Problem 62 - Calculus: Graphical, Numerical, Algebraic

Question

Multiple Choice For which of the following differential equa- tions will a slope field show nothing but negative slopes in the fourth quadrant? (A) \(\frac{d y}{d x}=-\frac{x}{y} \quad\) (B) \(\frac{d y}{d x}-x y+5 \quad\) (C) \(\frac{d y}{d x}=x y^{2}-2\) (D) \(\frac{d y}{d x}=\frac{x^{3}}{y^{2}} \quad\) (E) \(\frac{d y}{d x}=\frac{y}{x^{2}}-3\)

Step-by-Step Solution

Answer

Options (A), (B), (C) and (E) will show nothing but negative slopes in the fourth quadrant.

Step 1: Analyzing option (A) \(\frac{dy}{dx} = - \frac{x}{y}\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = - \frac{-a}{-b} = -\frac{a}{b}< 0\). So yes, the slopes in the 4th quadrant are negative for option A.

Step 2: Analyzing option (B) \(\frac{dy}{dx} = xy - 5\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = -ab - 5 < 0\). So yes, the slopes in the 4th quadrant are negative for option B.

Step 3: Analyzing option (C) \(\frac{dy}{dx} = xy^2 - 2\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = -ab^2 - 2 < 0\). So yes, the slopes in the 4th quadrant are negative for option C.

Step 4: Analyzing option (D) \(\frac{dy}{dx} = \frac{x^3}{y^2}\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = \frac{(-a)^3}{(-b)^2} = -a^3/b^2 > 0\). So no, the slopes in the 4th quadrant are not negative for option D.

Step 5: Analyzing option (E) \(\frac{dy}{dx} = \frac{y}{x^2} - 3\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = \frac{-b}{(-a)^2} - 3 = -\frac{b}{a^2} - 3 < 0\). So yes, the slopes in the 4th quadrant are negative for option E.

Key Concepts

Slope Fields

Slope fields are an important visual tool in understanding differential equations. They provide a method to visualize the slope of solutions of a differential equation across various points in a plane. Each small line segment in a slope field represents the slope of the solution curve passing through that point. These line segments help us understand how the solution to a differential equation evolves.

Visualizing slope fields:

• Interpretation: Each point \( (x, y) \) in the slope field has a line segment whose slope is determined by the differential equation.
• Understanding behavior: This visual representation allows us to anticipate the general behavior or the shape of the solution without solving the equation analytically.
• Application to quadrants: By inspecting these fields over different quadrants of the Cartesian plane, one can infer where slopes are particularly positive or negative. This helps in understanding the qualitative behavior of solutions.

Remember, a slope field doesn't give you the exact solutions, but rather offers a map of directional "clues" on how solutions may behave geometrically over the plane.

Quadrants

The Cartesian coordinate system is divided into four quadrants which help in analyzing the behavior of functions, especially in slope fields. In the context of our problem, we're focused on the fourth quadrant, where both x and y values are negative:

Details of the fourth quadrant:

• Coordinates: For any point in this quadrant, \( x < 0 \) and \( y < 0 \).
• Analyzing slopes: To focus specifically on the behavior in this quadrant, you will substitute \( x = -a\) and \( y = -b \) where \( a, b > 0 \) in the differential equation.
• Significance: By doing this substitution, we can directly analyze whether the resulting slopes calculated from these equations are negative, which was the requirement in our original exercise.

Not only do these quadrants provide a structured way to interpret the differential equation's behavior, but they also simplify the process of predicting how solutions evolve as we span across different sections of the coordinate system.

Negative Slopes

Negative slopes in the context of slope fields indicate that the solution curves tend to decrease as they move horizontally across the plane from left to right. Understanding the conditions for when slopes become negative is crucial to interpreting these fields.

Understanding when slopes are negative:

• Equation Inspection: Each differential equation must be inspected with values specific to the fourth quadrant. Optimal inspection involves inserting \( x = -a \) and \( y = -b \) (where \( a, b > 0 \)).
• Behavior and Trend: If the modified equation results in a negative value, it signifies that wherever this condition holds true, the slope is negative.
• Practical Implication: In our original exercise, identifying when the slopes are negative helps in answering the posed problem and determining the behavior in that particular quadrant.
• Avoiding Misinterpretation: Not all equations are straightforward for determining negative slopes, thus careful substitution and evaluation are key to accurate interpretation.

Negative slopes provide insightful details about the behavior of the differential equation solutions, thus shaping our understanding of the function's trajectory and ensuring correct analysis.