Question

In Exercises \(1-9\) find the orthogonal trajectories of each given family of curves. In each case sketch several members of the family and several of the orthogonal trajectories on the same set of axes. $$ y=e^{(x} . $$

Short answer

The orthogonal trajectories for the given family of curves \(y = e^x\) can be found by solving the differential equation \(\frac{dy}{dx} = -\frac{1}{e^x}\). After solving, we obtain the equation of the orthogonal trajectories as \(y = C (e^{-x})\), where C is an arbitrary constant.

Step-by-step solution

Find the derivative

Differentiate the given equation with respect to x: $$y = e^x$$ $$\frac{dy}{dx} = e^x$$

Replace the derivative with its negative reciprocal

Now take the negative reciprocal of the derivative to form the new differential equation for the orthogonal trajectories: $$\frac{dy}{dx} = -\frac{1}{e^x}$$

Solve the differential equation

Now we will solve the differential equation to find the equation for the orthogonal trajectories. Rewrite the differential equation as: $$dy = -\frac{1}{e^x} dx$$ Integrate both sides with respect to x: $$\int dy = -\int \frac{1}{e^x} dx$$ $$y = \int - e^{-x} dx +C_1$$ In order to find the integral of \(- e^{-x}\), we use substitution method where \(u = -x\), so \(du = - dx\). Now rewrite the integral as: $$y = \int e^u du +C_1$$ Integrate and resubstitute the value of u (-x): $$y = e^{-x}+C_1$$

Rearrange to get the final equation

Multiply both sides by the constant \(C_1\), and rewrite the equation to get the final equation of orthogonal trajectories for the family of curves: $$y = C (e^{-x})$$ where C is an arbitrary constant. Now, several members of the given family of curves can be plotted along with its orthogonal trajectories.

Key concepts

Differential Equations

Differential equations are essential mathematical tools for describing relationships between changing quantities. In basic terms, a differential equation involves an unknown function and its derivatives. They form the foundation for understanding aspects of calculus where change is involved.
When we explore orthogonal trajectories, we are often dealing with systems where two different families of curves are perpendicular to each other at points of intersection. In such cases, differential equations help us determine these trajectory paths by establishing a relationship between the curves through their derivatives.
For example, in the exercise, we're given a curve described by the function \( y = e^x \). The differential equation representing this is \( \frac{dy}{dx} = e^x \). This equation reflects the rate of change of \( y \) with \( x \), which is essential in finding orthogonal trajectories.

Family of Curves

The concept of a family of curves refers to a set of curves described by an equation containing one or more arbitrary constants. Each specific value of these constants produces a distinct curve within the family. Understanding how these curves interact and intersect is crucial, especially when finding trajectories that are orthogonal to each curve in the set.
In the initial problem, the family of curves is defined by the equation \( y = e^x \). Each curve in this family increases exponentially with \( x \), which reflects a consistent pattern or trend. To find orthogonal trajectories, we determine curves that intersect this family at right angles. This leads us to use various calculus techniques, like differentiation and finding negative reciprocals, to establish the needed orthogonal relationships.

Derivative

The derivative is a concept in calculus that measures how a function changes as its input changes. In simple terms, it gives us the slope of the tangent line to the curve of a function at any given point. Calculating the derivative is a vital step in exploring problems involving rates of change or movement, like the problem of orthogonal trajectories.
In our exercise, the derivative of the curve \( y = e^x \) is determined as \( \frac{dy}{dx} = e^x \). This represents the slope of the curve at any point \( x \). By understanding the slope, we can then find the curve orthogonal to it, requiring a slope that is the negative reciprocal of the original curve’s slope.

Integration

Integration is a process in calculus that allows us to find a function given its derivative, often used to calculate areas under curves, volumes, and other accumulative quantities. In solving differential equations for orthogonal trajectories, integration helps us find the trajectory equation from the slope information.
In the step-by-step solution, after identifying the differential equation \( dy = -\frac{1}{e^x} dx \), integration takes the form of \( \int dy = -\int \frac{1}{e^x} dx \). Solving these integrals gives us the relationship of the orthogonal trajectory compared to the original family of curves. Specifically, solving \( y = e^{-x} + C_1 \) involves the integration of known functions by manipulating variables to simplify the process.

Negative Reciprocal

The concept of a negative reciprocal is crucial when dealing with orthogonal lines or curves. In a Cartesian plane, if two lines are perpendicular, the product of their slopes equals \(-1\). Therefore, to find the slope of a line that is orthogonal to another, you take the negative reciprocal of the original slope.
In solving the differential equation for the orthogonal trajectory, the original slope of the curve \( y = e^x \) is \( \frac{dy}{dx} = e^x \). The negative reciprocal of this, which gives us the slope of the orthogonal trajectory, is \( \frac{dy}{dx} = -\frac{1}{e^x} \).
Understanding negative reciprocals is pivotal as it guides us through changing the perspective from one family of curves to another that intersect perpendicularly, ultimately leading to the solution of orthogonal trajectories.