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Changing Voltage The voltage \(V\) (volts), current \(I\) (amperes), and resistance \(R(\) ohms of an electric circuit like the one shown here are related by the equation \(V=I R .\) Suppose that \(V\) is increasing at the rate of 1 volt/sec while \(I\) is decreasing at the rate of 1\(/ 3\) amp/sec. Let \(t\) denote time in sec. (a) What is the value of \(d V / d t ?\) (b) What is the value of \(d I / d t ?\) (c) Write an equation that relates \(d R / d t\) to \(d V / d t\) and \(d I / d t\) (d) Writing to Learn Find the rate at which \(R\) is changing when \(V=12\) volts and \(I=2\) amp. Is \(R\) increasing, or decreasing? Explain.

Short Answer

Expert verified
The value of \(d V / d t\) is 1 volt/sec. The value of \(d I / d t\) is -1/3 amp/sec. The derived equation for \(d R / d t\) is \(\frac{1}{I} (\frac{d V}{d t} - R \frac{d I}{d t})\). At \(V=12\) volts and \(I=2\) amp, the rate at which \(R\) is changing is 1 ohm/sec and it is increasing.

Step by step solution

01

Find Value of \(d V / d t\)

The value of \(d V / d t\) is given as 1 volt/sec.
02

Find Value of \(d I / d t\)

The value of \(d I / d t\) is given as -1/3 amp/sec.
03

Derive the Equation for \(d R / d t\)

From the given equation, \(V = I R\). Differentiating both sides with respect to time, we have \(\frac{d V}{d t} = I \frac{d R}{d t} + R \frac{d I}{d t}\). Solving for \(d R / d t\) yields \(\frac{d R}{d t} = \frac{1}{I} (\frac{d V}{d t} - R \frac{d I}{d t})\)
04

Calculate Rate of Change of Resistance

At \(V=12\) volts and \(I=2\) amp, the resistance, \(R = V / I = 6\) ohms. Substituting these values into the derived equation of \(d R / d t\), we have \(d R / d t = \frac{1}{2} (1 - 6 * -1/3) = 1\) ohms/sec.
05

Determine Direction of Change of Resistance

As \(d R / d t\) is positive, the resistance \(R\) is increasing.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Voltage-Current Relationship

Understanding the relationship between voltage, current, and resistance is fundamental in analyzing electric circuits. This triad is described by Ohm's Law, which states that the voltage (V, in volts) is directly proportional to the current (I, in amperes) and inversely proportional to the resistance (R, in ohms) in a circuit. The formula is succinctly expressed as \(V = I \times R\). Essentially, this formula means that if we increase the current while keeping the resistance constant, the voltage across the circuit will increase proportionally. Conversely, for a constant current, increasing the resistance will result in a higher voltage.

In the context of the exercise, the students are meant to observe these relationships in a dynamic scenario where the voltage is increasing and the current is decreasing. The exercise encourages an understanding of how these variables interact over time, not just statically. This dynamic relationship is crucial for real-world applications where electrical circuits are subject to variable conditions.

Rate of Change in Circuits
Moving beyond static measurements, the rate of change in electrical circuits is crucial when conditions are constantly evolving. For instance, when considering a voltage increase over time, we are discussing the rate at which voltage changes—this is denoted as \(\frac{dV}{dt}\), where \(t\) represents time. Similarly, changes in current will be represented as \(\frac{dI}{dt}\).When studying circuits, grasping the rate of change is as important as understanding the magnitudes of voltage, current, and resistance. This understanding is applied in the exercise by assessing the rates at which the voltage is increasing and the current is decreasing. Knowing these rates, it is then possible to derive related rates for other quantities, such as resistance, by applying calculus - a procedure that the exercise walks students through step by step.Equations involving rates of change are not static - they represent how a circuit responds over time and are key to optimizing circuit design and function. For example, a circuit designer might calculate how quickly a resistor can adjust its resistance in response to a surge in voltage, thereby protecting sensitive circuit components from damage. This dynamic analysis is also vital for troubleshooting circuit behavior under live, changing conditions.
Related Rates Calculus
The calculus concept of related rates is about how different quantities change together over time. In calculus, when two or more quantities are changing with respect to time, and there's a known relationship between them, we can use derivatives to find the rates at which these quantities change. This becomes particularly crucial in describing phenomena in physics, engineering, and economics, where multiple variables interact dynamically.In the textbook exercise, the relationship is established by the equation \(V=IR\), with given rates of change \(\frac{dV}{dt}=1\, \text{volt/sec}\) and \(\frac{dI}{dt}=-\frac{1}{3}\, \text{amp/sec}\). The exercise involves finding the rate of change of resistance, \(\frac{dR}{dt}\), which requires differentiating with respect to time. This practice is a classic example of related rates in calculus.Solving for \(\frac{dR}{dt}\) can be done by employing the chain rule—a fundamental tool in differentiation. Once we have the related rate for resistance, we can then analyze how resistance changes in conjunction with voltage and current. This knowledge can be applied, for example, in predicting how temperature changes in a circuit may affect resistance, given that temperature can affect material properties and thus the resistance of components in a circuit. The exercise aims to solidify students' understanding of related rates by applying these concepts to a tangible scenario.

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Most popular questions from this chapter

Walkers \(A\) and \(B\) are walking on straight streets that meet at right angles. \(A\) approaches the intersection at 2 \(\mathrm{m} / \mathrm{sec}\) and \(B\) moves away from the intersection at 1 \(\mathrm{m} / \mathrm{sec}\) as shown in the figure. At what rate is the angle \(\theta\) changing when \(A\) is 10 \(\mathrm{m}\)from the intersection and \(B\) is 20 \(\mathrm{m}\) from the intersection? Express your answer in degrees per second to the nearest degree.

Multiple Choice Which of the following conditions would enable you to conclude that the graph of \(f\) has a point of inflection at \(x=c ?\) (A) There is a local maximum of \(f^{\prime}\) at \(x=c\) . (B) \(f^{\prime \prime}(c)=0 .\) (C) \(f^{\prime \prime}(c)\) does not exist. (D) The sign of \(f^{\prime}\) changes at \(x=c\) . (E) \(f\) is a cubic polynomial and \(c=0\)

Draining Conical Reservoir Water is flowing at the rate of 50 \(\mathrm{m}^{3} / \mathrm{min}\) from a concrete conical reservoir (vertex down) of base radius 45 \(\mathrm{m}\) and height 6 \(\mathrm{m} .\) (a) How fast is the water level falling when the water is 5 \(\mathrm{m}\) deep? (b) How fast is the radius of the water's surface changing at that moment? Give your answer in \(\mathrm{cm} / \mathrm{min.}\)

How We Cough When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the question of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v(\) in \(\mathrm{cm} / \mathrm{sec})\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in \(\mathrm{cm}\) and \(c\) is a positive constant whose value depends in part on the length of the trachea. (a) Show that \(v\) is greatest when \(r=(2 / 3) r_{0},\) that is, when the trachea is about 33\(\%\) contracted. The remarkable fact is that \(X\) -ray photographs confirm that the trachea contracts about this much during a cough. (b) Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see to the claim that \(v\) is a maximum when \(r=(2 / 3) r_{0}\) .

Upper Bounds Show that for any numbers \(a\) and \(b\) \(|\sin b-\sin a| \leq|b-a|\)

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