Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Problem 7.2.44b

Chapter 7, Problem 7.2.44b

A running model A model for the startup of a runner in a short race results in the velocity function v(t) = a(1 - e⁻ᵗ/ᶜ), where a and c are positive constants, t is measured in seconds, and v has units of m/s. (Source: Joe Keller, A Theory of Competitive Running, Physics Today, 26, Sep 1973)

b. Using the velocity in part (a) and assuming s(0) = 0, find the position function s(t), for t ≥ 0.

Verified step by step guidance

Recall that the position function $s(t)$ is the integral of the velocity function $v(t)$ with respect to time $t$. Mathematically, this is expressed as $s(t) = \int v(t) \, dt + C$, where $C$ is the constant of integration.

Substitute the given velocity function $v(t) = a(1 - e^{-t/c})$ into the integral to get $s(t) = \int a(1 - e^{-t/c}) \, dt + C$.

Split the integral into two parts: $s(t) = a \int 1 \, dt - a \int e^{-t/c} \, dt + C$.

Integrate each term separately: the integral of 1 with respect to $t$ is $t$, and the integral of $e^{-t/c}$ with respect to $t$ involves using the substitution $u = -t/c$, so $\int e^{-t/c} \, dt = -c e^{-t/c} + D$, where $D$ is another constant of integration.

Combine the results to write $s(t) = a \left( t + c e^{-t/c} \right) + C$. Use the initial condition $s(0) = 0$ to solve for $C$ by substituting $t=0$ into the expression and setting $s(0) = 0$.

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Key Concepts

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

Velocity and Position Relationship

Velocity is the rate of change of position with respect to time. Mathematically, velocity v(t) is the derivative of position s(t), so s(t) can be found by integrating v(t) over time. Given an initial position, integrating the velocity function yields the position function.

Integration of Exponential Functions

Integrating functions involving exponentials like e^(-t/c) requires understanding the antiderivative rules for exponential decay. The integral of e^(-kt) is (-1/k)e^(-kt) plus a constant. This is essential to find the position function from the given velocity.

Initial Conditions in Differential Equations

Initial conditions, such as s(0) = 0, are used to determine the constant of integration after integrating a velocity function. They ensure the position function accurately reflects the physical scenario, anchoring the solution to a specific starting point.

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Related Practice

Textbook Question

Theorem 7.8

Differentiate sinh⁻¹ x = ln (x + √(x² + 1)) to show that d/dx (sinh⁻¹ x) = 1 / √(x² + 1).

Textbook Question

Overtaking City A has a current population of 500,000 people and grows at a rate of 3%/yr. City B has a current population of 300,000 and grows at a rate of 5%/yr.

b. Suppose City C has a current population of y₀ < 500,000 and a growth rate of p > 3%/yr. What is the relationship between y₀ and p such that Cities A and C have the same population in 10 years?

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume x > 0 and y > 0.

b. ln 0 = 1

Textbook Question

Projection sensitivity

According to the 2014 national population projections published by the U.S. Census Bureau, the U.S. population is projected to be 334.4 million in 2020 with an estimated growth rate of 0.79%/yr.

b. Suppose the actual growth rate is instead 0.7%. What are the resulting doubling time and projected 2050 population?

Textbook Question

Depreciation of equipment A large die-casting machine used to make automobile engine blocks is purchased for \$2.5 million. For tax purposes, the value of the machine can be depreciated by 6.8% of its current value each year.

b. After how many years is the value of the machine 10% of its original value?

Textbook Question

Definitions of hyperbolic sine and cosine Complete the following steps to prove that when the x- and y-coordinates of a point on the hyperbola x² - y² = 1 are defined as cosh t and sinh t, respectively, where t is twice the area of the shaded region in the figure, x and y can be expressed as

x = cosh t = (eᵗ + e⁻ᵗ) / 2 and y = sinh t = (eᵗ - e⁻ᵗ) / 2.

b. In Chapter 8, the formula for the integral in part (a) is derived:

∫ √(z² − 1) dz = (z/2)√(z² − 1) − (1/2) ln|z + √(z² − 1)| + C.

Evaluate this integral on the interval [1, x], explain why the absolute value can be dropped, and combine the result with part (a) to show that:

t = ln(x + √(x² − 1)).