Problem 33 - Thomas Calculus

Question

Firing from \(\left(x_{0}, y_{0}\right) \quad\) Derive the equations $$\begin{aligned} x &=x_{0}+\left(v_{0} \cos \alpha\right) t \\ y &=y_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2} \end{aligned}$$ (see Equation \((7)\) in the text) by solving the following initial value problem for a vector \(r\) in the plane. Differential equation: $$\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}$$ Initial conditions:$$\mathbf{r}(0)=x_{0} \mathbf{i}+y_{0} \mathbf{j}\( \)\frac{d \mathbf{r}}{d t}(0)=\left(\boldsymbol{v}_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}$$

Step-by-Step Solution

Answer

The projectile's position is given by: \[x = x_0 + (v_0 \cos \alpha)t \] \[y = y_0 + (v_0 \sin \alpha)t - \frac{1}{2}gt^2\].

Step 1: Understand the Problem Statement

The problem requires us to derive the equations of motion for a projectile starting at \((x_0, y_0)\) using given initial conditions and a differential equation for the vector motion. The vector \(\mathbf{r}\) represents the position, and its second derivative is given in the plane.

Step 2: Analyze the Given Differential Equation

The differential equation is \(\frac{d^2 \mathbf{r}}{d t^2} = -g \mathbf{j}\). This implies the acceleration is constant and points in the negative y-direction due to gravity. \(\mathbf{j}\) indicates the y-axis unit vector, while the x-axis has no acceleration component.

Step 3: Integrate to Find Velocity

Integrate the acceleration equation to find velocity. Since acceleration \(\frac{d^2 \mathbf{r}}{d t^2} = -g \mathbf{j}\), integrating with respect to time gives: \[\frac{d \mathbf{r}}{d t} = -gt \mathbf{j} + \mathbf{C}\], where \(\mathbf{C}\) is a vector constant derived from initial conditions.

Step 4: Apply Initial Velocity Conditions

Use initial velocity conditions to solve for \(\mathbf{C}\). At \(t = 0\), the velocity is \(\left(v_{0} \cos \alpha\right) \mathbf{i} + \left(v_{0} \sin \alpha\right) \mathbf{j}\). So, \(\mathbf{C} = \left(v_{0} \cos \alpha\right) \mathbf{i} + \left(v_{0} \sin \alpha\right) \mathbf{j}\). Thus, velocity \(\frac{d \mathbf{r}}{d t} = \left(v_{0} \cos \alpha\right) \mathbf{i} + \left(v_{0} \sin \alpha - gt\right) \mathbf{j}\).

Step 5: Integrate to Find Position

Integrate the velocity equation to obtain the position \(\mathbf{r}(t)\). Integrating \(\frac{d \mathbf{r}}{d t} = \left(v_{0} \cos \alpha\right) \mathbf{i} + \left(v_{0} \sin \alpha - gt\right) \mathbf{j}\) yields: \[\mathbf{r}(t) = \left(v_0 \cos \alpha\right)t \mathbf{i} + \left((v_0 \sin \alpha)t - \frac{1}{2}gt^2\right) \mathbf{j} + \mathbf{C'}\], where \(\mathbf{C'}\) is another integration constant.

Step 6: Apply Initial Position Conditions

Apply the initial condition for position \(\mathbf{r}(0) = x_0 \mathbf{i} + y_0 \mathbf{j}\) to solve for \(\mathbf{C'}\). Substituting, we find \(\mathbf{C'} = x_0 \mathbf{i} + y_0 \mathbf{j}\). Therefore, the position function becomes \(\mathbf{r}(t) = \left(x_0 + (v_0 \cos \alpha)t\right) \mathbf{i} + \left(y_0 + (v_0 \sin \alpha)t - \frac{1}{2}gt^2\right) \mathbf{j}\).

Step 7: Summary of Resulting Equations

Summarize the results by converting vector notation into familiar component form. We derive: \[x(t) = x_0 + (v_0 \cos \alpha) t\] and \[y(t) = y_0 + (v_0 \sin \alpha) t - \frac{1}{2} gt^2\]. These equations describe the projectile's motion given initial conditions and constant acceleration due to gravity.

Key Concepts

Differential Equations

Differential equations are mathematical relations that express how a function's value changes in relation to another variable. In the context of projectile motion, these equations describe how the position and velocity of an object change over time. When dealing with projectile motion, a differential equation can precisely define how an object's position changes due to external forces, like gravity.

In this exercise, we start with a second-order differential equation: \(\frac{d^2 \mathbf{r}}{dt^2} = -g \mathbf{j}\). This equation tells us that the rate of change of the velocity (which is acceleration) is constant and directed downward (in the -y direction) because of gravity. "\(\mathbf{j}\)" represents the unit vector in the vertical direction, indicating no horizontal forces are acting.

To solve this problem, we perform integration, which is the reverse of differentiation. The solution to a differential equation often involves finding the original function by integrating the derivative. Here, integrating acceleration gives us the velocity equation, which we integrate again to determine the position of the projectile over time.

Initial Value Problems

An initial value problem (IVP) involves solving a differential equation banded with a set of initial conditions. These initial conditions specify the values of the function and its derivatives at a starting point, which simplifies finding the particular solution to the differential equation.

In the case of projectile motion, our initial conditions specify the starting position and velocity of the projectile. For example, the initial condition \(\mathbf{r}(0) = x_0 \mathbf{i} + y_0 \mathbf{j}\) gives the initial position, meaning the projectile starts at coordinates \((x_0, y_0)\) at time \(t = 0\).

Similarly, \(\frac{d \mathbf{r}}{dt}(0) = \left( v_0 \cos \alpha \right) \mathbf{i} + \left( v_0 \sin \alpha \right) \mathbf{j}\) provides the initial velocity, specifying both the speed and direction of travel. By combining these conditions with the differential equations, we derive expressions that allow us to track the projectile's future position and path.

Vector Calculus

Vector calculus is a field of mathematics that deals with vector fields and operations on vectors. It allows us to mathematically describe physical quantities that have both magnitude and direction, such as velocity and acceleration in projectile motion.

In this exercise, the vector \(\mathbf{r}(t)\) represents the position of the projectile as a vector, comprising \(x(t)\) and \(y(t)\) components. By working in vector notation, we can simplify complex motion equations and directly apply operations like differentiation and integration simultaneously to both components.
For projectile motion, breaking the motion into components along \(\mathbf{i}\) (horizontal) and \(\mathbf{j}\) (vertical) vectors, we solve for each separately using vector calculus techniques. This approach simplifies handling complex motions, such as those involving gravity, by accommodating them within a structured framework of vectors that clearly define directionality.
The vector approach not only aids in simplifying the integration process but also provides clarity in visualizing how different forces affect movement, which is crucial when deriving equations for motion in projectile problems.