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 7.2.50c

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 \(C\).

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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.

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