What is the inverse function of ln x, and what are its domain and range?
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- 0. Functions7h 54m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms36m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 41m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
0. Functions
Logarithmic Functions
Problem 7.3.95a
Textbook Question
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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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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