Problem 462 - Calculus Volume 1

Question

The following problems consider the historic average cost per gigabyte of RAM on a computer. $$\begin{array}{|c|c|c|}\hline \text { Year } & {5 \text { -Year Change (s) }} \\\ \hline 1980 & {0} \\ \hline 1985 & {-5,468,750} \\ \hline 1990 & {-755,495} \\ \hline 1995 & {-73,005} \\ \hline 2000 & {-29,768} \\ \hline 2005 & {-918} \\ \hline 2010 & {-177} \\ \hline\end{array}$$ The velocity of a bullet from a rifle can be approximated by \(v(t)=6400 t^{2}-6505 t+2686,\) where \(t\) is seconds after the shot and \(v\) is the velocity measured in feet per second. This equation only models the velocity for the first half-second after the shot: \(0 \leq t \leq 0.5\) . What is the total distance the bullet travels in 0.5 \(\sec ?\)

Step-by-Step Solution

Answer

The total distance the bullet travels in 0.5 seconds is approximately 859.875 feet.

Step 1: Writing the velocity function

We are given the velocity function for the bullet, which is \( v(t) = 6400t^2 - 6505t + 2686 \). This function describes how the velocity of the bullet changes over time between \( t = 0 \) and \( t = 0.5 \) seconds.

Step 2: Integrating the velocity function

To find the total distance the bullet travels, we need to integrate the velocity function over the interval from \( t = 0 \) to \( t = 0.5 \) seconds. The integral of velocity with respect to time will give us the displacement.\[\text{Distance} = \int_{0}^{0.5} (6400t^2 - 6505t + 2686) \, dt\]

Step 3: Calculating the integral

Perform the integration of each term separately:- \( \int_{0}^{0.5} 6400t^2 \, dt = \left[ \frac{6400}{3}t^3 \right]_{0}^{0.5} \)- \( \int_{0}^{0.5} -6505t \, dt = \left[ \frac{-6505}{2} t^2 \right]_{0}^{0.5} \)- \( \int_{0}^{0.5} 2686 \, dt = \left[ 2686t \right]_{0}^{0.5} \)

Step 4: Evaluating the definite integrals

Calculate the definite integrals for each term:- For \( \frac{6400}{3}t^3 \): \[ \left( \frac{6400}{3} \times (0.5)^3 \right) - \left( \frac{6400}{3} \times 0^3 \right) = \frac{6400}{3} \times 0.125 = \frac{800}{3} \]- For \( \frac{-6505}{2}t^2 \): \[ \left( \frac{-6505}{2} \times (0.5)^2 \right) - \left( \frac{-6505}{2} \times 0^2 \right) = \frac{-6505}{2} \times 0.25 = -813.125 \]- For \( 2686t \): \[ 2686 \times 0.5 - 2686 \times 0 = 1343 \]

Step 5: Summing the results

Add up the results of the three integrals to find the total displacement:\[\text{Total Distance} = \frac{800}{3} - 813.125 + 1343 = \frac{800}{3} + 1343 - 813.125 \approx 330 + 1343 - 813.125 = 859.875 \] feet.

Key Concepts

Velocity Function

A velocity function describes how the speed of an object changes over time. In this scenario, we are looking at the velocity of a bullet, which can be modeled by the quadratic equation:

• \( v(t) = 6400t^2 - 6505t + 2686 \)

Here, \( t \) is the time in seconds after the bullet is fired, and \( v(t) \) is its velocity in feet per second. This equation specifically represents the bullet's velocity for the first 0.5 seconds after it is shot.
In physics, velocity does not just tell us how fast something is moving, but also includes the direction. In this case, we are primarily concerned with the magnitude of speed as the bullet heads towards its target.

Integration

Integration is a fundamental concept in calculus used to find areas under curves, among other things. It comes in handy when you want to determine the accumulated total of a quantity, like distance, when given a rate of change, like velocity.
In the context of our exercise, we use integration to compute the displacement of the bullet from a velocity function.
By integrating the velocity function over a time interval, you determine how far the bullet has traveled during that time span.
The process involves forming an integral of the velocity function, expressed in mathematical notation as:

• \[ \text{Distance} = \int_{0}^{0.5} (6400t^2 - 6505t + 2686) \, dt \]

After setting up the integral, you solve it term by term to discover the total displacement.

Definite Integrals

A definite integral is a type of integral that calculates the net area under a curve between two points along the x-axis. It is used to evaluate the total change in a quantity.
In our scenario, we use a definite integral to determine the bullet's displacement over the interval from 0 to 0.5 seconds.
The expression

• \[ \int_{0}^{0.5} (6400t^2 - 6505t + 2686) \, dt \]

represents the definite integral. Evaluating this integral over the given interval will provide the bullet's total distance traveled in that time frame.
Remember, the results of definite integrals account not just for the speeds but also for the time duration, offering precise total displacement measurements.

Displacement Calculation

Displacement is the total change in position of an object, and in this context, it represents the total distance a bullet travels once fired.
To calculate displacement via integration, you perform a definite integration of the velocity function over the given interval.
Once you have integrated each part of the velocity equation, you evaluate those integrals.

• The results are summed up to acquire the total displacement.

For this problem:

• \( \text{Total Distance} = \frac{800}{3} + 1343 - 813.125 = 859.875 \) feet.

Through this process, the integral transforms abstract rate of change measurements into a quantifiable distance that highlights how integration links velocity with displacement.