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

Briggs 3rd Edition

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

Problema 7.2.50c

Capitolo 7, Problema 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.

Guida verificata passo dopo passo

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 \(C\).

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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.

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05:02

Intro to the Chain Rule

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.

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Percorso guidato

06:15

Derivatives Applied To Acceleration

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.

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06:06

Solving Separable Differential Equations

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