Problem 85 - Calculus Single Variable

Question

Suppose an object of mass \(m\) is propelled upwards from the surface of the earth with initial velocity \(v_{0}\). Suppose that the (downward) force of air resistance \(R(v)\) is proportional to the square of the speed: \(R(v)=-k \cdot v^{2},\) where \(k\) is a positive constant that carries the units of mass/ length. (This is the quadratic drag law.) Solve the initial value problem for motion: $$ m \frac{d v}{d t}=-k v^{2}-m g, \quad v(0)=v_{0} $$

Step-by-Step Solution

Answer

The velocity as a function of time is derived considering quadratic air resistance and gravity.

Step 1: Understand the Differential Equation

The given differential equation is \(m \frac{d v}{d t} = -k v^2 - mg\). The initial condition provided is \(v(0) = v_0\). This equation describes the velocity \(v\) of an object considering the forces acting on it: gravitational force \(mg\) and air resistance \(R(v) = -k v^2\).

Step 2: Rearrange the Equation

To simplify, divide the entire differential equation by \(m\) to isolate \(\frac{d v}{d t}\) on one side. This yields:\[ \frac{d v}{d t} = -\frac{k}{m} v^2 - g \]

Step 3: Separate Variables

Separate the variables by moving all terms involving \(v\) to one side and \(t\) to the other side:\[ \frac{d v}{g + \frac{k}{m} v^2} = -dt \]

Step 4: Integrate Both Sides

Integrate both sides with respect to their variables:\[ \int \frac{d v}{g + \frac{k}{m} v^2} = -\int dt \]

Step 5: Integrate the Left Side

The left integral is a standard form. Using integration strategies (such as substitution), it evaluates to:\[ \frac{1}{\sqrt{g}} \tan^{-1}\left(\frac{\sqrt{k}}{\sqrt{mg}}v\right) \]

Step 6: Integrate the Right Side

The integral on the right side with respect to \(t\) is straightforward:\[ -t + C \] where \(C\) is the constant of integration.

Step 7: Solve for the Constant

Apply the initial condition \(v(0) = v_0\) to find \(C\). Substituting these into the integrated equation, we have:\[ \frac{1}{\sqrt{g}} \tan^{-1}\left(\frac{\sqrt{k}}{\sqrt{mg}}v_0\right) = C \]

Step 8: Solve for Velocity as a Function of Time

Now solve the resulting equation for \(v(t)\):\[ \frac{1}{\sqrt{g}} \tan^{-1}\left(\frac{\sqrt{k}}{\sqrt{mg}}v\right) = -t + \frac{1}{\sqrt{g}} \tan^{-1}\left(\frac{\sqrt{k}}{\sqrt{mg}}v_0\right) \]Isolate \(v\) on one side using inverse trigonometric functions.

Step 9: Interpret the Result

The expression for \(v(t)\) after isolating for \(v\) gives the velocity of the object as a function of time considering quadratic air resistance and gravity. This concludes the solution for the initial value problem.

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with a specific starting condition, allowing us to determine a unique solution. In this exercise, we are introduced to such a problem where we need to find the velocity of an object over time, given an initial velocity.

• The initial condition is denoted as \( v(0) = v_0 \), meaning at time \( t = 0 \), the velocity is \( v_0 \).
• It gives us the "starting point" for solving the differential equation, providing a unique trajectory for the solution based on the forces acting on the object.
• An initial value problem becomes particularly handy when dealing with real-world situations, like motion, where knowing the initial state is critical for predicting future states.

By incorporating the given initial velocity, we can solve the differential equation and find a precise expression for the velocity as time progresses.

Quadratic Drag Law

The quadratic drag law describes a scenario where the force of air resistance acting on an object is proportional to the square of its velocity. In this exercise:

• The air resistance force is represented as \( R(v) = -k \cdot v^2 \), where \( k \) is a constant that depends on factors like the shape and texture of the object and the properties of the air.
• This form of air resistance becomes significant at higher speeds, unlike linear drag, because the force increases rapidly with velocity.
• Quadratic drag is crucial for understanding objects moving at significant speeds, such as vehicles, aircraft, or even skydivers.

Incorporating this into our differential equation helps us to accurately model more realistic scenarios of motion through air.

Gravitational Force

Gravitational force is a constant force acting on an object due to gravity. It plays a vital role in this differential equation:

• The force is represented as \( mg \), where \( m \) is the mass of the object, and \( g \) is the gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \) on Earth).
• Gravity tries to pull the object down, counteracting the upward motion that the initial velocity might provide.
• In the context of the equation, gravitational force is a continual downward pull that must be considered alongside the air resistance.

The balance of gravitational force and air resistance helps in determining the behavior of the object's velocity through time.

Integration Techniques

Integration techniques are essential for solving the differential equation in this problem. Let's explore what this entails:

• To find the velocity function, we rearrange and separate variables in the differential equation, enabling us to integrate both sides independently.
• We utilize substitution and inverse trigonometric functions to solve the resulting integral, particularly using \( \int \frac{d v}{g + \frac{k}{m} v^2} = - \int dt \).
• The solution involves recognizing standard integral forms, enabling easier computation, and determining the integral of each side with respect to their respective variables.

By applying these techniques, we shape the expression that gives a velocity through time, demonstrating how methods of integration are critical tools in solving complex differential equations.