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Chapter 10: Problem 9

In a simple electric circuit, Ohm's law states that \(V=I R,\) where \(\mathrm{V}\) is the voltage in volts, I is the current in amperes, and \(\mathrm{R}\) is the resistance in ohms. Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.02 ohms per second. When the resistance is 100 ohms and the current is 0.02 amperes, at what rate is the current changing? ___________ amperes per second .

Short Answer

Expert verified
The rate at which the current is changing when the resistance is 100 ohms and the current is 0.02 amperes is approximately \(-0.000304 \, \text{amperes/sec}\).

Step by step solution

01

Write down the given rate of changes

We are given the following rates of change: - Voltage (V): -0.03 volts/second - Resistance (R): 0.02 ohms/second
02

Write down the given values for when we need to find the rate

We are asked to find the rate at which the current (I) is changing when: - Resistance (R): 100 ohms - Current (I): 0.02 amperes
03

Differentiate the Ohm's law equation with respect to time t

Differentiate the equation \(V = IR\) with respect to time (t) to obtain the rate of change of current: \[\frac{dV}{dt} = \frac{d(IR)}{dt}\] Keep in mind that \(I, R, \) and \(V\) are all functions of time, so we'll use the Chain Rule to differentiate the product \(IR\): \[\frac{dV}{dt} = R \frac{dI}{dt} + I \frac{dR}{dt}\]
04

Plug in the given values and rates of change

Now, we can substitute the given values and rates of change into the equation: \[-0.03 = 100 \cdot \frac{dI}{dt} + 0.02 \cdot 0.02\]
05

Solve for dI/dt

Next, solve for the rate of the change of the current, \(\frac{dI}{dt}\): \[-0.03 = 100 \cdot \frac{dI}{dt} + 0.0004\] \[-0.0304 = 100 \cdot \frac{dI}{dt}\] \[\frac{dI}{dt} = \frac{-0.0304}{100}\] \[\frac{dI}{dt} = -0.000304\]
06

Write the final answer

Thus, the rate at which the current is changing when the resistance is 100 ohms and the current is 0.02 amperes is: \[-0.000304 \, \text{amperes/sec}\]

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Key Concepts

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

Differential Calculus
Differential calculus is a subfield of calculus concerned with the study of how things change. It allows us to calculate the rate of change of quantities. In the exercise, we applied differential calculus to determine how quickly the current was changing in an electric circuit. By taking the derivative of Ohm's Law \(V = IR\) with respect to time, we are able to find this rate of change, known as \(\frac{dI}{dt}\).

This involves applying the Chain Rule, a fundamental technique in calculus used to differentiate a function of a function, because both current \(I\) and resistance \(R\) are variables that change over time. The Chain Rule is expressed as \(\frac{df(g(t))}{dt} = f'(g(t)) \cdot g'(t)\), and in the context of the given problem, it means taking the derivative of each term independently while considering the other as a constant, and then summing the results.
Electric Circuits
Electric circuits are pathways through which electricity flows. Ohm鈥檚 Law, a fundamental principle in the study of electric circuits, illustrates the relationship between voltage \(V\), current \(I\), and resistance \(R\) in a simple circuit. According to this law, \(V = IR\), meaning that the voltage (the force driving the current) is equal to the product of the current (the flow rate of charge) and the resistance (which opposes the flow).

In practical terms, if you alter the resistance or the voltage in a circuit, the current will change accordingly. This change is what students were asked to find in the exercise. Understanding how any of these variables affect one another is crucial for students who wish to work with electric circuits, whether it be in designing them or troubleshooting existing systems.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of more than one variable. In the context of the provided exercise, voltage, current, and resistance are all functions of time, representing a multivariable situation because the rate of change of one is affected by the rates of change of the others.

When applying multivariable calculus, we often work with partial derivatives, sine each variable's rate of change is considered with the understanding that other variables are also changing. However, in this problem, we are using the time derivative of Ohm鈥檚 Law because we are examining how the current changes over time due to changes in both voltage and resistance. This is an example of how multivariable calculus principles can be narrowed down to handle real-world problems involving several changing variables.

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Most popular questions from this chapter

(a) Check the local linearity of \(f(x, y)=e^{-x} \cos (y)\) near \(x=1, y=1.5\) by filling in the following table of values of \(f\) for \(x=0.9,1,1.1\) and \(y=1.4,1.5,1.6 .\) Express values of \(f\) with 4 digits after the decimal point. (b) Next, fill in the table for the values \(x=0.99,1,1.01\) and \(y=\) 1.49,1.5,1.51 , again showing 4 digits after the decimal point. Notice if the two tables look nearly linear, and whether the second looks more linear than the first (in particular, think about how you would decide if they were linear, or if the one were more closely linear than the other). (c) Give the local linearization of \(f(x, y)=e^{-x} \cos (y)\) at (1,1.5) : Using the second of your tables: \(f(x, y) \approx\) ____________. Using the fact that \(f_{x}(x, y)=-e^{-x} \cos (y)\) and \(f_{y}(x, y)=-e^{-x} \sin (y):\) \(f(x, y) \approx\) ___________.

In the following questions, we determine and apply the linearization for several different functions. a. Find the linearization \(L(x, y)\) for the function \(f\) defined by \(f(x, y)=\) \(\cos (x)\left(2 e^{2 y}+e^{-2 y}\right)\) at the point \(\left(x_{0}, y_{0}\right)=(0,0) .\) Use the linearization to estimate the value of \(f(0.1,0.2)\). Compare your estimate to the actual value of \(f(0.1,0.2)\) b. The Heat Index, \(I,\) (measured in apparent degrees \(\mathrm{F}\) ) is a function of the actual temperature \(T\) outside (in degrees \(\mathrm{F}\) ) and the relative humidity \(H\) (measured as a percentage). A portion of the table which gives values for this function, \(I=I(T, H),\) is provided in Table 10.4 .13 $$\begin{array}{ccccc}\hline T \downarrow \backslash H \rightarrow & 70 & 75 & 80 & 85 \\ \hline 90 & 106 & 109 & 112 & 115 \\\\\hline 92 & 112 & 115 & 119 & 123 \\ \hline 94 & 118 & 122 & 127 & 132 \\\\\hline 96 & 125 & 130 & 135 & 141 \\\\\hline\end{array}$$ Suppose you are given that \(I_{T}(94,75)=3.75\) and \(I_{H}(94,75)=0.9\). Use this given information and one other value from the table to estimate the value of \(I(93.1,77)\) using the linearization at (94,75) . Using proper terminology and notation, explain your work and thinking. c. Just as we can find a local linearization for a differentiable function of two variables, we can do so for functions of three or more variables. By extending the concept of the local linearization from two to three variables, find the linearization of the function \(h(x, y, z)=e^{2 x}(y+\) \(z^{2}\) ) at the point \(\left(x_{0}, y_{0}, z_{0}\right)=(0,1,-2) .\) Then, use the linearization to estimate the value of \(h(-0.1,0.9,-1.8)\).

Find the maximum and minimum values of the function \(f(x, y, z)=\) \(x^{2} y^{2} z^{2}\) subject to the constraint \(x^{2}+y^{2}+z^{2}=64 .\) Maximum value is ____________, occuring at points (positive integer or "infinitely many"). Minimum value is ____________, occuring at points (positive integer or "infinitely many").

The temperature on an unevenly heated metal plate positioned in the first quadrant of the \(x y\) -plane is given by $$C(x, y)=\frac{25 x y+25}{(x-1)^{2}+(y-1)^{2}+1}$$ Assume that temperature is measured in degrees Celsius and that \(x\) and \(y\) are each measured in inches. (Note: At no point in the following questions should you expand the denominator of \(C(x, y) .)\) a. Determine \(\left.\frac{\partial C}{\partial x}\right|_{(x, y)}\) and \(\left.\frac{\partial C}{\partial y}\right|_{(x, y)}\). b. If an ant is on the metal plate, standing at the point (2,3) , and starts walking in the direction parallel to the positive \(y\) axis, at what rate will the temperature the ant is experiencing change? Explain, and include appropriate units. c. If an ant is walking along the line \(y=3\) in the positive \(x\) direction, at what instantaneous rate will the temperature the ant is experiencing change when the ant passes the point (1,3)\(?\) d. Now suppose the ant is stationed at the point (6,3) and walks in a straight line towards the point (2,0) . Determine the average rate of change in temperature (per unit distance traveled) the ant encounters in moving between these two points. Explain your reasoning carefully. What are the units on your answer?

Find all the first and second order partial derivatives of \(f(x, y)=3 \sin (2 x+\) \(y)-4 \cos (x-y)\) A. \(\frac{\partial f}{\partial x}=f_{x}=\) ________. B. \(\frac{\partial f}{\partial y}=f_{y}=\) ________. C. \(\frac{\partial^{2} f}{\partial x^{2}}=f_{x x}=\) ________. D. \(\frac{\partial^{2} f}{\partial y^{2}}=f_{y y}=\) ________. E. \(\frac{\partial^{2} f}{\partial x \partial y}=f_{y x}=\) ________. F. \(\frac{\partial^{2} f}{\partial y \partial x}=f_{x y}=\) ________.
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