Problem 52
Question
The current in a wire is defined as the derivative of the charge: \( I(t) = Q'(t) \). (See Example 3.7.3.) What does \( \displaystyle \int^b_a I(t) \, dt \) represent?
Step-by-Step Solution
Verified Answer
\( \int_a^b I(t) \, dt \) represents the change in charge from time \( a \) to time \( b \).
1Step 1: Understand the Definition of Current
Current \( I(t) \) is defined as the derivative of the charge \( Q(t) \) with respect to time. That is, \( I(t) = Q'(t) \).
2Step 2: Interpret the Integral Expression
The integral \( \int_a^b I(t) \, dt \) represents the accumulation of the current over time from \( t = a \) to \( t = b \). By the Fundamental Theorem of Calculus, it will give us the net change in the charge over this time interval.
3Step 3: Apply the Fundamental Theorem of Calculus
By applying the Fundamental Theorem of Calculus, we know that \( \int_a^b I(t) \, dt = Q(b) - Q(a) \), where \( I(t) = Q'(t) \). This calculation finds the difference in the charge from time \( a \) to time \( b \).
4Step 4: Summarize the Result
The integral \( \int_a^b I(t) \, dt \) represents the total change in the charge \( Q(t) \) between times \( t = a \) and \( t = b \).
Key Concepts
Integrals in CalculusDerivative of ChargeCurrent and Charge Relationship
Integrals in Calculus
Integrals are a fundamental concept in calculus, serving as a powerful tool to determine the accumulation of quantities. When we talk about integrals, we usually refer to the process of finding the area under a curve. This area represents the accumulated total from a certain starting point to an endpoint.
In mathematical terms, if you have a function like the current in a wire, denoted as \( I(t) \), the integral \( \int_a^b I(t) \, dt \) describes the accumulated current between times \( t = a \) and \( t = b \). The notation \( a \) and \( b \) denote the limits of integration, which specify the interval over which you accumulate the current.
The Fundamental Theorem of Calculus bridges the gap between derivatives, which tell us about rates of change, and integrals, which tell about accumulation. It affirms that if you take the integral of the derivative of a function, you return to the original function, evaluated at the bounds. This concept is crucial when looking at how quantities accumulate over time, such as charge in an electric circuit.
In mathematical terms, if you have a function like the current in a wire, denoted as \( I(t) \), the integral \( \int_a^b I(t) \, dt \) describes the accumulated current between times \( t = a \) and \( t = b \). The notation \( a \) and \( b \) denote the limits of integration, which specify the interval over which you accumulate the current.
The Fundamental Theorem of Calculus bridges the gap between derivatives, which tell us about rates of change, and integrals, which tell about accumulation. It affirms that if you take the integral of the derivative of a function, you return to the original function, evaluated at the bounds. This concept is crucial when looking at how quantities accumulate over time, such as charge in an electric circuit.
Derivative of Charge
In calculus, the derivative of a function captures how the function is changing at any given point. When considering a scenario like the flow of electric charge, we describe the rate at which the charge \( Q(t) \) changes over time with a derivative, denoted \( Q'(t) \).
Derivatives provide us with the instantaneous rate of change. In the context of the given exercise, it lets us see how the charge is increasing or decreasing at any single moment. For a wire carrying current, this rate of change of charge is defined as the current \( I(t) \).
Therefore, knowing the derivative of the charge with respect to time is equivalent to understanding the behavior of the current. If you can determine \( I(t) \), then you are essentially observing how rapidly the charge is flowing through the wire at any instant.
Derivatives provide us with the instantaneous rate of change. In the context of the given exercise, it lets us see how the charge is increasing or decreasing at any single moment. For a wire carrying current, this rate of change of charge is defined as the current \( I(t) \).
Therefore, knowing the derivative of the charge with respect to time is equivalent to understanding the behavior of the current. If you can determine \( I(t) \), then you are essentially observing how rapidly the charge is flowing through the wire at any instant.
Current and Charge Relationship
The relationship between current and charge is a key concept in understanding electrical circuits. Current is defined as the rate at which charge moves through a conductor. In formula terms, for a time-dependent charge \( Q(t) \), the current \( I(t) \) is \( Q'(t) \), or the derivative of the charge.
This relationship tells us that if you know how the current varies with time, you can understand how the total charge changes over an interval. Integrating the current over a specific period gives you the net change in charge. This is how the integral \( \int_a^b I(t) \, dt \) becomes crucial. It computes the total amount of electrical charge that has been transferred from one point in time to another.
Thus, the interplay between the current and the accumulated charge showcases how fast or slow the charge is building up or reducing, providing a comprehensive view of the electric flow in a circuit over time.
This relationship tells us that if you know how the current varies with time, you can understand how the total charge changes over an interval. Integrating the current over a specific period gives you the net change in charge. This is how the integral \( \int_a^b I(t) \, dt \) becomes crucial. It computes the total amount of electrical charge that has been transferred from one point in time to another.
Thus, the interplay between the current and the accumulated charge showcases how fast or slow the charge is building up or reducing, providing a comprehensive view of the electric flow in a circuit over time.
Other exercises in this chapter
Problem 51
Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area. \( y = \sin x \), \( 0 \le x \le \pi
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Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take \( C = 0 \)). \
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Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area. \( y = \sec^2 x \), \( 0 \le x \le \
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Evaluate the definite integral. \( \displaystyle \int^1_0 \cos(\pi t/2) \, dt \)
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