Problem 37 - Thomas Calculus

Question

Linear drag Derive the equations $$\begin{aligned} x &=\frac{v_{0}}{k}\left(1-e^{-k l}\right) \cos \alpha \\ y &=\frac{v_{0}}{k}\left(1-e^{-k t}\right)(\sin \alpha)+\frac{g}{k^{2}}\left(1-k t-e^{-k t}\right) \end{aligned}$$ by solving the following initial value problem for a vector \(r\) in the plane. $$ \text{Differential equation:}\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}-k \mathbf{v}=-g \mathbf{j}-k \frac{d \mathbf{r}}{d t}$$ $$\text{Initial conditions:}\begin{aligned} \mathbf{r}(0) &=\mathbf{0} \\\\\left.\frac{d \mathbf{r}}{d t}\right|_{t=0} &=\mathbf{v}_{0}=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(\boldsymbol{v}_{0} \sin \alpha\right) \mathbf{j} \end{aligned}$$ 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

The derived equations for the motion are correct as they satisfy the given initial value problem.

Step 1: Set Up the First-Order System

We start by rewriting the second-order differential equation for \( \mathbf{r} \) as a system of first-order differential equations. Set \( \mathbf{v} = \frac{d\mathbf{r}}{dt} \). Then, we have two equations: \( \frac{d\mathbf{r}}{dt} = \mathbf{v} \) and \( \frac{d\mathbf{v}}{dt} = -g \mathbf{j} - k \mathbf{v} \).

Step 2: Solve for Velocity Function \( \mathbf{v}(t) \)

Introduce components \( v_x(t) \) and \( v_y(t) \) so that \( \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} \). The differential equation for velocity becomes \( \frac{dv_x}{dt} = -k v_x \) and \( \frac{dv_y}{dt} = -g - k v_y \). Solve these to obtain \( v_x(t) = v_0 \cos(\alpha) e^{-kt} \) and \( v_y(t) = \left(v_0 \sin(\alpha) + \frac{g}{k}\right)e^{-kt} - \frac{g}{k} \).

Step 3: Integrate to Find Position Function \( \mathbf{r}(t) \)

Now integrate \( v_x(t) \) and \( v_y(t) \) to find \( x(t) \) and \( y(t) \) respectively. For \( x(t) \), integrate \( v_x(t) = v_0 \cos(\alpha) e^{-kt} \) to get \( x(t) = \frac{v_0 \cos(\alpha)}{k}(1 - e^{-kt}) \).

Step 4: Solve for \( y(t) \) with Gravity Component

Integrate \( v_y(t) = \left(v_0 \sin(\alpha) + \frac{g}{k}\right)e^{-kt} - \frac{g}{k} \) for \( y(t) \). The integral leads to \( y(t) = \frac{v_0 \sin(\alpha)}{k} (1 - e^{-kt}) + \frac{g}{k^2}(1 - kt - e^{-kt}) \).

Step 5: Verify Initial Conditions and Conclusion

Check that \( \mathbf{r}(0) = \mathbf{0} \) and \( \frac{d\mathbf{r}}{dt} \bigg|_{t=0} = \mathbf{v}_0 \) hold true with the derived \( x(t) \) and \( y(t) \). The obtained solutions for \( x \) and \( y \) satisfy both the differential equations and initial conditions.

Key Concepts

Differential Equations

Differential equations play a crucial role in understanding the dynamics of projectile motion with linear drag. These are mathematical equations that involve derivatives, showing how one or more variables change with respect to others. In our exercise, we are dealing with a second-order differential equation. This particular equation describes the motion of a projectile under the influence of gravity and air resistance (linear drag).

For our problem, the differential equation is \[ \frac{d^{2} \mathbf{r}}{d t^{2}} = -g \mathbf{j} - k \mathbf{v} \] where \( \mathbf{r} \) is the position vector and \( \mathbf{v} \) is the velocity. The term \(-g \mathbf{j}\) represents the gravitational force acting downward, while \(-k \mathbf{v}\) accounts for the linear drag.

To solve this, we first break it down into a system of first-order equations. By setting \( \mathbf{v} = \frac{d\mathbf{r}}{dt} \), the original second-order equation can be split into two linked first-order differential equations:

• \( \frac{d\mathbf{r}}{dt} = \mathbf{v} \)
• \( \frac{d\mathbf{v}}{dt} = -g \mathbf{j} - k \mathbf{v} \)

This transformation makes it easier to analyze and solve for the motion of the projectile.

Linear Drag

Linear drag is a force that opposes the motion of an object through a fluid (like air). It's called 'linear' because it is directly proportional to the velocity of the object. In mathematical terms, the drag force is expressed as \[ F_{drag} = -k \mathbf{v} \] where \( k \) is a positive constant known as the drag coefficient, and \( \mathbf{v} \) is the velocity vector. Linear drag reduces the speed of the object, and its effect becomes more noticeable as velocity increases.

In the context of our exercise, linear drag makes the differential equation non-trivial compared to ideal projectile motion (which disregards air resistance). This force affects both horizontal and vertical components of velocity differently:

• The drag reduces horizontal velocity over time, necessitating the use of exponential terms when solving for position \( x(t) \).
• Similarly, vertical velocity is influenced by both gravity and drag, complicating the equations for \( y(t) \).

Precisely accounting for linear drag helps predict both the range and time of flight of the projectile with greater accuracy.

Initial Value Problem

An initial value problem (IVP) involves solving a differential equation with given initial conditions. These initial conditions specify the starting point of the system's motion—both position and velocity. For our projectile problem, the initial conditions ensure we solve the equations in a way that reflects the actual scenario:

\[ \mathbf{r}(0) = \mathbf{0} \] and \[ \left.\frac{d \mathbf{r}}{d t}\right|_{t=0} = \mathbf{v}_{0} = (v_{0} \cos \alpha) \mathbf{i} + (v_{0} \sin \alpha) \mathbf{j} \]

• \( \mathbf{r}(0) = \mathbf{0} \) states that the projectile starts from the origin.
• \( \mathbf{v}_{0} \) provides the launch velocity, showing both speed and direction.

Solving this IVP means finding a function that represents the position of the object at any time \( t \) while satisfying these initial conditions. Verifying the final solutions against these conditions helps confirm the accuracy of derived functions for \( x(t) \) and \( y(t) \).

Integration Techniques

Integration is a fundamental technique used in solving differential equations, especially when converting velocity functions to position functions. In our context, once the velocities, \( v_x(t) = v_0 \cos(\alpha) e^{-kt} \) and \( v_y(t) = \left(v_0 \sin(\alpha) + \frac{g}{k}\right)e^{-kt} - \frac{g}{k} \), are obtained, integration helps us find the corresponding positions \( x(t) \) and \( y(t) \).

The integration proceeds by treating each velocity component separately:

• For horizontal motion, integrate \( v_x(t) \) to derive: \[ x(t) = \frac{v_0 \cos(\alpha)}{k}(1 - e^{-kt}) \]


• For vertical motion, integrate \( v_y(t) \) to obtain: \[ y(t) = \frac{v_0 \sin(\alpha)}{k} (1 - e^{-kt}) + \frac{g}{k^2}(1 - kt - e^{-kt}) \]

These integration results show how the projectile's position evolves over time while considering both gravity and air resistance. Each integration is carefully performed, respecting the exponential nature of linear drag and the constant gravity acting on the projectile.