Problem 10 - Thomas Calculus

Question

In Exercises \(5 - 10 ,\) find the orthogonal trajectories of the family of curves. Sketch several members of each family. $$ y = e ^ { k x } $$

Step-by-Step Solution

Answer

The orthogonal trajectories are given by \( \frac{y^2}{2} \ln(y) - \frac{y^2}{4} = -\frac{x^2}{2} + C \).

Step 1: Identify the Family of Curves

The given family of curves is represented by the equation \( y = e^{kx} \), where \( k \) is a parameter. These are exponential growth (or decay) curves depending on the value of \( k \).

Step 2: Derive the Differential Equation

To find the differential equation representing this family, we differentiate the family equation with respect to \( x \). This gives us \[ \frac{dy}{dx} = k e^{kx} \]By expressing \( k \) from the original equation \( y = e^{kx} \), we get \( k = \frac{ ext{ln}(y)}{x} \). Substitute \( k \) in the differential equation:\[ \frac{dy}{dx} = \frac{y \cdot \text{ln}(y)}{x} \] which simplifies by using the expression of \( y \).

Step 3: Setup Orthogonality Condition

Orthogonal trajectories occur when their tangent lines are perpendicular at the points of intersection. Therefore, the product of slopes of the curves and their orthogonal trajectories is \( -1 \). Setting up, we have: \[ \frac{dy}{dx}_{trajectory} = -\frac{1}{\frac{dy}{dx}_{original}} = -\frac{x}{y \cdot \ln(y)} \]

Step 4: Solve the Differential Equation for Orthogonal Trajectories

The equation for orthogonal trajectories is \[ \frac{dy}{dx} = -\frac{x}{y \cdot \ln(y)} \]First, simplify by separating variables: \[ y \cdot \ln(y) \, dy = -x \, dx \]Integrate both sides:- Integrate \( y \cdot \ln(y) \, dy \) using substitution;- Integrate \(-x \, dx\).We get:\[ \frac{y^2}{2} \ln(y) - \frac{y^2}{4} = -\frac{x^2}{2} + C \].

Step 5: Sketch the Curves

Sketch several curves of the original family \( y = e^{kx} \) which are exponential curves for different values of \( k \). Then, overlay curves calculated for the orthogonal trajectories, accountable for symmetry and perpendicular intersection at points.

Key Concepts

Family of Curves

When we discuss a "family of curves," we're referring to a set of curves that share a common feature or equation but differ by one or more parameters. In this exercise, the family of curves is defined by the equation \( y = e^{kx} \). Here, \( k \) is the parameter that changes, and it determines specific members of this family. These curves can depict exponential growth or decay depending on the value assigned to \( k \).

• If \( k > 0 \), the curves demonstrate exponential growth.
• If \( k < 0 \), the curves exhibit exponential decay.
• If \( k = 0 \), the curve is horizontal, and \( y = 1 \).

Understanding a family of curves allows us to visualize how changing a parameter affects the shape and behavior of these curves. This concept provides the groundwork for exploring more complex interactions, like orthogonal trajectories.

Differential Equations

Differential equations play a crucial role in understanding relationships within a family of curves. They represent the connection between the variables and their rates of change. For the family \( y = e^{kx} \), differentiating with respect to \( x \) gives the differential equation \( \frac{dy}{dx} = k e^{kx} \).

However, to find orthogonal trajectories, we need a differential equation without parameters. To do this, we express \( k \) in terms of \( y \) and \( x \) from the original equation, leading to \( k = \frac{\ln(y)}{x} \). By substituting this back into the differentiated form, we obtain \( \frac{dy}{dx} = \frac{y \cdot \ln(y)}{x} \).

This equation captures the relationship between \( y \) and \( x \) fully, representing the curves' behavior without any unspecified parameters. Differential equations are the backbone of analyzing how curves evolve and intersect.

Exponential Growth

Exponential growth occurs when quantities increase at rates proportional to their current value, often visualized through curves defined by equations like \( y = e^{kx} \). In such scenarios:

• With a positive \( k \), the graph rises steeply as \( x \) increases.
• These curves illustrate rapid escalation, mirroring many real-world phenomena like population growth.
• An understanding of these curves aids in predicting how rapidly systems can change under certain conditions.

Exponential decay can also be observed if \( k \) is negative, where curves instead taper off as \( x \) increases, representing phenomena such as radioactive decay. Recognizing the structure of exponential curves helps grasp the contrasting dynamics of growth and decay.

Perpendicular Slopes

The concept of perpendicular slopes is essential for finding orthogonal trajectories. If two curves intersect at a point and their tangent lines are perpendicular, their slopes multiply to \(-1\). This is the fundamental condition for orthogonality.

In the exercise, the original slope from the curves is given by \( \frac{dy}{dx} \). Hence, the slope of the orthogonal trajectory at intersection points is \( -\frac{1}{\frac{dy}{dx}} \). For the family \( y = e^{kx} \), this translates to the perpendicular slope \( \frac{dy}{dx}_{trajectory} = -\frac{x}{y \cdot \ln(y)} \).

This requirement ensures that at every intersection, the trajectories intersect at right angles. The orthogonal trajectory and original curve's meeting point, and their respective angles, highlight essential geometric relationships that serve many applications in physics and engineering.