Problem 16 - Calculus: Early Transcendentals

Question

The displacement (in feet) of a particle moving in a straight line is given by \( s = \frac{1}{2}t^2 - 6t + 23 \), where \( t \) is measured in seconds. (a) Find the average velocity over each time interval: (i) \( [4, 8] \) (ii) \( [6, 8] \) (iii) \( [8, 10] \) (iv) \( [8, 12] \) (b) Find the instantaneous velocity when \( t = 8 \). (c) Draw the graph of \( s \) as a function of \( t \) and draw the secant lines whose slopes are the average velocities in part (a). Then draw the tangent line whose slope is the instantaneous velocity in part (b).

Step-by-Step Solution

Answer

(a) Average velocities: (i) -2.25 ft/s, (ii) -1 ft/s, (iii) 2 ft/s, (iv) 3 ft/s. (b) Instantaneous velocity at t=8: 2 ft/s.

Step 1: Calculate Displacements for Part (a) Intervals

For part (a), we need to calculate the displacement values at each endpoint of the given intervals using the displacement function.Calculate \( s(4) \):\[ s(4) = \frac{1}{2}(4)^2 - 6(4) + 23 = 8 \]Calculate \( s(8) \):\[ s(8) = \frac{1}{2}(8)^2 - 6(8) + 23 = -1 \]Calculate \( s(6) \):\[ s(6) = \frac{1}{2}(6)^2 - 6(6) + 23 = -1 \]Calculate \( s(10) \):\[ s(10) = \frac{1}{2}(10)^2 - 6(10) + 23 = 3 \]Calculate \( s(12) \):\[ s(12) = \frac{1}{2}(12)^2 - 6(12) + 23 = 11 \]

Step 2: Calculate Average Velocity for Part (a)(i)

The average velocity over an interval \([a, b]\) is given by \( \frac{s(b) - s(a)}{b - a} \).For interval \([4, 8]\):\[ \text{Average Velocity} = \frac{s(8) - s(4)}{8 - 4} = \frac{-1 - 8}{4} = -2.25 \text{ ft/s} \]

Step 3: Calculate Average Velocity for Part (a)(ii)

Using the same method for interval \([6, 8]\):\[ \text{Average Velocity} = \frac{s(8) - s(6)}{8 - 6} = \frac{-1 + 1}{2} = -1 \text{ ft/s} \]

Step 4: Calculate Average Velocity for Part (a)(iii)

For interval \([8, 10]\):\[ \text{Average Velocity} = \frac{s(10) - s(8)}{10 - 8} = \frac{3 + 1}{2} = 2 \text{ ft/s} \]

Step 5: Calculate Average Velocity for Part (a)(iv)

For interval \([8, 12]\):\[ \text{Average Velocity} = \frac{s(12) - s(8)}{12 - 8} = \frac{11 + 1}{4} = 3 \text{ ft/s} \]

Step 6: Find Instantaneous Velocity (Part b)

The instantaneous velocity is the derivative of the displacement function evaluated at \( t = 8 \).The derivative of \( s(t) = \frac{1}{2}t^2 - 6t + 23 \) is:\[ s'(t) = t - 6 \]Evaluate at \( t = 8 \):\[ s'(8) = 8 - 6 = 2 \text{ ft/s} \]

Step 7: Graph the Function and Lines (Part c)

To complete part (c), graph the displacement function \( s(t) = \frac{1}{2}t^2 - 6t + 23 \) along with its secant lines for intervals specified in part (a) and the tangent line at \( t = 8 \).- Draw the curve based on the equation.- For each interval (e.g., [4, 8]; [6, 8]; [8, 10]; [8, 12]), draw secant lines using the calculated average velocities.- Draw a tangent line at \( t = 8 \) using the slope of 2 ft/s from part (b).

Key Concepts

Displacement Function

The displacement function is a mathematical expression that tells us the position of a moving object at a given time. In this exercise, the displacement function is given by: \[ s(t) = \frac{1}{2} t^2 - 6t + 23 \] This equation describes the particle's position as it moves along a straight line. The variable \( t \) represents time, measured in seconds. Understanding how this function behaves over time helps us predict where the particle will be at any point.

• The term \( \frac{1}{2}t^2 \) indicates that as time progresses, the displacement due to this part of the equation increases quadratically.
• The term \(-6t\) shows a linear decrease in displacement as time increases.
• The constant \(+23\) shifts the entire graph vertically.

When analyzing this function, it's essential to calculate specific displacement values at particular time intervals to understand the movement better.

Average Velocity

Average velocity helps us determine how fast an object moves over a time period. It is calculated using the formula: \[ \text{Average Velocity} = \frac{s(b) - s(a)}{b - a} \] This formula takes the displacement at the end and beginning of a time interval and divides by the time duration.

• For interval \([4, 8]\), the average velocity came out as \(-2.25\) feet per second.
• For \([6, 8]\), it is \(-1\) feet per second.
• For \([8, 10]\), the average velocity is \(2\) feet per second.
• For \([8, 12]\), it is \(3\) feet per second.

The negative velocities imply the particle is moving backward, while the positive velocities suggest forward motion.

Instantaneous Velocity

Instantaneous velocity refers to the speed of the particle at a specific point in time. Unlike average velocity, which is over an interval, instantaneous velocity is a derivative at a single instant.To find it, we differentiate the displacement function to get the velocity function: \[ s'(t) = t - 6 \] Evaluating \( s'(t) \) at a specific time gives us the instantaneous velocity. For \( t = 8 \), we calculate:\[ s'(8) = 8 - 6 = 2 \text{ feet per second} \] This means at exactly 8 seconds, the particle's velocity is 2 feet per second. This velocity is positive, indicating the particle is moving forward at this instant.

Derivative

In calculus, a derivative represents the rate of change of a function concerning one of its variables. Here, the derivative of the displacement function \( s(t) \) in terms of time \( t \) is the velocity function.The derivative captures how quickly the displacement function changes as time progresses. So, if we have:\[ s(t) = \frac{1}{2}t^2 - 6t + 23 \] The derivative is given by:\[ s'(t) = t - 6 \] By finding the derivative, we directly obtain the velocity at any given time \( t \). This way, we can analyze how the particle's speed varies across different times and identify points of acceleration or deceleration.

Graphing Secant and Tangent Lines

Understanding the graphical representation of functions is critical in calculus. Secant lines provide a visual of the average velocity over an interval, while tangent lines show the instantaneous velocity at a specific point.

• Secant Lines: These lines are drawn across two points on the displacement graph, representing average velocities. The slope of each secant line corresponds to the average velocity over that particular interval.
• Tangent Line: This line touches the graph at only one point and represents the instantaneous velocity. Its slope is the derivative value at that specific time point.

For example, for interval \([4, 8]\), a secant line connects the points corresponding to \( s(4) \) and \( s(8) \). For \( t = 8 \), the tangent line's slope of 2 feet per second accurately depicts the exact rate of change at that moment.