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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblem 7.3.96a
Chapter 7, Problem 7.3.96a

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).

Verified step by step guidance
1
Recall that the position function for the falling body is given by \(d(t)\), and the velocity function \(v(t)\) is the derivative of the position function with respect to time, i.e., \(v(t) = d'(t)\).
Identify the constants given: mass \(m = 75\) kg and drag coefficient \(k = 0.2\). Also, gravitational acceleration \(g\) is typically \(9.8 \ \text{m/s}^2\) unless otherwise specified.
Express the velocity function in terms of \(m\), \(k\), and \(g\). The problem states that \(v(t) = \sqrt{\frac{mg}{k}} \tanh \left( \sqrt{\frac{kg}{m}} t \right)\), so we need to confirm this by differentiating the position function \(d(t)\).
Use the chain rule to differentiate \(d(t)\), which likely involves hyperbolic functions due to the presence of \(\tanh\) in \(v(t)\). The derivative of \(\tanh(x)\) is \(\text{sech}^2(x)\), and the derivative inside the argument must be accounted for.
After differentiating, simplify the expression to show that it matches the given velocity formula \(v(t) = \sqrt{\frac{mg}{k}} \tanh \left( \sqrt{\frac{kg}{m}} t \right)\), confirming the velocity function.

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Key Concepts

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

Position and Velocity Functions

The position function d(t) describes the location of a falling body at time t, while the velocity function v(t) is its derivative d'(t), representing the rate of change of position. Understanding how to differentiate position functions is essential to find velocity.
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Using The Velocity Function

Hyperbolic Functions and Their Derivatives

Hyperbolic functions like tanh(x) often appear in solutions to differential equations involving drag forces. Knowing the properties and derivatives of tanh(x) helps verify velocity expressions derived from position functions.
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Asymptotes of Hyperbolas

Modeling Drag Force in Falling Bodies

The parameters m (mass) and k (drag coefficient) model the effect of air resistance on a falling body. The velocity formula involving √(mg/k) and tanh(√(kg/m) t) arises from solving the motion equation with drag proportional to velocity.
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Exponential Growth & Decay
Related Practice
Textbook Question

61–62. Points of intersection and area

a. Sketch the graphs of the functions f and g and find the x-coordinate of the points at which they intersect.


f(x) = sech x, g(x) = tanh x; the region bounded by the graphs of f, g, and the y-axis

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Textbook Question

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.

a. Find an exponential decay function V₁(t) that equals the total volume of the quiescent cells in the tumor t days after treatment.

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Textbook Question

Zero net area Consider the function f(x) = (1 − x)/x

a. Are there numbers 0 < a < 1 such that ∫₁₋ₐ¹⁺ᵃ f(x) dx = 0?

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Textbook Question

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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Textbook Question

A power line is attached at the same height to two utility poles that are separated by a distance of 100 ft; the power line follows the curve ƒ(x) = a cosh x/a. Use the following steps to find the value of a that produces a sag of 10 ft midway between the poles. Use a coordinate system that places the poles at x = ±50.


a. Show that a satisfies the equation cosh 50/a − 1 = 10/a.

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Textbook Question

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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