Problem 3 - Calculus Early Transcendentals

Question

Given the velocity function \(v\) of an object moving along a line, explain how definite integrals can be used to find the displacement of the object.

Step-by-Step Solution

Answer

Answer: Definite integrals can be used to find the displacement of an object by calculating the accumulated change in displacement over a specified time interval. This is achieved by integrating the velocity function over the time interval, resulting in the net change in displacement.

Step 1: Introduce the relationship between velocity and displacement

Velocity is the rate of change of displacement with respect to time. In mathematical terms, we can write that as: $$ v(t) = \frac{d(s(t))}{dt} $$ Here, \(v(t)\) represents the velocity function of an object as a function of time, and \(s(t)\) is the displacement of an object as a function of time.

Step 2: Define definite integrals

A definite integral is a mathematical operation that calculates the accumulated value of a function (like velocity) over an interval. The definite integral can be thought of as the signed area under a curve (the function). The definite integral of a function \(f(x)\) from \(a\) to \(b\) is denoted as: $$ \int_{a}^{b} f(x) dx $$ The result of the definite integral is a number that represents the accumulated change of the function between the interval a and b.

Step 3: Connect the definite integral with displacement

To find the displacement, we want to know how the position of the object has changed over a given time interval. Since the velocity function \(v(t)\) tells us the rate of change of displacement, we can use the definite integral to find the accumulated change in displacement between two points in time. In other words, the definite integral of the velocity function can be used to determine the net change in displacement (\(\Delta s\)) over a specified time interval \([a, b]\): $$ \Delta s = \int_{a}^{b} v(t) dt $$ This connection is a result of the fundamental theorem of calculus, which links differentiation and integration.

Step 4: Example

Suppose we are given a velocity function \(v(t) = t^2 - 4t + 3\). To find the displacement of the object between the time interval \([1, 2]\), we can apply the definite integral as follows: $$ \Delta s = \int_{1}^{2} (t^2 - 4t + 3) dt $$ After finding the antiderivative function, we get: $$ \Delta s = \Big[\frac{1}{3}t^3-2t^2 + 3t\Big]_1^2 $$ Then, apply the fundamental theorem of calculus: $$ \Delta s = \Big(\frac{1}{3}(2)^3-2(2)^2 + 3(2)\Big)-\Big(\frac{1}{3}(1)^3-2(1)^2 + 3(1)\Big) $$ $$ \Delta s = (-1) - (0) = -1 $$ So, the displacement of the object between the time interval \([1, 2]\) is \(-1\).