Problem 77 - Calculus Single Variable

Question

The trajectory of a fly ball is such that the height in feet above ground is \(H(t)=4+72 t-16 t^{2}\) when \(t\) is measured in seconds. a. Compute the average velocity in the following time intervals: i. [2,3] iii. [2,2.01] ii. [2,2.1] iv. [2,2.001] b. Compute the instantaneous velocity at \(t=2\).

Step-by-Step Solution

Answer

Average velocities: 8, ~5.6, ~5.4, ~5.44 ft/s; Instantaneous velocity: 8 ft/s at t=2.

Step 1: Average Velocity Formula

To find the average velocity over a time interval [a, b], use the formula \( v_{avg} = \frac{H(b) - H(a)}{b - a} \). This represents the change in height over the change in time.

Step 2: Calculate Average Velocity for [2,3]

For the interval [2,3], compute \( H(2) = 4 + 72(2) - 16(2)^2 \) which gives \( H(2) = 68 \). Compute \( H(3) = 4 + 72(3) - 16(3)^2 \) which gives \( H(3) = 76 \). Now find \( v_{avg} = \frac{76 - 68}{3 - 2} = 8 \).

Step 3: Calculate Average Velocity for [2,2.1]

For the interval [2,2.1], we already have \( H(2) = 68 \). Compute \( H(2.1) = 4 + 72(2.1) - 16(2.1)^2 \) which gives \( H(2.1) \approx 68.56 \). Now find \( v_{avg} = \frac{68.56 - 68}{2.1 - 2} \approx 5.6 \).

Step 4: Calculate Average Velocity for [2,2.01]

For the interval [2,2.01], use \( H(2) = 68 \). Compute \( H(2.01) = 4 + 72(2.01) - 16(2.01)^2 \) giving \( H(2.01) \approx 68.144 \). Then \( v_{avg} = \frac{68.144 - 68}{2.01 - 2} \approx 5.4 \).

Step 5: Calculate Average Velocity for [2,2.001]

For the interval [2,2.001], already knowing \( H(2) = 68 \), compute \( H(2.001) = 4 + 72(2.001) - 16(2.001)^2 \) yielding \( H(2.001) \approx 68.0144 \). Then \( v_{avg} = \frac{68.0144 - 68}{2.001 - 2} \approx 5.44 \).

Step 6: Instantaneous Velocity Calculation

To find the instantaneous velocity at \( t = 2 \), take the derivative of \( H(t) \). \( H(t) = 4 + 72t - 16t^2 \), so \( H'(t) = 72 - 32t \). Plug in \( t = 2 \): \( H'(2) = 72 - 32(2) = 72 - 64 = 8 \).

Key Concepts

Average Velocity

Average velocity is a measure of how quickly something changes its position over a given time period. It's calculated using the formula:

• \( v_{avg} = \frac{H(b) - H(a)}{b - a} \)

Here, \( H(a) \) and \( H(b) \) are the heights at the starting and ending times, \( a \) and \( b \), respectively. Average velocity represents the slope of the secant line between two points on the trajectory equation, providing a simple way to understand motion over finite time intervals.
For example, in the interval [2,3], the average velocity would be determined by the difference in height at times \( t = 3 \) and \( t = 2 \), divided by the time interval (1 second in this case). This straightforward calculation allows us to grasp how the velocity evolves over different intervals.

Derivative

In physics and calculus, a derivative serves as a tool to find the rate at which one quantity changes concerning another. Within the context of motion, it gives us the instantaneous rate of change, or velocity, at a specific point in time.
The derivative of the height function, represented by \( H(t) = 4 + 72t - 16t^2 \), is calculated as:

• \( H'(t) = 72 - 32t \)

This formula tells us how fast and in which direction the ball's height changes at any given moment. By analyzing the derivative, we can determine the instantaneous velocity, offering deeper insights into the motion's dynamics. At \( t = 2 \), this results in an instantaneous velocity of 8 feet per second.

Trajectory Equation

The trajectory equation describes the path of a moving object, specifically how the object's position changes over time. For the given problem, the trajectory equation is:

• \( H(t) = 4 + 72t - 16t^2 \)

This quadratic equation captures the essence of projectile motion, showing how height depends on time. The different terms reflect various aspects of motion, such as initial height (4 feet), initial velocity (72 feet per second), and the influence of gravity through the acceleration (-16 feet per second squared).
Understanding the components of these equations helps in visualizing how objects move in real-life scenarios, whether on sports fields or in physics experiments.

Change in Height Over Time

The change in height over time in this context refers to how the fly ball's vertical position varies as time progresses. The significance comes from being able to measure this change to better understand the object's motion.
To evaluate this change, we use the trajectory equation \( H(t) = 4 + 72t - 16t^2 \) and compute differences in height at specific time values, such as 2 seconds and 3 seconds, giving insights into the rate of ascent or descent.
Consider an interval like [2,2.1]: the change in height is the difference between the values of \( H(t) \) at \( t = 2 \) and \( t = 2.1 \). Through these calculations, students learn about motion in a practical setting, aiding deeper comprehension of similar principles in various physics applications.