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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.6.56c

{Use of Tech} Power and energy 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 / 9kWh where t=0 corresponds to midnight.
c. Graph the power function and interpret the graph. What are the units of power in this case?

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To find the power function, we need to determine the rate of change of energy with respect to time. This involves taking the derivative of the energy function E(t) with respect to t.
The given energy function is E(t) = 100t + 4t² - t³ / 9. To find the power function P(t), compute the derivative: P(t) = dE/dt.
Differentiate each term of E(t) with respect to t: The derivative of 100t is 100, the derivative of 4t² is 8t, and the derivative of -t³/9 is -3t²/9 or -t²/3.
Combine these derivatives to get the power function: P(t) = 100 + 8t - t²/3.
Graph the power function P(t) = 100 + 8t - t²/3. The units of power in this case are kilowatts (kW), as the energy function E(t) is given in kilowatt-hours (kWh) and time t is in hours.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Energy vs. Power

Energy is the capacity to do work or produce heat, measured in joules (J) or calories (Cal). Power, on the other hand, is the rate at which energy is consumed or produced, measured in watts (W), where 1 W equals 1 J/s. Understanding the distinction between these two concepts is crucial for analyzing how energy is utilized over time.
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The Power Rule

Cumulative Energy Function

The cumulative energy function E(t) = 100t + 4t² - t³ represents the total energy consumed over time, where t is measured in hours. This function allows us to calculate the energy used at any given time and is essential for determining the power function by differentiating E(t) with respect to time.
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Graphing Power Function

To graph the power function, we first derive it from the cumulative energy function, which gives us the rate of energy consumption at any time t. The resulting power function will be plotted against time, allowing us to visualize how power usage changes throughout the day. The units of power in this context will be in kilowatts (kW), reflecting the energy consumption rate.
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Related Practice
Textbook Question

62–65. {Use of Tech} Graphing f and f'

c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.

f(x)=(sec^−1 x)/x on [1,∞)

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Textbook Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 64 ft/s from a height of 32 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t²+64t+32.

c. What is the height of the stone at the highest point?

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Textbook Question

Derivatives from a graph If possible, evaluate the following derivatives using the graphs of f and f'. <IMAGE>

c. (f^-1)'(f(2))

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Textbook Question

Derivatives of sin^n x Calculate the following derivatives using the Product Rule.

c. d/dx (sin⁴ x)

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Textbook Question

Vibrations of a spring Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position y=0y=0 when the mass hangs at rest. Suppose you push the mass to a position y0y_0 units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system), the position y of the mass after t seconds is y=y0cos(tkm)y=y_0\(\cos\]\left\)(t\(\sqrt{\frac{k}{m}\)}\(\right\)), where k>0k>0 is a constant measuring the stiffness of the spring (the larger the value of kk, the stiffer the spring) and yy is positive in the upward direction.

Use equation (4) to answer the following questions.

c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness (kk is increased by a factor of 44)?

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Textbook Question

Finding derivatives from a table Find the values of the following derivatives using the table. <IMAGE>


c. d/dx ((f(x)g(x)) |x=3

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