Problem 38 - Calculus Early Transcendentals

Question

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 that the initial velocity and position are given by \(v(0)=10\) and \(s(0)=0,\) respectively. a. Use \(a(t)=v^{\prime}(t)\) to find the velocity of the object as a function of time. b. Use \(v(t)=s^{\prime}(t)\) to find the position of the object as a function of time. c. Use the fact that \(d v / d t=(d v / d s)(d s / d t)\) (by the Chain Rule) to find the velocity as a function of position.

Step-by-Step Solution

Answer

Question: Given the acceleration function \(a(t) = -kv(t)\) with initial velocity \(v(0)=10\) and initial position \(s(0)=0\), find the velocity function, the position function, and the velocity as a function of position. Answer: The velocity function is \(v(t) = 10e^{-kt}\), the position function is \(s(t) = -\frac{10}{k}e^{-kt} + \frac{10}{k}\), and the velocity as a function of position is \(v(s) = -\frac{k}{10}s + 10\).

Step 1: a. Finding the velocity function

Integrate the acceleration function \(a(t) = -kv(t)\) with respect to time: 1. Separate the variables by dividing both sides by \(v(t)\) and multiplying both sides by \(dt\): \(\frac{1}{v(t)} \, dv = -k \, dt\). 2. Integrate both sides: \(\int \frac{1}{v(t)} \, dv = \int -k \, dt\). 3. Apply the integration: \(\ln|v(t)| = -kt + C_1\). 4. Solve for \(v(t)\): \(v(t) = e^{-kt + C_1}\). 5. Since \(v(0) = 10\), plug in the value and solve for \(C_1\): \(10 = e^0 \cdot e^{C_1}\), \(C_1 = \ln(10)\). 6. The velocity function is \(v(t) = 10e^{-kt}\).

Step 2: b. Finding the position function

Integrate the velocity function \(v(t) = 10e^{-kt}\) with respect to time: 1. Integrate the velocity function with respect to time: \(\int 10e^{-kt} \, dt\). 2. Apply integration: \(s(t) = -\frac{10}{k}e^{-kt} + C_2\). 3. Since \(s(0) = 0\), plug in the value and solve for \(C_2\): \(0 = -\frac{10}{k}e^0 + C_2\), \(C_2 = \frac{10}{k}\). 4. The position function is \(s(t) = -\frac{10}{k}e^{-kt} + \frac{10}{k}\).

Step 3: c. Finding the velocity as a function of position

Follow the chain rule: \(\frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt}\) and substitute the provided expressions: 1. Replace \(\frac{dv}{dt}\) with the acceleration function \(a(t) = -kv(t)\) and replace \(\frac{ds}{dt}\) with the velocity function \(v(t) = 10e^{-kt}\): \(-kv(t) = \frac{dv}{ds} \cdot 10e^{-kt}\). 2. Divide both sides by \(10e^{-kt}\), which simplifies to: \(-\frac{k}{10} = \frac{dv}{ds}\). 3. Integrate both sides with respect to \(s\): \(\int -\frac{k}{10} \, ds = \int dv\). 4. Apply the integration: \(-\frac{k}{10}s = v + C_3\). 5. Since we already found that \(v(0) = 10\), plug in the value and solve for \(C_3\): \(-\frac{k}{10} \cdot 0 = 10 + C_3\), \(C_3 = 10\). 6. The velocity as a function of position is \(v(s) = -\frac{k}{10}s + 10\).

Key Concepts

Acceleration and Velocity

Understanding the relationship between acceleration (\(a(t)\)) and velocity (\(v(t)\)) is crucial in calculus, particularly when analyzing the motion of an object along a straight line. Acceleration is defined as the rate of change of velocity with respect to time. In the given exercise, acceleration is proportional to the negative of velocity, indicating a resistance proportional to the velocity, like air resistance or friction.

When approaching problems involving acceleration and velocity, it's essential to start by stating the relationship between them: acceleration is the derivative of velocity with respect to time, or mathematically, \(a(t) = v'(t)\). To find the velocity as a function of time when given the acceleration function, we must perform integration—the reverse process of differentiation. Here, the negative sign and the presence of the velocity term present a case of a separable differential equation, which leads us into a process where we can separate the variables and integrate each side individually.

After integrating and applying the initial condition (\(v(0) = 10\)), we find the velocity as a function of time: \(v(t) = 10e^{-kt}\). This equation expresses velocity decaying exponentially over time, with \(k\) playing a role in the rate of decay, making it foundational for further analysis in the exercise.

Chain Rule Integration

The Chain Rule is a fundamental theorem in calculus that allows us to differentiate composite functions. But it also plays a pivotal role in integration, especially when variables are entwined. The given exercise leverages the Chain Rule distinctly by expressing the derivative of velocity with respect to time in terms of position (\(s\)).

The Chain Rule implies that \(\frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt}\), binding together velocity, acceleration, and the derivative of velocity with respect to position. To find the velocity as a function of position, we treat \(\frac{dv}{dt}\) as \(a(t)\) and \(\frac{ds}{dt}\) as \(v(t)\), thereby tying together acceleration and velocity in terms of position.

Through integration and applying \(v(0) = 10\), we derive the formula \(v(s) = -\frac{k}{10}s + 10\), which is monumental in tracking how velocity changes with position. This connection is not only crucial for understanding motion but also illustrates the elegant interplay between different aspects of calculus in modeling real-world scenarios.

Separation of Variables

Separation of variables is a method used to solve differential equations, where we can 'separate' the variables involved into opposite sides of an equation. It is one of the simplest and most intuitive methods for solving first-order ordinary differential equations. This technique is particularly useful when dealing with equations where the derivative of a function is equal to the product of a function of the dependent variable and a function of the independent variable.

In the case of our exercise, separation of variables is employed to dissociate the velocity term from the time term in the acceleration function. The process involves dividing both sides of the acceleration equation by the velocity \(v(t)\), thereby isolating it, and then integrating both sides with respect to their respective variables.

After the separation, integration of both sides yields a logarithmic function due to the integrand \(1/v(t)\), which is then exponentiated to solve for \(v(t)\) explicitly as a function of time. The initial condition \(v(0) = 10\) ensures a specific solution, leading to the exponential decay model of velocity over time. This simple yet powerful technique provides a clear path from an acceleration function to a defined velocity function, demonstrating a core aspect of solving differential equations in calculus.