Problem 57 - Calculus Early Transcendentals

Question

A projectile with mass \(m\) is launched into the air on a parabolic trajectory. For \(t \geq 0,\) its horizontal and vertical coordinates are \(x(t)=u_{0} t\) and \(y(t)=-\frac{1}{2} g t^{2}+v_{0} t\), respectively, where \(u_{0}\) is the initial horizontal velocity, \(v_{0}\) is the initial vertical velocity, and \(g\) is the acceleration due to gravity. Recalling that \(u(t)=x^{\prime}(t)\) and \(v(t)=y^{\prime}(t)\) are the components of the velocity, the energy of the projectile (kinetic plus potential) is \(E(t)=\frac{1}{2} m\left(u^{2}+v^{2}\right)+m g y\). Use the Chain Rule to compute \(E^{\prime}(t)\) and show that \(E^{\prime}(t)=0\), for all \(t \geq 0 .\) Interpret the result.

Step-by-Step Solution

Answer

#Answer#: The total mechanical energy of the projectile remains constant throughout its trajectory, as shown by the derivative \(E^{\prime}(t)=0\). This is a result of the conservation of energy principle for the projectile in motion, which is only affected by gravity as the external force.

Step 1: Find u(t) and v(t)

First, we need to find the horizontal and vertical velocity components, u(t) and v(t), which are the derivatives of x(t) and y(t) respectively. We have: \(x(t) = u_0 t\) \(y(t) = -\frac{1}{2}gt^2 + v_0 t\) Taking the derivatives: \(u(t) = \frac{dx(t)}{dt} = u_0\) \(v(t) = \frac{dy(t)}{dt} = -gt + v_0\)

Step 2: Differentiate the energy expression

Next, we will differentiate the expression for the energy with respect to time using the Chain Rule: \(E(t) = \frac{1}{2} m(u^2 + v^2) + mgy\) We have: \(\frac{dE(t)}{dt} = m\left(u\frac{du}{dt} + v\frac{dv}{dt}\right) + mgy'\) where y' denotes the derivative of y(t) with respect to t.

Step 3: Substitute the expressions for u(t), v(t), and y'(t)

Now, we will substitute the expressions for u(t), v(t), and y'(t) that we found in Step 1: \(\frac{dE(t)}{dt} = m\left(u_0\cdot 0 + (-gt+v_0)\cdot(-g)\right) + mg(-gt + v_0)\) Simplify the equation: \(\frac{dE(t)}{dt} = m(-gv_0 + g^2t) + mg(-gt + v_0)\)

Step 4: Show that E'(t) = 0

Now we will show that the equation is equal to 0: \(\frac{dE(t)}{dt} = m(-gv_0 + g^2t + (-gt + v_0)g) = m(-gv_0 -gtg +v_0 g) = m(-gv_0 -gtg + v_0 g)\) \(\frac{dE(t)}{dt} = m(g(-v_0 -t + v_0)) = mg(-t) + mgt = 0\) Thus, \(E^{\prime}(t)=0\) for all \(t \geq 0\).

Step 5: Interpret the result

Since the derivative of the energy \(E'(t)\) with respect to time is equal to 0, this means that the total mechanical energy of the projectile remains constant throughout its trajectory. This is a result of the conservation of energy principle for the projectile in motion, which is only affected by gravity as the external force.

Key Concepts

Kinematics in Projectile Motion

When we talk about kinematics, we are referring to the branch of mechanics that deals with the motion of objects, without considering the forces that cause this motion. In the case of projectile motion, kinematics helps us describe how a projectile travels through the air. We use different equations and concepts to determine parameters like velocity, acceleration, and displacement.

For projectile motion, we split the motion into horizontal and vertical components:

• The horizontal motion is given by the equation: \( x(t) = u_0t \), where \( u_0 \) is the initial horizontal velocity.
• The vertical motion is represented by: \( y(t) = -\frac{1}{2}gt^2 + v_0t \), where \( v_0 \) is the initial vertical velocity, and \( g \) is the acceleration due to gravity.

These equations allow us to compute the position of the projectile in both directions at any given time \( t \).

The derivatives of these equations provide the velocity components:

• Horizontal velocity: \( u(t) = u_0 \)
• Vertical velocity: \( v(t) = -gt + v_0 \)

Understanding these components is crucial for predicting the path and timing of a projectile's trajectory.

Energy Conservation in Projectile Motion

Energy conservation is a fundamental principle in physics that states that energy cannot be created or destroyed in an isolated system — it only changes forms. For projectile motion, this principle means that the sum of the kinetic and potential energy remains constant if we ignore air resistance.

The expression for the total energy \( E(t) \) is given by:
\[ E(t) = \frac{1}{2} m (u^2 + v^2) + mgy \]
Here, kinetic energy is expressed as \( \frac{1}{2} m(u^2 + v^2) \), derived from the horizontal and vertical components of velocity, while potential energy is \( mgy \), dependent entirely on the height of the projectile.

In the original problem, when calculating \( E'(t) \), i.e., the change in energy over time, and showing it equals zero, it demonstrates that energy is conserved throughout the projectile's flight. This result implies that while kinetic energy and potential energy might exchange forms (e.g., as the projectile rises, kinetic energy converts into potential energy), their total remains unchanged, allowing the projectile to maintain its motion consistently.

Understanding the Chain Rule in Calculus

The Chain Rule is a vital concept in calculus used to differentiate composite functions. In simple terms, if a function y is composed of another function u, the Chain Rule helps us find the derivative of y with respect to an independent variable, say t.

Consider a function \( E(t) \), which is constructed as a composition of other functions of \( t \), such as \( u(t) \), \( v(t) \), and \( y(t) \). The Chain Rule helps break down \( \frac{dE}{dt} \) into manageable parts by differentiating each internal function separately and then combining results:

• \( \frac{d}{dt}[u^2] = 2u \frac{du}{dt} \)
• \( \frac{d}{dt}[v^2] = 2v \frac{dv}{dt} \)
• \( \frac{d}{dt}[y] = y' \)

These parts are vital when solving derivative problems involving composite functions, like finding \( E'(t) = 0 \) in the exercise. By employing the Chain Rule, you can simplify complex expressions and arrive at insights such as confirming energy conservation in projectile motion.