ; s(0)=1$$

Step-by-Step Solution

Answer

Answer: The object moves in the positive direction on the intervals [(1π)/4, (3π)/4] and [(5π)/4, (7π)/4]. The object moves in the negative direction when it is not in these intervals.

Step 1: Graph the velocity function \(v(t) = 3\sin \pi t\) on the interval \([0, 4]\)

To do this, simply sketch the graph of \(v(t) = 3\sin \pi t\) using a graphing calculator or an online plotting tool. Note the x-axis represents the time \(t\) measured in seconds, while the y-axis represents the velocity in meters per second.

Step 2: Determine when the object is moving in the positive and negative directions

An object moves in the positive direction when the velocity is positive, and in the negative direction when the velocity is negative. To determine when the velocity is positive or negative, we analyze the sign of \(v(t) = 3\sin \pi t\). By considering the period of \(\sin \pi t\), notice that the velocity is positive between \({(4n + 1)\pi}/{4} \leq t \leq {(4n + 3)\pi}/{4}\) where \(n\) is an integer. For the interval \([0,4]\), that would be \([(1\pi)/4 \leq t \leq (3\pi)/4]\) and \([(5\pi)/4 \leq t \leq (7\pi)/4].\) #b. Determine the position function using both the antiderivative method and the Fundamental Theorem of Calculus#

Step 3: Determine the position function using the antiderivative method

To find the position function \(s(t)\), we first need to find the antiderivative of \(v(t) = 3\sin \pi t.\) The antiderivative is given by $$s(t) = -\frac{3}{\pi}\cos\pi t + C$$ where \(C\) is the integration constant. Since the initial position \(s(0)=1\), we can plug \(t = 0\) into the position function to solve for the constant \(C\). This gives us $$1 = -\frac{3}{\pi}\cos 0 + C\Rightarrow C = 1 + \frac{3}{\pi}$$ So, the position function is given by $$s(t) = -\frac{3}{\pi}\cos\pi t + 1 + \frac{3}{\pi}$$

Step 4: Check the position function using the Fundamental Theorem of Calculus

We have that $$v(t) = 3\sin \pi t$$ and using the Fundamental Theorem of Calculus, we can find the position function as $$s(t) = s(0) + \int_0^t v(u) du = 1 + \int_0^t (3\sin \pi u) du$$ Now, integrating \(v(t)\) with respect to \(u\), $$s(t) = 1 -\frac{3}{\pi}\left[\cos(\pi u)\right]_0^t = 1 - \frac{3}{\pi}(\cos\pi t - \cos0) = -\frac{3}{\pi}\cos\pi t + 1 + \frac{3}{\pi}$$ We can see that the position function we found using the Fundamental Theorem of Calculus is the same as the one found using the antiderivative method. Their agreement confirms that the position function is correct. #c. Graph the position function on the given interval#

Step 5: Graph the position function \(s(t) = -\frac{3}{\pi}\cos\pi t + 1 + \frac{3}{\pi}\)

Plot the position function \(s(t) = -\frac{3}{\pi}\cos\pi t + 1 + \frac{3}{\pi}\) on the interval \([0,4].\) You can use a graphing calculator or an online plotting tool to do this. Note the x-axis represents the time \(t\) measured in seconds, while the y-axis represents the position in meters.

Key Concepts

Velocity Function

When you're examining how an object moves along a line, the term "velocity function" is essential. This function tells us the rate at which the position of an object changes with respect to time. In our example, the velocity function is given by \(v(t) = 3 \sin \pi t\), which describes how fast and in what direction the object moves at any given time \(t\).
A velocity function can be positive, negative, or zero.

• Positive velocity means the object is moving forward.
• Negative velocity indicates backward motion.
• Zero velocity means the object is momentarily at rest.

In the interval \([0, 4]\), we observe where the sine function is positive or negative to determine the object's direction of motion.

Position Function

The position function provides the exact location of an object at any time \(t\). It is the accumulation of the velocity over time and starts from an initial position. In this problem, we began by knowing the object's initial position, \(s(0) = 1\).
The goal is to find a position function, \(s(t)\), that accurately describes the object's location at any time \(t \geq 0\).
By integrating the velocity function \(v(t) = 3 \sin \pi t\), and using the initial position to solve for the constant \(C\), we derive:

• \(s(t) = -\frac{3}{\pi}\cos\pi t + 1 + \frac{3}{\pi}\)

This equation not only gives the position but also confirms how the velocity informs the object's movement over time.

Antiderivative Method

The antiderivative method is a straightforward way to find the position function from a given velocity function. Essentially, it involves finding the integral, or antiderivative, of the velocity function.
In our example, by integrating \(v(t) = 3 \sin \pi t\), we calculate the indefinite integral:

• \(\int v(t) \, dt = -\frac{3}{\pi} \cos \pi t + C\)

where \(C\) is the constant of integration.
By substituting the initial condition \(s(0) = 1\), we solved for \(C\). Once we find \(C\), the position function is complete, providing our target information for any time \(t\).
This method is beneficial for confirming results and understanding the relationship between position, velocity, and their derivatives.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, two main operations in calculus. It affirms that integration can "undo" differentiation and vice versa.
This theorem is crucial when deriving a position function from a velocity function. It states that the position function \(s(t)\) can be calculated as:

• \(s(t) = s(0) + \int_0^t v(u) \, du\)

For our function \(v(t) = 3 \sin \pi t\), we integrate from 0 to \(t\) to find how the object's position changes over time.
The result matches exactly with the antiderivative method, reaffirming the validity of our calculated position function. This demonstration shows how powerful calculus can be in moving from velocities to an accurate description of motion.