Problem 41 - 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 \(r\) in the plane. Differential equation: $$\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}-k \mathbf{v}=-g \mathbf{j}-k \frac{d \mathbf{r}}{d t}$$ 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}$$ 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 are obtained by integrating the velocity equation under the influence of gravity and drag, following the initial conditions.

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{d^{2} \mathbf{r}}{dt^{2}} = -g \mathbf{j} - k \frac{d \mathbf{r}}{dt} \). This equation describes the motion of a projectile subject to gravity and linear drag. The term \(-g \mathbf{j}\) represents the gravitational force, and \(-k \frac{d \mathbf{r}}{dt}\) represents the drag, which is proportional to the velocity.

Step 2: Convert the Second Order ODE to First Order ODEs

To solve the equation, reduce it to two first order equations. Let \( \mathbf{v} = \frac{d \mathbf{r}}{dt} \), then the second derivative becomes \( \frac{d \mathbf{v}}{dt} = -g \mathbf{j} - k \mathbf{v} \). Thus, we have two equations: \( \frac{d \mathbf{r}}{dt} = \mathbf{v} \) and \( \frac{d \mathbf{v}}{dt} = -g \mathbf{j} - k \mathbf{v} \).

Step 3: Solve the Velocity Equation

The velocity equation \( \frac{d \mathbf{v}}{dt} = -g \mathbf{j} - k \mathbf{v} \) can be solved separately for \( \mathbf{i} \) and \( \mathbf{j} \) components. For the \( j \)-component, solve \( \frac{d v_{y}}{dt} = -g - k v_{y} \) by integrating to find \( v_{y}(t) = \frac{g}{k} + (v_{0} \sin \alpha - \frac{g}{k})e^{-kt} \). For \( i \)-component, solve \( \frac{d v_{x}}{dt} = -k v_{x} \) and integrate to find \( v_{x}(t) = v_{0} \cos \alpha \ e^{-kt} \).

Step 4: Integrate Velocity to Get Position

Integrate the expressions for \( v_{x}(t) \) and \( v_{y}(t) \) with respect to time to find \( x(t) \) and \( y(t) \). For \( x(t) \), the integral of \( v_{x}(t) \) is \( x(t) = \frac{v_{0}}{k}(1 - e^{-kt}) \cos \alpha \). For \( y(t) \), integrating \( v_{y}(t) \) gives \( y(t) = \frac{v_{0} \sin \alpha}{k}(1 - e^{-kt}) + \frac{g}{k^2}(1 - kt - e^{-kt}) \).

Step 5: Verify Initial Conditions

Ensure that the solutions satisfy the initial conditions \( \mathbf{r}(0) = \mathbf{0} \) and \( \frac{d \mathbf{r}}{dt} \big|_{t=0} = \mathbf{v}_{0} \). At \( t=0 \), both results \( x(0) \) and \( y(0) \) are zero, matching \( \mathbf{r}(0) = \mathbf{0} \). The expressions for velocity components also revert to the initial velocity \( \mathbf{v}_{0} \) at \( t=0 \), confirming the conditions are met.

Key Concepts

Differential Equations

Understanding differential equations is key to solving many physical problems, like analyzing the motion of a projectile under linear drag. A differential equation expresses a relationship between a function and its derivatives. In this exercise, the differential equation \( \frac{d^{2} \mathbf{r}}{dt^{2}} = -g \mathbf{j} - k \frac{d \mathbf{r}}{dt} \) describes how both gravity and drag affect a projectile. Gravity is represented by the term \(-g \mathbf{j}\), acting downward, while the drag force, \(-k \frac{d \mathbf{r}}{dt}\), acts opposite to the velocity. Differentiating a vector or scalar function like this helps predict how its speed and direction change over time. In practice, converting such equations into solvable forms usually involves addressing the components separately, often transforming higher order differential equations into sets of first-order equations for simpler handling.

• The equation considers both drag and gravitational forces.
• Converts higher order equations into two first-order equations.

Through solving, one can find the projectile's path over time, factoring in the resistance it encounters as a result of air drag.

Projectile Motion

Projectile motion describes the curved trajectory that an object follows when launched into the air. This exercise involves analyzing projectile motion with added linear drag, making calculations a bit more complex than assuming motion in a vacuum. Linear drag represents the air resistance encountered by the moving object, which is proportional to its velocity.
The trajectory itself is influenced by three major factors:

• The initial speed \( v_0 \), which dictates initial kinetic energy.
• The launch angle \( \alpha \), determining the initial directional components.
• Gravitational pull \(g\), which constantly acts downwards, pulling the projectile towards the earth.

These elements work together to define how the projectile will rise and fall. Solving for these conditions with drag resistance helps visualize actual real-world motions which are inherently more complex and less predictable than idealized no-drag scenarios.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation paired with conditions specified at the start, or initially, to find a unique solution. In this exercise, the IVP is presented through the equation and the following initial conditions:

• \(\mathbf{r}(0)=\mathbf{0}\).
• \(\frac{d \mathbf{r}}{d t}\big|_{t=0}=\mathbf{v}_{0}\).

These conditions state that at time \(t=0\), the position is zero and the velocity equals the initial velocity \(\mathbf{v}_{0}\) given as: \(\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}\).
Initial conditions are crucial as they determine the specific path of the solution among many possibilities. Without them, the results would be vague, leading to several theoretical projections rather than a definitive motion track. The goal of initial value problems is to make the scenario tangible for predictive analytics, such as determining where a projectile will land.

Gravity

Gravity plays a significant role in projectile motion as a constant force acting downward. It's typically represented by \(g\), the acceleration due to gravity, which is approximately \(9.8 \, m/s^2\) on the surface of the Earth. In this problem, gravity influences the downward path of the projectile, shown by \(-g \mathbf{j}\).
This constant pull by gravity combines with the initial velocity and angle to decide the projectile's vertically-aligned (\(\mathbf{j}\)-component) movement. The initial upward thrust imparted by the velocity separates slowly because of gravity, which reduces the projectile's height incrementally until it descends.

• Gravity acts constantly and uniformly on the projectile.
• The force ensures that the projectile eventually returns to the ground.

By ingraining gravity into the motion equations, we ensure accurate predictions that match observed physical behavior.

Velocity Integration

Velocity integration is the technique used to find the trajectory path of the projectile motion described earlier. Once we have expressions for the velocity components, \(v_x(t)\) and \(v_y(t)\), integrating those gives us the position functions \(x(t)\) and \(y(t)\). This is a crucial step as it transitions from knowing speed to precisely predicting location over time.
Specifically, integration involves:

• Converting velocity-time functions to position-time equations.
• Acknowledging the exponential nature due to the drag component \(e^{-kt}\).

In the problem, solving the two components separately is necessary:

• \(x(t) = \frac{v_{0}}{k}(1 - e^{-kt}) \cos \alpha\)
• \(y(t) = \frac{v_{0} \sin \alpha}{k}(1 - e^{-kt}) + \frac{g}{k^2}(1 - kt - e^{-kt})\)

These functions merge both the linear and exponential changes over time, allowing observation of how drag slows the component speeds gradually, affecting the overall arcs and landing points of the projectile's path.