Question

Find the orthogonal trajectories of the given family of curves. $$ x^{2}+2 y^{2}=c^{2} $$

Short answer

Answer: The equation of the orthogonal trajectories for the given family of curves is \(Y = \pm g(x^2 + 2y^2)^{\frac{1}{4}}\), where \(g = e^C\).

Step-by-step solution

Find the Differential Equation of the Given Family of Curves

Differentiate the given equation implicitly with respect to x. $$ \frac{d}{dx}(x^2 + 2y^2) = \frac{d}{dx}(c^2) $$ Differentiating both sides, we get: $$ 2x + 4y\frac{dy}{dx} = 0 $$

Find the Condition for Orthogonality

Since the tangent lines of the orthogonal trajectories should be perpendicular, we have the following condition: $$ \frac{dy}{dx}\frac{dY}{dX} = -1 $$ Where \(\frac{dy}{dx}\) represents the slope of the tangent line of the given family of curves and \(\frac{dY}{dX}\) represents the slope of the orthogonal trajectories.

Replace the Slopes of the Curves with the Condition for Orthogonality

From Step 1, we find the slope of the given family of curves: $$ \frac{dy}{dx} = -\frac{x}{2y} $$ Replace the slope in the orthogonality condition to find the slope of the orthogonal trajectories: $$ -\frac{x}{2y}\frac{dY}{dX} = -1 $$ Solve for \(\frac{dY}{dX}\): $$ \frac{dY}{dX} = \frac{x}{2y} $$

Integrate the Differential Equation of Orthogonal Trajectories

Now, we have the differential equation that represents the orthogonal trajectories: $$ \frac{dY}{dX} - \frac{x}{2y} = 0 $$ Rearrange and integrate with respect to x: $$ \frac{dY}{dX} = \frac{x}{2y} $$ $$ \frac{dY}{Y} = \frac{1}{2y} dx $$ Integrate both sides: $$ \int\frac{dY}{Y} = \int\frac{1}{2y} dx $$ $$ \ln|Y| = \frac{1}{4}\ln|x^2 + 2y^2| + C $$

Solve for the Equation of the Orthogonal Trajectories

Take the exponential of both sides to obtain the equation: $$ |Y| = e^{\frac{1}{4}\ln|x^2 + 2y^2| + C} $$ $$ |Y| = g(x^2 + 2y^2)^{\frac{1}{4}} $$ Where \(g = e^C\). We have derived the orthogonal trajectories for the given family of curves: $$ Y = \pm g(x^2 + 2y^2)^{\frac{1}{4}} $$

Key concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In this context, we're dealing with the family of curves defined by the equation \(x^{2} + 2y^{2} = c^{2}\).
The goal is to find another family of curves that intersect the given ones at right angles, known as orthogonal trajectories. To achieve this, we must first express the relationship between the variables and their rates of change (derivatives).
This is done by differentiating the given equation with respect to \(x\), resulting in a relation between \(x\), \(y\), and \(\frac{dy}{dx}\). The expression \(2x + 4y\frac{dy}{dx} = 0\) is formed, representing the rate equation for the given family of curves.

Implicit Differentiation

Implicit differentiation is a technique used when you need to find the derivative of a function, in scenarios where the function is not explicitly solved for one variable in terms of another.
In this problem, the equation \(x^{2} + 2y^{2} = c^{2}\) is not solved explicitly for \(y\). To differentiate it, you treat \(y\) as a function of \(x\), even though it's not isolated.
By applying implicit differentiation, we calculate the derivative with respect to \(x\), giving us \(2x + 4y\frac{dy}{dx} = 0\). This allows us to see how changes in \(x\) and \(y\) relate through their derivatives.

Perpendicular Curves

Perpendicular curves intersect at right angles, signifying a particular symmetry and mathematical relationship between them.
To find curves that are orthogonal to a given family, we need to make use of the condition of perpendicularity. For two curves represented by slopes \(\frac{dy}{dx}\) and \(\frac{dY}{dX}\), the product of these slopes should equal \(-1\).
In our scenario, after calculating the slope for the given curves as \(\frac{dy}{dx} = -\frac{x}{2y}\), we substitute this into the perpendicular condition. Keeping the condition \(\frac{dy}{dx} \times \frac{dY}{dX} = -1\) ensures the resulting curve is orthogonal, deriving \(\frac{dY}{dX} = \frac{x}{2y}\). This new slope defines the orthogonal trajectories.

Integration

Integration is a fundamental calculus operation used to find the function that corresponds to a given derivative, effectively reversing differentiation.
Once the differential equation for the orthogonal trajectories is established, \(\frac{dY}{dX} = \frac{x}{2y}\), integration allows us to "undo" the differentiation, retrieving the family of curves as a result of its changes along \(x\).
In this case, we separate variables, \(\frac{dY}{Y} = \frac{1}{2y} dx\), and integrate both sides. This leads to \(\ln|Y| = \frac{1}{4}\ln|x^2 + 2y^2| + C\). By solving these integrals, the equation \(Y = \pm g(x^2 + 2y^2)^{\frac{1}{4}}\) is obtained, revealing the orthogonal trajectories in their explicit form.