Question

The following problems consider the historic average cost per gigabyte of RAM on a computer. $$\begin{array}{|l|l|} \hline \text { Year } & \text { 5-Year Change (\$) } \\ \hline 1980 & 0 \\ \hline 1985 & -5,468,750 \\ \hline 1990 & -755,495 \\ \hline 1995 & -73,005 \\ \hline 2000 & -29,768 \\ \hline 2005 & -918 \\ \hline 2010 & -177 \\ \hline \end{array}$$ 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 \(0.5 \mathrm{sec} ?\)

Short answer

The bullet travels approximately 663.2 feet in 0.5 seconds.

Step-by-step solution

Identify the problem

We need to calculate the total distance a bullet travels in 0.5 seconds, given the velocity function of the bullet over time.

Recall the formula for distance from velocity

The distance traveled can be calculated by integrating the velocity function over the given time period. Specifically, we will integrate the velocity function from 0 to 0.5 seconds.

Set up the integral

The velocity function is given by \[ v(t) = 6400 t^2 - 6505 t + 2686 \]To find the distance, we need to solve the integral: \[ \int_{0}^{0.5} (6400 t^2 - 6505 t + 2686) \, dt \]

Integrate the function

Calculate the integral \[ \int (6400 t^2 - 6505 t + 2686) \, dt = \frac{6400}{3} t^3 - \frac{6505}{2} t^2 + 2686 t + C \] where \(C\) is the constant of integration. However, because we are calculating a definite integral, the constant \(C\) can be omitted.

Evaluate the definite integral

Evaluate the integral from 0 to 0.5:\[ \left. \left( \frac{6400}{3} t^3 - \frac{6505}{2} t^2 + 2686 t \right) \right|_{0}^{0.5} \] This results in evaluating:\[ \frac{6400}{3} (0.5)^3 - \frac{6505}{2} (0.5)^2 + 2686(0.5) \]Minus the evaluation at \(t=0\), which is 0.

Calculate the final result

First, calculate the value at \(t=0.5\):\[ \frac{6400}{3} (0.5)^3 = 133.33 \]\[ \frac{6505}{2} (0.5)^2 = 813.125 \]\[ 2686 \times 0.5 = 1343 \]Now calculate the sum:\[133.33 - 813.125 + 1343 \approx 663.205\]Thus, the total distance traveled by the bullet in 0.5 seconds is approximately 663.2 feet.

Key concepts

Integral Calculus

Integral calculus is a fascinating area of mathematics focused on the process of integration. Integration is essentially the reverse of differentiation. It's about finding a function whose derivative is a given function. In simpler terms, it's like piecing together small bits of information (like how fast you were going along a road) to find a whole (like the entire journey distance).
In practical situations, integral calculus is used to compute areas under curves, among other things. The key idea is to sum up many small quantities to get a total amount. This includes finding the total distance traveled when the velocity at any particular time is known. This definitely plays a big role in solving physical problems where rates of change are involved.
Using integral calculus, we can turn a velocity function into a distance function by integrating over the period of interest. Understanding this concept helps us link the change in position over time.

Velocity and Distance

Velocity is the rate at which an object changes its position. It is a vector quantity, meaning it has both magnitude and direction. In the context of our exercise, velocity is defined by a polynomial function, which helps provide the speed of a bullet at any given time.

The relationship between velocity and distance is foundational in physics. If you know an object’s velocity over time, you can determine the distance it moves within that timespan by integrating the velocity function. This is because distance is the integral of velocity.

• When velocity is known as a function of time, the total distance can be calculated by the definite integral over the given time interval.
• This approach is particularly useful in analyzing motion where the speed varies over time, such as with bullets, cars, or even athletes.

Having the velocity function allows you to look back and determine not just the speed at different points in time, but the path traveled over the whole duration too.

Definite Integral

A definite integral is a type of integral calculus that calculates the accumulation of quantities, such as areas under curves, between two points on the x-axis. It provides a precise way to quantify how a quantity changes over a specific interval.

In the exercise, the definite integral is set up to find how far a bullet travels from time zero to half a second. The formula used is:

• \[\int_{0}^{0.5} (6400 t^2 - 6505 t + 2686) \, dt\], making use of definite limits of integration, 0 and 0.5, which specify the period during which we observe the bullet's behavior.
• To solve this integral, we evaluate the antiderivative at the upper limit and subtract the antiderivative evaluated at the lower limit.

Through this process, we ensure a calculation of precise values reflective of real-world observations.
Definite integrals play a major role in a variety of scientific and engineering fields, where they are used to calculate things like total growth, energy consumption, and travel distances.

Polynomial Functions

Polynomial functions are mathematical expressions involving sums of powers of a variable, such as the speed of a bullet in our exercise. They are one of the most simple yet versatile forms of functions in calculus.

In this exercise, the velocity function is a polynomial given by:\[v(t)=6400 t^{2}-6505 t+2686\]

• Each term of the polynomial represents a part of the velocity's behavior over time, with coefficients affecting the speed variation rate and direction.
• Higher-order polynomials involve quadratic terms \(t^2\), which influence acceleration, and even cubic terms \(t^3\), which affect changes in acceleration.

The beauty of polynomial functions is their simplicity in structure yet depth in application. They can be differentiated and integrated smoothly, which makes them invaluable tools for solving problems related to motion, growth, and decay.
Understanding how polynomial functions work helps demystify complex motions like a bullet’s path and showcases the power of mathematical modeling.