Textbook Question
17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.
lim_x→ -1 (x³ - x² - 5x - 3)/(x⁴ + 2x³ - x² -4x -2)
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Ch. 4 - Applications of the Derivative
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 4, Problem 3.1.48
Interpreting the derivative Find the derivative of each function at the given point and interpret the physical meaning of this quantity. Include units in your answer.
When a faucet is turned on to fill a bathtub, the volume of water in gallons in the tub after t minutes is V(t)=3t. Find V′(12).
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First, identify the function given: V(t) = 3t, which represents the volume of water in the bathtub in gallons as a function of time in minutes.
To find the derivative V'(t), apply the basic rule of differentiation for a linear function. The derivative of V(t) = 3t with respect to t is V'(t) = 3.
Evaluate the derivative at the given point t = 12. Since V'(t) = 3, V'(12) = 3.
Interpret the physical meaning of the derivative V'(12) = 3. This means that the rate at which the volume of water in the bathtub is increasing is 3 gallons per minute.
Include units in your interpretation: The derivative V'(12) = 3 gallons per minute indicates that at t = 12 minutes, the water is being added to the bathtub at a constant rate of 3 gallons per minute.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative
The derivative of a function measures the rate at which the function's value changes with respect to changes in its input. In this context, it represents the instantaneous rate of change of the volume of water in the bathtub with respect to time. Mathematically, it is defined as the limit of the average rate of change as the interval approaches zero.
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Derivatives
Physical Interpretation of Derivatives
In applied contexts, the derivative can be interpreted as a physical quantity. For the function V(t) = 3t, the derivative V′(t) indicates how quickly the volume of water is increasing at a specific time. This can be understood as the flow rate of water into the bathtub, typically measured in gallons per minute.
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Units of Measurement
When calculating derivatives in real-world scenarios, it is essential to include appropriate units to convey the meaning of the result. In this case, since V(t) is measured in gallons and t in minutes, the derivative V′(t) will have units of gallons per minute, providing a clear understanding of the rate at which water is being added to the bathtub.
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Related Practice
Textbook Question
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
x⁴+y⁴ = 2;(1,−1)
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Textbook Question
Sketch the graph of a function continuous on the given interval that satisfies the following conditions.
ƒ is continuous on the interval [-4, 4] ; f'(x) = 0 for x = -2, 0, and 3; ƒ has an absolute minimum at x = 3; ƒ has a local minimum at x = -2 ; ƒ has a local maximum at x = 0; ƒ has an absolute maximum at x = -4.
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