Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. The variable y = t + 1 doubles in value whenever t increases by 1 unit.
Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
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
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Briggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblem 7.3.95a
Briggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblem 7.3.95aChapter 7, Problem 7.3.95a
Falling body When an object falling from rest encounters air resistance proportional to the square of its velocity, the distance it falls (in meters) after t seconds is given by d(t) = (m/k) ln (cosh (√(kg/m) t)), where m is the mass of the object in kilograms, g = 9.8 m/s² is the acceleration due to gravity, and k is a physical constant.
a. A BASE jumper (m = 75 kg) leaps from a tall cliff and performs a ten-second delay (she free-falls for 10 s and then opens her chute). How far does she fall in 10 s? Assume k = 0.2.
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Identify the given function for the distance fallen: \(d(t) = \frac{m}{k} \ln \left( \cosh \left( \sqrt{\frac{kg}{m}} \, t \right) \right)\), where \(m\) is mass, \(k\) is a constant, \(g = 9.8\) m/s², and \(t\) is time in seconds.
Substitute the known values into the formula: \(m = 75\), \(k = 0.2\), \(g = 9.8\), and \(t = 10\) seconds.
Calculate the term inside the square root: \(\sqrt{\frac{kg}{m}} = \sqrt{\frac{0.2 \times 9.8}{75}}\).
Evaluate the argument of the hyperbolic cosine function: \(\sqrt{\frac{kg}{m}} \times t = \left( \text{value from previous step} \right) \times 10\).
Compute the distance fallen by plugging the value into the formula: \(d(10) = \frac{75}{0.2} \times \ln \left( \cosh \left( \text{value from previous step} \right) \right)\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Modeling Air Resistance in Free Fall
Air resistance affects falling objects by opposing motion, often modeled as proportional to velocity squared for high speeds. This nonlinear drag force changes the acceleration and velocity over time, making the motion differ from simple free fall. Understanding this helps interpret the given distance formula involving hyperbolic functions.
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Exponential Growth & Decay
Hyperbolic Functions and Their Properties
Hyperbolic functions like cosh(x) and sinh(x) arise in solutions to differential equations involving quadratic velocity terms. The function cosh(x) = (e^x + e^{-x})/2 grows exponentially and appears in the distance formula, reflecting the balance between gravity and air resistance in the falling motion.
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Applying Given Formulas with Physical Constants
To find the distance fallen, substitute the known values (mass m, gravity g, constant k, and time t) into the formula d(t) = (m/k) ln(cosh(√(kg/m) t)). This requires careful calculation of the square root term and the natural logarithm, ensuring units are consistent and the physical meaning is preserved.
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