Problem 47 - Calculus Single Variable

Question

The drag on an airplane at a given altitude is given by \(\left(a+b / v^{4}\right) v^{2}\) where \(a\) and \(b\) are positive constants, and \(v\) is the velocity of the plane. At what speed is drag minimized?

Step-by-Step Solution

Answer

The drag is minimized at the velocity \( v = \sqrt[4]{\frac{b}{a}} \).

Step 1: Identify the Function

The function representing the drag on the airplane is given by \( D(v) = \left( a + \frac{b}{v^{4}} \right) v^{2} \). Our goal is to find the value of \( v \) that minimizes this function.

Step 2: Differentiate the Function

To find the critical points of the function, we need to differentiate it with respect to \( v \). The function can be expanded to \( D(v) = a v^{2} + \frac{b}{v^{2}} \). The derivative \( D'(v) \) is then calculated as follows: \( D'(v) = \frac{d}{dv} (a v^{2} + \frac{b}{v^{2}}) = 2a v - \frac{2b}{v^{3}} \).

Step 3: Find Critical Points

To find where the drag reaches a minimum, set the derivative equal to zero: \( 2a v - \frac{2b}{v^{3}} = 0 \). Simplifying, \( 2a v = \frac{2b}{v^{3}} \). Solving for \( v \), multiply both sides by \( v^3 \) to get \( 2a v^4 = 2b \), leading to \( v^4 = \frac{b}{a} \).

Step 4: Solve for Velocity

Take the fourth root of both sides to find \( v \). Thus, \( v = \sqrt[4]{\frac{b}{a}} \). This is the velocity at which the drag is minimized.

Step 5: Confirm Minimum with Second Derivative

To ensure this value of \( v \) is indeed a minimum, we check the second derivative, \( D''(v) = \frac{d}{dv} (2a v - \frac{2b}{v^{3}}) = 2a + \frac{6b}{v^{4}} \). Since both terms are positive for positive \( v \), \( D''(v) > 0 \). Therefore, \( v = \sqrt[4]{\frac{b}{a}} \) is a minimum.

Key Concepts

Critical Points

In calculus, determining critical points is a crucial part of finding where functions reach their extremes, like minimum or maximum values. Critical points occur where the derivative of a function is zero or undefined. For the drag function in our problem, expressed as \( D(v) = \left( a + \frac{b}{v^{4}} \right) v^{2} \), we need to find the critical points to identify where drag is minimized. By differentiating the function and setting the derivative \( D'(v) \) to zero, we pinpoint the critical points. In this case, setting \( 2a v - \frac{2b}{v^{3}} = 0 \) helps us solve for \( v \), and leads us to the formula \( v^4 = \frac{b}{a} \). By finding \( v \) from this critical point, we can solve the problem effectively.

Second Derivative Test

The second derivative test is a handy tool to confirm whether a critical point is a minimum or maximum. Once we find a critical point from the first derivative, we should check the second derivative.In our exercise, the second derivative \( D''(v) = 2a + \frac{6b}{v^{4}} \) is calculated.

• If \( D''(v) > 0 \) at the critical point, like it is here, it's a minimum.
• If \( D''(v) < 0 \), it's a maximum point.

Since both terms in the second derivative are positive for the calculated \( v \), we have confirmed that \( v = \sqrt[4]{\frac{b}{a}} \) brings the minimal drag. This check reassures us that our solution indeed pinpoints the minimum drag condition.

Differentiation

At the heart of calculus, differentiation involves finding the derivative of a function to determine rates of change. This process lets us explore how a function behaves, including identifying slopes and points of inflection.In our specific problem, we differentiated the drag function \( D(v) = a v^{2} + \frac{b}{v^{2}} \) to find the first derivative \( D'(v) = 2a v - \frac{2b}{v^{3}} \).

• This first derivative represents the rate at which drag changes with velocity.
• Setting the derivative to zero allows us to determine critical points.

Mastering differentiation enables us to analyze curves effectively and, as seen here, solve optimization problems by determining such key points.

Velocity and Drag

Understanding the relationship between velocity and drag is critical in aerodynamics. Drag force, a type of resistance experienced by an object moving through a fluid (like air), is influenced by the velocity.In this exercise, drag is given by the equation \( D(v) = \left( a + \frac{b}{v^{4}} \right) v^{2} \). This signifies how drag depends on velocity, \( v \).

• The term \( a v^{2} \) accounts for drag increasing with the square of velocity, while
• \( \frac{b}{v^{2}} \) adds complexity as it causes decreases with higher velocities.

By optimizing, or finding the velocity at which drag is minimized, we balance these effects. Recognizing this minimizes resistance and enhances efficiency, a fundamental concept in improving performance of airplanes and other vehicles moving through air.