Question

Finding Orthogonal Trajectories In Exercises \(41-46,\) find the orthogonal trajectories of the family. Use a graphing utility to graph several members of each family. $$ y=C e^{x} $$

Short answer

The orthogonal trajectories of the given family of curves \(y=C e^{x}\) is \(2x = -y^2 + 2A\).

Step-by-step solution

Finding the Derivative

Differentiate the given equation \(y=C e^{x}\) with respect to \(x\). To do this, use the basic rule of differentiation for \(e^{x}\), which is the derivative of \(e^{x}\) is \(e^{x}\). So, the derivative \(dy/dx\) will be \(C e^{x}\).

Eliminating the Constant

In order to eliminate the arbitrary constant, substitute \(y = C e^{x}\) into the first step to get \(dy/dx = y\).

Inverse of Derivative

To find the orthogonal trajectories, replace \(dy/dx\) with negative inverse of derivative \(-1/(dy/dx)\) or \(-1/y\).

Solve the Differential Equation

Now, we have a new differential equation \(dx/dy = -1/y\). We can rewrite it as \(y dx = -dy\). By integrating both sides, we get \(y^2/2 = -x + A\), where \(A\) is the constant of integration.

The Final Equation

Rearrange the equation from step 4 to isolate \(x\). Thus, the equation is \(x = -y^2/2 + A\). Multiply by -2 and rearrange to get \(2x = -y^2 + 2A\).

Key concepts

Differentiation

Differentiation is a key calculus concept that deals with finding the rate at which a function is changing at any given point. It involves calculating the derivative of a function, which represents the slope of the tangent to a curve at a specific point.
For the function given in the exercise, \(y = Ce^{x}\), where \(C\) is a constant, we need to differentiate with respect to \(x\). The derivative of an exponential function like \(e^{x}\) is the function itself, \(e^{x}\). Thus, the differentiation process yields \(\frac{dy}{dx} = Ce^{x}\).

In plain terms, this result tells us that the rate of change of the function \(y\) with respect to \(x\) is proportional to its value at any point \(x\). This intrinsic relationship between exponential functions and their derivatives becomes a crucial foundation when solving problems in calculus.

Differential Equations

Differential equations involve equations that include functions and their derivatives. Solving them typically involves finding a function that satisfies the given relationship between variables and their derivatives.
In this exercise, our task involves finding orthogonal trajectories for the given family \(y = Ce^{x}\). An orthogonal trajectory of a family of curves is a path that intersects every member of that family at right angles.

By differentiating, we determine that \(\frac{dy}{dx} = Ce^{x}\). We then eliminate the constant \(C\) by expressing the derivative as \(\frac{dy}{dx} = y\). The orthogonal trajectory is then found by swapping \(\frac{dy}{dx}\) with its negative reciprocal, giving rise to the differential equation \(\frac{dx}{dy} = -\frac{1}{y}\).
Rewriting and integrating this yields \(ydx = -dy\), which on integration transforms into \(y^2/2 = -x + A\), with \(A\) as an integration constant. This new equation describes the orthogonal trajectories to the original curve family. Solving such differential equations gives us new sets of functions that describe possible solutions.

Exponential Function

The exponential function, denoted as \(e^{x}\), holds a special place in mathematics due to its unique properties, notably its constant rate of growth. The base \(e\) is approximately 2.718, and it's an irrational number.
Exponential functions grow rapidly as their variable increases. In this exercise, \(y = Ce^{x}\) is an exponential function where \(C\) represents a constant that adjusts the function's initial value but does not affect its rate of growth.

A remarkable property of exponential functions is that their derivative is proportional to the function itself. This means the slope of \(e^{x}\) at any point is the same as the value of \(e^{x}\) at that point, making these functions central to calculus and many real-world applications, such as compound interest, population growth, and radioactive decay.
Thus, exponential functions are not just theoretical constructs—they provide a vital toolset for modeling and understanding complex systems that involve growth and change.