Problem 56 - Single Variable Calculus

Question

Range of a Projectile The range of an artillery shell fired at an angle of \(\theta^{\circ}\) with the horizontal is $$ R=\frac{v_{0}^{2}}{g} \sin 2 \theta $$ feet, where \(v_{0}\) is the muzzle velocity of the shell in feet per second, and \(g\) is the constant of acceleration due to gravity \(\left(32 \mathrm{ft} / \mathrm{sec}^{2}\right) .\) Find the angle of elevation of the gun that will give it a maximum range.

Step-by-Step Solution

Answer

The angle of elevation \(\theta\) that will give the maximum range of the projectile is \(45^{\circ}\).

Step 1: Understand the variables and the formula

The given formula for the range of a projectile is $$ R = \frac{v_{0}^2}{g} \sin{2 \theta} $$ where \(R\) is the range, \(v_0\) is the initial velocity, \(g\) is the gravitational constant, and \(\theta\) is the angle of elevation. Step 2: Take the derivative

Step 2: Compute the derivative with respect to \(\theta\)

To find when the range is maximized, we need to find the derivative of the range function with respect to the angle \(\theta\). Using the chain rule, we have $$ \frac{dR}{d\theta} = \frac{v_{0}^2}{g} \cos{2 \theta} \cdot \frac{d(2 \theta)}{d\theta} $$ Simplifying further, we get $$ \frac{dR}{d\theta} = \frac{2v_{0}^2 \cos{2 \theta}}{g} $$ Step 3: Find critical points

Step 3: Set the derivative equal to zero and solve

To find the maximum range, we must determine when the derivative of the range function is zero or undefined. The derivative is defined everywhere, so we only need to find when the derivative is zero. $$ \frac{2v_{0}^2 \cos{2 \theta}}{g} = 0 $$ This will only happen when \(\cos{2 \theta} = 0\). This occurs when \(2 \theta = 90^\circ\), which corresponds to $$ \theta = 45^\circ $$ Step 4: State the final answer

Step 4: Angle of elevation that gives the maximum range

The angle of elevation \(\theta\) that will give the maximum range of the projectile is \(45^{\circ}\).

Key Concepts

Derivative Optimization

When working with functions in calculus, a common problem is to find the maximum or minimum value that the function can take. This is known as optimization. A powerful tool we use to solve such problems is the derivative, which gives us the rate at which the function is changing at each point.

In the context of projectile motion, we can optimize the range by finding the angle that provides the maximal distance. We do this by first deriving the equation that describes the range with respect to the angle, much like in the steps provided above.

Once we have the derivative, we look for critical points—places where the derivative is either zero or undefined. These points are the candidates for local maxima or minima. For a clear-cut maximum or minimum, we need to further analyze these points, sometimes by using the second derivative test or by comparing the values of the function at these points.

In the case of projectile range, we found that the derivative equals zero when the angle is 45 degrees. This angle often turns out to be the answer in ideal conditions because it nicely balances the vertical and horizontal components of the projectile's initial velocity.

Angle of Elevation

The angle of elevation is a geometrical concept used to describe the angle above the horizontal at which an observer or a projectile is aimed. It is critically important in the study of projectile motion, astronomy, surveying, and many other fields.

In the exercise, the angle of elevation directly affects the distance the projectile travels, for a given initial velocity and under constant gravity. A higher angle of elevation increases the height the projectile reaches but reduces its horizontal range since more of the initial kinetic energy is funneled into fighting gravity rather than moving horizontally. Conversely, a lower angle increases the horizontal distance but gives the projectile less time in the air.

The exercise aims to strike the perfect balance between height and distance to achieve the maximum range. The optimal angle, unfettered by air resistance or any other external factors, happens to be 45 degrees, where the sin function reaches its peak for the double angle identity used in the range formula.

Projectile Motion Calculus

Applying calculus to projectile motion allows us to extract precise information about the behavior of a projectile over time, such as its position, velocity, and acceleration at any given instant. The motion of a projectile is affected by gravity, which results in a parabolic trajectory under ideal conditions.

The key to understanding projectile motion is breaking it down into two components: horizontal and vertical. The horizontal motion does not change over time if we ignore air resistance, as there is no horizontal acceleration. The vertical motion, however, is directly affected by gravity.

By using calculus, particularly differentiation and integration, we can derive formulas to predict the range, maximum height, and duration of flight for the projectile. Taking derivatives helps us understand how the projectile's velocity and position change over time, while integrals can tell us the total distance traveled. These calculus tools are fundamental in optimizing outcomes for projectile motion, as seen in the exercise. Here, we used the derivative to maximize the projectile's range, finding the best angle of elevation to achieve this.