Problem 5 - Calculus Early Transcendentals

Question

If a ball is thrown into the air with a velocity of 40 \(\mathrm{ft} / \mathrm{s}\) , its height in feet t seconds later is given by \(\mathrm{y}=40 \mathrm{t}-16 \mathrm{t}^{2}\) . (a) Find the average velocity for the time period beginning when \(t=2\) and lasting $$\begin{array}{ll}{\text { (i) } 0.5 \text { second }} & {\text { (il) } 0.1 \text { second }} \\ {\text { (iii) } 0.05 \text { second }} & {\text { (iv) } 0.01 \text { second }}\end{array}$$ (b) Estimate the instantaneous velocity when \(t=2\)

Step-by-Step Solution

Answer

For (a), average velocities are -26, -25.6, -27.2, -16 ft/s. For (b), the instantaneous velocity at \(t=2\) is estimated as \(-16\) ft/s.

Step 1: Understanding the Problem

To solve this problem, we first need to find the average velocity over a given time interval, which is defined as the change in height divided by the time duration. Then, we estimate the instantaneous velocity at a specific time by considering smaller intervals around the time point.

Step 2: Average Velocity Formula

The formula for average velocity over a time interval from \(t_1\) to \(t_2\) is:\[avg.\ velocity = \frac{y(t_2) - y(t_1)}{t_2 - t_1}\]where \(y(t) = 40t - 16t^2\).

Step 3: Calculate y-values

Compute the height at the start and end of each interval for \(t=2, t=2.5, t=2.1, t=2.05, t=2.01\):- \(y(2) = 40 \times 2 - 16 \times 2^2 = 48\)- \(y(2.5) = 40 \times 2.5 - 16 \times (2.5)^2 = 35\)- \(y(2.1) = 40 \times 2.1 - 16 \times (2.1)^2 = 45.44\)- \(y(2.05) = 40 \times 2.05 - 16 \times (2.05)^2 = 46.64\)- \(y(2.01) = 40 \times 2.01 - 16 \times (2.01)^2 = 47.84\).

Step 4: Calculate Average Velocities

Using the formula from Step 2, calculate the average velocities:- For (i): \(t=2 \) to \(t=2.5\): \[ avg.\ velocity = \frac{35 - 48}{2.5 - 2} = \frac{-13}{0.5} = -26 \text{ ft/s}\]- For (ii): \(t=2 \) to \(t=2.1\): \[ avg.\ velocity = \frac{45.44 - 48}{2.1 - 2} = \frac{-2.56}{0.1} = -25.6 \text{ ft/s}\]- For (iii): \(t=2 \) to \(t=2.05\): \[ avg.\ velocity = \frac{46.64 - 48}{2.05 - 2} = \frac{-1.36}{0.05} = -27.2 \text{ ft/s}\]- For (iv): \(t=2 \) to \(t=2.01\): \[ avg.\ velocity = \frac{47.84 - 48}{2.01 - 2} = \frac{-0.16}{0.01} = -16 \text{ ft/s}\].

Step 5: Estimate Instantaneous Velocity

The instantaneous velocity at \(t=2\) can be estimated by using smaller intervals around \(t=2\). With intervals decreasing, the average velocity approaches the instantaneous velocity: \(-16\) ft/s.

Key Concepts

Average Velocity

When understanding motion, average velocity is a crucial concept. It's what tells us how much an object's position changes over a specific time period. In simpler terms, average velocity is the change in distance divided by the change in time. If you have a math problem, like the height of a ball thrown up into the air, you can find how fast its height is changing by using this formula for average velocity: \[ \text{Average Velocity} = \frac{y(t_2) - y(t_1)}{t_2 - t_1} \] where \( t_1 \) and \( t_2 \) are start and end times, and \( y(t) \) is the position function. This formula compares the height at different times to give a notion of speed over an interval. Calculating average velocity helps us see a bigger picture of motion, unlike instantaneous velocity which focuses on a specific moment.

Projectile Motion

Projectile motion deals with objects thrown into the air, like balls or arrows. When these objects move under the influence of gravity, their path can be plotted as a trajectory, usually a curved path. This path is influenced by initial velocity, angle of release, and gravity.Key elements to remember about projectile motion include:

• Initial Velocity: The speed and direction when the object is first thrown.
• Gravity: Pulls the object back to Earth, affecting its speed as it rises and falls.

For the ball in our exercise, its height is given by \( y = 40t - 16t^2 \). This equation helps us find its position at any time \( t \). Using this, we can also determine speeds like average and instantaneous velocity. By recognizing projectile motion, you gain a better understanding of how factors like gravity influence motion.

Quadratic Functions

Quadratic functions play a central role in our understanding of projectile motion. A quadratic function is a polynomial of degree 2, generally expressed as \( ax^2 + bx + c \). The parabola shape that these functions form describes many natural processes, including our ball's flight.For the equation \( y = 40t - 16t^2 \):

• \(40t \)
• \(-16t^2\)

are elements of a quadratic equation describing the ball’s height over time. The role of the \( t^2 \) term is significant. It reflects the gravitational pull that acts downward, forming the characteristic parabolic path. Understanding quadratic functions can demystify why the ball moves as it does and enable predictions about its future height.

Limit Process

In calculus, the limit process is fundamental for understanding instantaneous velocity. While average velocity gives us a broad look, instantaneous velocity zooms in to see how fast an object moves at a precise moment.To find instantaneous velocity, we use the **limit** as the intervals in time become smaller. When time \( \Delta t \) approaches zero, average velocity becomes more accurate in pinpointing the object's speed at a specific time.This is why, as in our exercise, the average velocity for smaller and smaller intervals around \( t = 2 \) helps us estimate the instantaneous velocity. Eventually, this estimate converges to the derivative of the position function at \( t = 2 \), which can be found through calculus.