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

Briggs 3rd Edition

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

Problem 7.2.50c

Chapter 7, Problem 7.2.50c

Acceleration, velocity, position Suppose the acceleration of an object moving along a line is given by a(t) = -k v(t), where k is a positive constant and v is the object's velocity. Assume the initial velocity and position are given by v(0) = 10 and s(0) = 0, respectively.
c. Use the fact that dv/dt = (dv/ds)(ds/dt) (by the Chain Rule) to find the velocity as a function of position.

Verified step by step guidance

Start with the given acceleration equation: \(a(t) = -k v(t)\), where \(a(t) = \frac{dv}{dt}\) and \(v(t)\) is the velocity.

Use the Chain Rule relationship: \(\frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt}\). Since \(\frac{ds}{dt} = v\), rewrite acceleration as \(a = \frac{dv}{dt} = v \frac{dv}{ds}\).

Substitute \(a = -k v\) into the Chain Rule expression to get: \(v \frac{dv}{ds} = -k v\).

Divide both sides by \(v\) (assuming \(v \neq 0\)) to simplify: \(\frac{dv}{ds} = -k\).

Integrate both sides with respect to \(s\) to find \(v\) as a function of \(s\): \(\int dv = \int -k \, ds\), which leads to \(v(s) = -k s + C\). Use the initial conditions to solve for the constant \(C\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Chain Rule in Calculus

The Chain Rule allows differentiation of composite functions by relating rates of change. In this problem, it connects acceleration a(t) = dv/dt to velocity and position via dv/dt = (dv/ds)(ds/dt), enabling us to express velocity as a function of position instead of time.

Relationship Between Velocity, Position, and Acceleration

Velocity is the rate of change of position with respect to time (v = ds/dt), and acceleration is the rate of change of velocity with respect to time (a = dv/dt). Understanding these relationships is essential to rewrite acceleration in terms of velocity and position, facilitating the solution.

Solving First-Order Differential Equations

The problem leads to a separable differential equation involving velocity and position. Solving such equations requires isolating variables and integrating both sides, which yields velocity as a function of position, given initial conditions.

Recommended video:

06:06

Solving Separable Differential Equations

Related Practice

Textbook Question

Properties of exp(x) Use the inverse relations between ln x and exp(x), and the properties of ln x, to prove the following properties:

c. (exp(x))ᵖ = exp(px), p rational

views

Textbook Question

Energy consumption On the first day of the year (t=0), a city uses electricity at a rate of 2000 MW. That rate is projected to increase at a rate of 1.3% per year.

b. Find the total energy (in MW-yr) used by the city over four full years beginning at t=0.

views

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

c. ln(1 + √2) = −ln(−1 + √2)

views

Textbook Question

Evaluating hyperbolic functions Use a calculator to evaluate each expression or state that the value does not exist. Report answers accurate to four decimal places to the right of the decimal point.

c. csch⁻¹ 5

views

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)).

views

Textbook Question

Power lines 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.

c. Use your answer in part (b) to find a, and then compute the length of the power line.

views