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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblema 7.3.97a
Capitolo 7, Problema 7.3.97a

Terminal velocity Refer to Exercises 95 and 96.


a. Compute a jumper’s terminal velocity, which is defined as lim t → ∞ v(t) = lim t → ∞ √(mg/k) tanh (√(kg/m) t).

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Identify the given velocity function: \(v(t) = \sqrt{\frac{mg}{k}} \tanh \left( \sqrt{\frac{kg}{m}} t \right)\), where \(m\) is mass, \(g\) is acceleration due to gravity, and \(k\) is a constant related to air resistance.
Recall the definition of terminal velocity as the limit of \(v(t)\) as time \(t\) approaches infinity: \(\lim_{t \to \infty} v(t)\).
Focus on the behavior of the hyperbolic tangent function \(\tanh(x)\) as \(x \to \infty\). Remember that \(\tanh(x)\) approaches 1 when \(x\) becomes very large.
Apply this limit to the velocity function: replace \(\tanh \left( \sqrt{\frac{kg}{m}} t \right)\) with 1 as \(t \to \infty\).
Conclude that the terminal velocity is \(\lim_{t \to \infty} v(t) = \sqrt{\frac{mg}{k}} \times 1 = \sqrt{\frac{mg}{k}}\).

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Limit of a Function as t Approaches Infinity

The limit of a function as t approaches infinity describes the behavior of the function as the input grows without bound. In this problem, evaluating lim t → ∞ v(t) helps find the steady-state velocity, known as terminal velocity, where acceleration ceases and velocity stabilizes.
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Hyperbolic Tangent Function (tanh)

The hyperbolic tangent function, tanh(x), is a smooth, continuous function that approaches 1 as x approaches infinity and -1 as x approaches negative infinity. Understanding tanh's limiting behavior is crucial to simplifying the velocity expression for large time values.
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Slopes of Tangent Lines

Terminal Velocity in Physics

Terminal velocity is the constant speed an object reaches when the force of gravity is balanced by the drag force, resulting in zero net acceleration. Mathematically, it is the limit of the velocity function as time goes to infinity, representing the maximum velocity attainable during free fall.
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Derivatives Applied To Velocity
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Chemotherapy In an experimental study at Dartmouth College, mice with tumors were treated with the chemotherapeutic drug Cisplatin. Before treatment, the tumors consisted entirely of clonogenic cells that divide rapidly, causing the tumors to double in size every 2.9 days. Immediately after treatment, 99% of the cells in the tumor became quiescent cells which do not divide and lose 50% of their volume every 5.7 days. For a particular mouse, assume the tumor size is 0.5 cm³ at the time of treatment.

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Velocity of falling body Refer to Exercise 95, which gives the position function for a falling body. Use m = 75 kg and k = 0.2.


a. Confirm that the BASE jumper’s velocity t seconds after jumping is v(t) = d'(t) = √(mg/k) tanh (√(kg/m) t).

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Shallow-water velocity equation

a. Confirm that the linear approximation to ƒ(x) = tanh x at a = 0 is L(x) = x.

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Evaluating hyperbolic functions Evaluate each expression without using a calculator or state that the value does not exist. Simplify answers as much as possible.


a. cosh 0

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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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