Problem 29 - Thomas Calculus

Question

Linear drag Derive the equations $$ \begin{aligned} x &=\frac{v_{0}}{k}\left(1-e^{-k t}\right) \cos \alpha \\ y &=\frac{v_{0}}{k}\left(1-e^{-k t}\right)(\sin \alpha)+\frac{g}{k^{2}}\left(1-k t-e^{-k t}\right) \end{aligned} $$ by solving the following initial value problem for a vector \(\mathbf{r}\) in the plane. $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}-k \mathbf{v}=-g \mathbf{j}-k \frac{d \mathbf{r}}{d t}} \\ {\text { Initial conditions: }} & {\mathbf{r}(0)=\mathbf{0}} \\ {} & {\left.\frac{d \mathbf{r}}{d t}\right|_{t=0}=\mathbf{v}_{0}=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}}\end{array} $$ The drag coefficient \(k\) is a positive constant representing resistance due to air density, \(v_{0}\) and \(\alpha\) are the projectile's initial speed and launch angle, and \(g\) is the acceleration of gravity.

Step-by-Step Solution

Answer

Integrate velocity components with respect to time to find the position equations for \( x(t) \) and \( y(t) \).

Step 1: Formulate the Differential Equation

Start with the given differential equation for the vector \( \mathbf{r} \): \[ \frac{d^{2} \mathbf{r}}{dt^{2}} = -g \mathbf{j} - k \frac{d \mathbf{r}}{dt} \] where \( \mathbf{j} \) represents the vertical direction. This equation considers gravity and linear drag.

Step 2: Solve for Velocity Components

Let \( \frac{d \mathbf{r}}{dt} = \mathbf{v} \). Then use the equation \( \frac{d^{2} \mathbf{r}}{dt^{2}} = \frac{d \mathbf{v}}{dt} \). Therefore, \( \frac{d \mathbf{v}}{dt} = -g \mathbf{j} - k \mathbf{v} \). This separates into two components: horizontal and vertical, yielding \( \frac{dv_x}{dt} = -k v_x \) and \( \frac{dv_y}{dt} = -g - k v_y \).

Step 3: Solve for Horizontal Velocity

For the horizontal direction, solve \( \frac{dv_x}{dt} = -k v_x \). The solution is obtained by separating variables \[ \int \frac{1}{v_x} dv_x = -k \int dt \] resulting in \( v_x(t) = v_{0} \cos \alpha \cdot e^{-kt} \).

Step 4: Solve for Vertical Velocity

For the vertical component, solve \( \frac{dv_y}{dt} = -g - kv_y \). Using an integrating factor or similar approach, solve to find \( v_y(t) = (v_{0} \sin \alpha + \frac{g}{k}) e^{-kt} - \frac{g}{k} \).

Step 5: Integrate Velocity to Find Position

Integrating the velocity components to get position. For horizontal position \[ x(t) = \int v_x(t) dt = \int v_{0} \cos \alpha \cdot e^{-kt} dt \] gives \( x(t) = \frac{v_{0}}{k}(1 - e^{-kt}) \cos \alpha \).

Step 6: Integrate Vertical Velocity for Position

Integrate \( v_y(t) \) to find \( y(t) \). Integrations involve \[ y(t) = \int \left((v_{0} \sin \alpha + \frac{g}{k}) e^{-kt} - \frac{g}{k}\right) dt \] resulting in \( y(t) = \frac{v_{0}}{k} (1 - e^{-kt}) \sin \alpha + \frac{g}{k^2}(1 - kt - e^{-kt}) \).

Step 7: Apply Initial Conditions

Verify initial conditions \( \mathbf{r}(0) = \mathbf{0} \) and \( \mathbf{v}(0) = \mathbf{v}_0 \). When substituted, the equations hold: \( x(0) = 0 \) and \( y(0) = 0 \), as well as initial velocities matching \( v_{0} \cos \alpha \) and \( v_{0} \sin \alpha \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They play a crucial role in describing various physical phenomena, including projectile motion with air resistance, as discussed here. Essentially, these equations allow us to depict how a function changes over time or space. In this problem, the differential equation given is \[\frac{d^{2} \mathbf{r}}{dt^{2}} = -g \mathbf{j} - k \frac{d \mathbf{r}}{dt}\]This represents the motion of a projectile, accounting for both gravity and drag due to air resistance. The term \(-g \mathbf{j}\)depicts the gravitational force acting downward, while\(-k \frac{d \mathbf{r}}{dt}\)accounts for the linear drag force opposing the motion.

Projectile Motion

Projectile motion refers to the motion of an object thrown into the air, subject to gravitational acceleration and other forces. In this scenario, the projectile's path is influenced by the initial velocity, launch angle, gravity, and air resistance. Generally, without air resistance, projectile motion follows a parabolic trajectory. However, linear drag complicates the trajectory, causing deviations from the ideal path. A projectile is propelled with an initial velocity\(v_{0}\)at an angle\(\alpha\)from the horizontal. The horizontal and vertical components of the initial velocity are

• Horizontal: \( v_{0} \cos \alpha \)
• Vertical: \( v_{0} \sin \alpha \)

These components help in deducing the motion equations that predict the projectile's path over time.

Air Resistance

Air resistance, or drag, is a force that opposes the motion of objects through the air. In this exercise, linear drag, represented by the coefficient\(k\), is considered. Linear drag is proportional to the velocity of the object and acts in the opposite direction. It can be expressed as\(-k \mathbf{v}\), where \(\mathbf{v}\) is the velocity vector of the projectile.The presence of air resistance modifies the trajectory of the projectile by reducing its speed continuously. This happens because part of the kinetic energy is lost in overcoming the air resistance. As a result, the equations governing the projectile's motion include terms that account for this energy loss, providing a more realistic description of its path compared to idealized models with no air resistance.

Initial Value Problem

An initial value problem comprises a differential equation coupled with specified values at the start of the problem. These specified values are initial conditions. In this problem, the initial conditions are given as

• \(\mathbf{r}(0)=\mathbf{0}\), which states that the starting position of the projectile is at the origin (0,0).
• \(\left.\frac{d \mathbf{r}}{d t}\right|_{t=0}=\mathbf{v}_{0}\), ensuring that the initial velocity of the projectile,\( \mathbf{v}_0\), matches the given values of speed and angle.

These initial conditions are crucial in finding unique solutions to the differential equations, ensuring that the calculated path and velocity of the projectile precisely match real-world scenarios.