Textbook Question
A capacitor is a device in an electrical circuit that stores charge. In one particular circuit, the charge on the capacitor Q varies in time as shown in the figure. <IMAGE>
b. Is Q′ positive or negative for t≥0?
175
views
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 55a
A capacitor is a device in an electrical circuit that stores charge. In one particular circuit, the charge on the capacitor Q varies in time as shown in the figure. <IMAGE>
a. At what time is the rate of change of the charge Q' the greatest?
Verified step by step guidance1
Step 1: Understand that the rate of change of the charge Q with respect to time is represented by the derivative Q'(t). We are looking for the time at which this derivative is the greatest.
Step 2: Analyze the graph of Q(t) provided in the image. The rate of change, Q'(t), is represented by the slope of the tangent line to the curve at any point.
Step 3: Identify the point on the graph where the slope of the tangent line is steepest. This point corresponds to the maximum value of Q'(t).
Step 4: If the graph is not available, consider the mathematical function Q(t) if provided. Take the derivative of Q(t) to find Q'(t) and determine where this derivative reaches its maximum value.
Step 5: Verify the time at which Q'(t) is greatest by checking the concavity of Q(t) around this point, ensuring it is indeed a maximum.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Rate of Change
The rate of change refers to how a quantity changes over time, often represented mathematically as a derivative. In the context of the capacitor's charge Q, the rate of change Q' indicates how quickly the charge is increasing or decreasing at any given moment. Understanding this concept is crucial for determining when the charge is changing most rapidly.
Recommended video:
04:16
Intro To Related Rates
Derivatives
Derivatives are fundamental tools in calculus that measure how a function changes as its input changes. For the charge Q as a function of time, the derivative Q' provides insight into the instantaneous rate of change of charge. Analyzing the derivative helps identify critical points where the rate of change reaches its maximum.
Recommended video:
05:44Derivatives
Graphical Analysis
Graphical analysis involves interpreting the visual representation of a function to understand its behavior. In this case, examining the graph of charge Q over time can reveal where the slope (representing Q') is steepest, indicating the greatest rate of change. This method is essential for visually identifying key points in the function's behavior.
Recommended video:
05:02Determining Differentiability Graphically
Related Practice
Textbook Question
Where is the function continuous? Differentiable? Use the graph of f in the figure to do the following. <IMAGE>
c. Sketch a graph of f'.
146
views
Textbook Question
Robert Boyle (1627–1691) found that for a given quantity of gas at a constant temperature, the pressure P (in kPa) and volume V of the gas (in m³) are accurately approximated by the equation V=k/P, where k>0 is constant. Suppose the volume of an expanding gas is increasing at a rate of 0.15 m³/min when the volume V=0.5 m³ and the pressure is P=50 kPa. At what rate is pressure changing at this moment?
203
views
Textbook Question
Evaluate and simplify y'.
x = cos (x−y)
286
views
Textbook Question
An object oscillates along a vertical line, and its position in centimeters is given by y(t)=30(sint - 1), where t ≥ 0 is measured in seconds and y is positive in the upward direction.
At what times and positions is the velocity zero?
216
views
Textbook Question
An object oscillates along a vertical line, and its position in centimeters is given by y(t) = 30(sin t - 1), where t ≥ 0 is measured in seconds and y is positive in the upward direction.
The acceleration of the oscillator is a(t) = v′(t). Find and graph the acceleration function.
220
views