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?
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Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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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 55f
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.
Verified step by step guidance1
To find the acceleration function, we first need to determine the velocity function v(t). The velocity is the derivative of the position function y(t) with respect to time t. So, we start by differentiating y(t) = 30(sin t - 1).
The derivative of y(t) = 30(sin t - 1) with respect to t is v(t) = 30 * cos(t). This is because the derivative of sin(t) is cos(t), and the constant -1 becomes 0 when differentiated.
Now, to find the acceleration function a(t), we need to differentiate the velocity function v(t) = 30 * cos(t) with respect to t.
The derivative of v(t) = 30 * cos(t) is a(t) = -30 * sin(t). This is because the derivative of cos(t) is -sin(t).
To graph the acceleration function a(t) = -30 * sin(t), note that it is a sinusoidal function with amplitude 30, period 2π, and it oscillates between -30 and 30. The graph will be a sine wave starting at 0 when t = 0, going downwards initially because of the negative sign.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Differentiation
Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative represents the rate of change of a function with respect to its variable. In this context, we need to differentiate the position function y(t) to find the velocity v(t) and then differentiate again to find the acceleration a(t).
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Finding Differentials
Acceleration
Acceleration is defined as the rate of change of velocity with respect to time. In this problem, the acceleration function a(t) is derived from the velocity function v(t), which is itself obtained by differentiating the position function y(t). Understanding how to compute and interpret acceleration is crucial for analyzing the motion of the oscillating object.
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06:15Derivatives Applied To Acceleration
Graphing Functions
Graphing functions involves plotting the values of a function on a coordinate system to visualize its behavior. For the acceleration function a(t), we will need to calculate its values over a range of t and then plot these points to observe how the acceleration changes over time. This visual representation helps in understanding the dynamics of the oscillating object.
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5:53Graph of Sine and Cosine Function
Related Practice
Textbook Question
The energy (in joules) released by an earthquake of magnitude M is given by the equation E = 25,000 ⋅ 101.5M. (This equation can be solved for M to define the magnitude of a given earthquake; it is a refinement of the original Richter scale created by Charles Richter in 1935.)
Compute the energy released by earthquakes of magnitude 1, 2, 3, 4, and 5. Plot the points on a graph and join them with a smooth curve.
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Textbook Question
Power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up. It is measured in units of joules or Calories, where 1 Cal=4184 J. One hour of walking consumes roughly 10⁶J, or 240 Cal. On the other hand, power is the rate at which energy is used, which is measured in watts, where 1 W = 1 J/s. Other useful units of power are kilowatts (1 kW=10³ W) and megawatts (1 MW=10⁶ W). If energy is used at a rate of 1 kW for one hour, the total amount of energy used is 1 kilowatt-hour (1 kWh = 3.6×10⁶ J) Suppose the cumulative energy used in a large building over a 24-hr period is given by E(t)=100t + 4t² − (t³ / 9) kWh where t = 0 corresponds to midnight.
The power is the rate of energy consumption; that is, P(t) = E′(t) Find the power over the interval 0 ≤ t ≤ 24.
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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?
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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>
a. At what time is the rate of change of the charge Q' the greatest?
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Textbook Question
Calculate the derivative of the following functions.
y = cos7/4(4x3)
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