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: To find the displacement of an object with a given velocity function, follow these steps: 1. Understand that by integrating the velocity function with respect to time, you can find the displacement. 2. Set up a definite integral for the velocity function with the given time interval as limits. 3. Compute the definite integral using the antiderivative of the velocity function (position function). 4. Calculate the displacement by finding the difference between the position function values at the limits of the time interval.

Step 1: Understand the relationship between velocity and displacement

The velocity function, \(v(t)\), represents the rate at which an object's position changes with respect to time. Displacement, on the other hand, is the overall change in the object's position during a given time interval. By integrating the velocity function with respect to time, we can find the total displacement of the object during the specified time interval.

Step 2: Set up the definite integral

To find the displacement of the object during a specific time interval, say, from \(t=a\) to \(t=b\), set up a definite integral of the velocity function \(v(t)\) with respect to \(t\), with lower and upper limits \(a\) and \(b\). This can be written as: $$\int_a^b v(t) dt$$

Step 3: Compute the definite integral

Compute the definite integral by using the antiderivative of the velocity function, also known as the position function \(s(t)\). If you already know the antiderivative of \(v(t)\), use that; if not, use integration techniques such as substitution, integration by parts, or partial fractions to find the antiderivative, depending on the form of the velocity function.

Step 4: Calculate the displacement

Once you have the position function, \(s(t)\), use the fundamental theorem of calculus to compute the definite integral: $$\int_a^b v(t) dt = s(b) - s(a)$$ The result is the displacement of the object from time \(t=a\) to \(t=b\). If the result is positive, it indicates that the object moved in the positive direction; if negative, the object moved in the negative direction. Keep in mind that displacement is different from the total distance traveled, as displacement only considers the net change in position, not the total path length.

Key Concepts

Definite Integrals

A definite integral is a concept from calculus that helps us calculate the accumulated change over an interval. Imagine you have a function that describes a rate of change, like the velocity of an object. The definite integral computes the total change over a specific time period, giving us an overall understanding of how much change occurred.
For instance, if we consider the velocity function, the definite integral allows us to determine the total displacement of the object from start to end. The notation is often written as \( \int_a^b v(t) \, dt \), where \( a \) and \( b \) are the time limits and \( v(t) \) is the velocity function.

• Lower and upper limits \( a \) and \( b \) define the interval.
• The function \( v(t) \) describes velocity at time \( t \).
• The process gives the net change in position—displacement.

Thus, understanding definite integrals is crucial for solving real-world problems where changes over time or distance need to be calculated.

Velocity Function

The velocity function \( v(t) \) plays a central role in calculating displacement through calculus. It represents the speed and direction at which an object travels at any given moment. By analyzing how velocity changes over time, we can determine how an object moves.
For example, if you know how fast a car is traveling at every moment (its velocity function), you can map out its journey. The velocity function tells you more than just speed; it also includes direction, which is crucial when calculating where the car ends up after a trip.

• \( v(t) \) gives the rate of change of position at time \( t \).
• Influences how fast or slow the integral values compile over time.
• Integral of \( v(t) \) helps convert rate information into aggregate change—displacement.

Understanding the velocity function is essential to grasp how displacement interrelates with definite integrals.

Displacement Calculation

Displacement, as opposed to total distance traveled, refers to the overall change in position of an object. When calculating displacement using calculus, one of the most effective methods is to use a definite integral of the velocity function over a specific time interval.
To calculate the displacement, evaluate \( \int_a^b v(t) \, dt \). This integral gives you \( s(b) - s(a) \), where \( s(t) \) is the antiderivative, or position function, indicating the net change in position from time \( a \) to \( b \).

• Displacement indicates net position change over a time interval.
• Contrast with total distance, which measures total ground covered.
• Displacement uses the definite integral of velocity to find the position change.

Understanding displacement helps in areas ranging from physics to engineering, providing the knowledge to determine how far and in what direction an object has moved.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a critical link between differential and integral calculus. It provides a systematic way to solve definite integrals and, by extension, aids in calculating displacement.
This theorem states that if a function \( f \) is continuous on \([a, b]\) and \( F \) is the antiderivative of \( f \), then
\[ \int_a^b f(x) \, dx = F(b) - F(a) \]
In the context of displacement, if \( v(t) \) is the velocity function, then \( s(t) \), the position function, serves as the antiderivative. Therefore, the displacement calculation becomes straightforward by evaluating \( s(b) - s(a) \).

• The theorem interconnects differentiation and integration processes.
• Makes computing definite integrals feasible by evaluating antiderivatives.
• Transforms rate functions into aggregate changes, like displacement.

By mastering this theorem, you can effectively use integrals to solve various practical problems, including determining how far an object moves over time.