Problem 17 - 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 anti- derivative method and the Fundamental Theorem of Calculus (Theorem 1). Check for agreement between the two methods. c. Graph the position function on the given interval. $$v(t)=9-t^{2} \text { on }[0,4] ; s(0)=-2$$

Step-by-Step Solution

Answer

Answer: Using the position function \(s(t) = 9t - \frac{t^3}{3} - 2\), we can find the position of the object at time t=2 as follows: $$s(2) = 9(2) - \frac{(2)^3}{3} - 2$$ $$s(2) = 18 - \frac{8}{3} - 2$$ $$s(2) = \frac{46}{3}$$ So, the position of the object at time t=2 is \(\frac{46}{3}\).

Step 1: Graph the velocity function and analyze the motion

To begin, let us first graph the given velocity function \(v(t) = 9 - t^2\). By analyzing the graph, we can determine when the object is moving in the positive direction (v(t) > 0) and when it is moving in the negative direction (v(t) < 0). From the graph, we can observe that the object is moving in the positive direction when \(t\in[0, 3]\) and in the negative direction when \(t\in[3, 4]\).

Step 2: Calculate the position function using antiderivative method

To find the position function, we need to find the antiderivative of the velocity function and apply the initial condition \(s(0) = -2\). The antiderivative of \(v(t)\) is: $$s(t) = \int (9 - t^2) dt = 9t - \frac{t^3}{3} + C$$ Now, let's apply the initial condition \(s(0) = -2\): $$-2 = 9(0) - \frac{(0)^3}{3} + C$$ Solving for C, we find that \(C = -2\). Therefore, our position function is: $$s(t) = 9t - \frac{t^3}{3} - 2$$

Step 3: Calculate the position function using the Fundamental Theorem of Calculus

Using the Fundamental Theorem of Calculus, we can find the position function by evaluating the integral of the velocity function over the interval \([0, t]\). We can also apply the initial condition: $$s(t) = s(0) + \int_{0}^{t} (9 - u^2) du$$ Now, let's evaluate the integral: $$s(t) = -2 + \int_{0}^{t} (9 - u^2) du$$ $$s(t) = -2 + [9u - \frac{u^3}{3}]_{0}^{t}$$ $$s(t) = -2 + (9t - \frac{t^3}{3})$$ Which simplifies to: $$s(t) = 9t - \frac{t^3}{3} - 2$$

Step 4: Check for agreement between the two methods

From Step 2 and Step 3, we can see that both methods resulted in the same position function: $$s(t) = 9t - \frac{t^3}{3} - 2$$ Thus, there is agreement between the antiderivative method and the Fundamental Theorem of Calculus.

Step 5: Graph the position function

Finally, let's graph the position function \(s(t) = 9t - \frac{t^3}{3} - 2\) on the given interval of \([0, 4]\). By looking at the graph, we can conclude that the position function starts at the initial position of \(s(0) = -2\) and increases while the object is moving in a positive direction (until \(t=3\)), then it starts to decrease while the object is moving in the negative direction (from \(t=3\) to \(t=4\)).

Key Concepts

Velocity Function

In this exercise, the velocity function represents the speed of an object moving along a line with respect to time, specifically given by the equation \(v(t) = 9 - t^2 \). This equation provides insight into how fast the object is moving and in which direction. By graphing this function over the interval \([0, 4]\), we can determine when the object is moving positively or negatively.

• When \(v(t) > 0\), the object moves in the positive direction, indicating an increase in position value.
• Conversely, when \(v(t) < 0\), it signifies movement in the negative direction, suggesting a decrease in position value.

In this scenario, the velocity function indicates that the object moves positively from \(t = 0\) to \(t = 3\), and then negatively from \(t = 3\) to \(t = 4\). This change in direction is crucial for understanding the object’s total displacement over time.

Anti-derivative Method

The anti-derivative method is a technique used to find the position function by integrating the velocity function. This process involves finding an indefinite integral or anti-derivative of the velocity function \(v(t)\). In this problem, the velocity function is \(v(t) = 9 - t^2\).To determine the position function \(s(t)\), we integrate \(v(t)\): \[s(t) = \int (9 - t^2) \, dt = 9t - \frac{t^3}{3} + C\]Here, \(C\) is the constant of integration, which is determined using the initial condition provided in the problem, \(s(0) = -2\). Substituting the initial condition: \[-2 = 9(0) - \frac{(0)^3}{3} + C\]Solving for \(C\), we find \(C = -2\). Thus, the position function becomes: \[s(t) = 9t - \frac{t^3}{3} - 2\]This method is useful in linking the velocity of an object to its position over time.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a powerful concept that connects differentiation with integration. It provides a way to calculate the area under a curve, which in this context orients around determining the object's position at any given time.To find the position function \(s(t)\), we'll evaluate the integral of the velocity function \(v(t)\) from \(0\) to \(t\), considering the initial position.Using the theorem:\[s(t) = s(0) + \int_{0}^{t} (9 - u^2) du\]Calculate the integral:\[s(t) = -2 + \left [9u - \frac{u^3}{3} \right ]_{0}^{t}\]Evaluating gives:\[s(t) = -2 + (9t - \frac{t^3}{3})\]This simplifies to:\[s(t) = 9t - \frac{t^3}{3} - 2\]The theorem reassures the process's validity and checks the derived position function. Both the anti-derivative method and the Fundamental Theorem provide matching results, confirming accuracy.

Graphing Functions

Graphing functions is an effective tool to visualize the behavior of any function over a given interval. For this exercise, both the velocity and position functions can be graphed to provide insights into the motion dynamics.For the velocity function \(v(t) = 9 - t^2\), seeing its graph indicates where the velocity is positive or negative, helping us determine different phases of motion.By graphing the position function \(s(t) = 9t - \frac{t^3}{3} - 2\), we can visualize how the object’s position changes over time from \([0, 4]\):

• Initially, the position increases as the velocity is positive (from \(t=0\) to \(t=3\)).
• Afterwards, it begins to decrease as the velocity turns negative (from \(t=3\) to \(t=4\)).

This graphical representation is crucial for understanding the kinematic behavior of the object, enhancing comprehension beyond theoretical derivations.