Question

Find the orthogonal trajectories of the family of curves : (i) \(\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{c}^{2}\) (ii) \(y^{2}=4 c x\) (iii) \(\mathrm{y}=\frac{\mathrm{C}}{\mathrm{x}^{2}}\) (iv) \(y=C \sqrt{x}\)

Short answer

In summary, 1. The family of orthogonal trajectories for the curve family \(x^2 - y^2 = c^2\) is: $$\bar{y} = -y\frac{\ln x}{x} + kx$$ 2. For the curve family \(y^2 = 4cx\), the family of orthogonal trajectories is: $$\bar{y}^2 = -4kx$$ 3. For the curve family \(y = \frac{C}{x^2}\), the family of orthogonal trajectories is: $$\bar{y} = k\bar{y}^3$$ 4. For the curve family \(y = C\sqrt{x}\), the family of orthogonal trajectories is: $$\bar{y} = k\bar{y}^{-1/2}$$

Step-by-step solution

Differentiate the given equation with respect to x

Differentiate \(\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{c}^{2}\) with respect to x using implicit differentiation to get the slope of the tangent line for a given curve in the family: $$2x - 2y\frac{dy}{dx} = 0$$

Solve for dy/dx

From the equation in step 1, we solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{x}{y}$$

Find the orthogonal trajectory equation

Since we need the slope of the orthogonal trajectory to be the negative reciprocal of the given curve, we'll have: $$\frac{d\bar{y}}{dx} = -\frac{y}{x}$$ Now we need to solve this differential equation for \(\bar{y}\). Integrate both sides with respect to x: $$\int\frac{d\bar{y}}{dx}dx = -\int \frac{y}{x}dx$$ We notice that the left side contains only terms in \(\bar{y}\) and the right side contains terms in x and y. So, let \(v = \frac{\bar{y}}{x}\), then that means: $$x\frac{dv}{dx} + v = -\frac{y}{x}$$ Now integrate both sides with respect to x: $$\int(x\frac{dv}{dx} + v)dx = -\int \frac{y}{x}dx$$ Integrate: $$xv = -y\ln x + k$$ Rewriting in terms of \(\bar{y}\): $$\bar{y} = -y\frac{\ln x}{x} + kx$$ This is the family of orthogonal trajectories for curve family \(\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{c}^{2}\). For the other curve families, follow the same steps as 1, 2 and 3 to find their orthogonal trajectories. The final results are: (ii) For the curve \(y^{2}=4 c x\): $$\bar{y}^2 = -4kx$$ (iii) For the curve \(\mathrm{y}=\frac{\mathrm{C}}{\mathrm{x}^{2}}\): $$\bar{y} = k\bar{y}^3$$ (iv) For the curve \(y=C \sqrt{x}\): $$\bar{y} = k\bar{y}^{-1/2}$$

Key concepts

Implicit Differentiation

Implicit differentiation is an essential technique used when dealing with equations of curves that define y implicitly as a function of x, rather than explicitly. Unlike explicit functions where y is given as a direct function of x (like y = x^2), implicit functions involve y in a more complex relationship with x (such as in x^2 + y^2 = 1).

When finding the derivative of an implicit function, we differentiate both sides with respect to x, treating y as a function of x. In doing so, we apply the chain rule whenever y is differentiated, introducing the term \(dy/dx\) into our equation. For example, if we differentiate x^2 - y^2 with respect to x, we get 2x - 2y * (dy/dx).

Here's why this is crucial: we're trying to find the slope of the curve at any given point without explicitly solving for y. Knowing the slope of the tangent line \(dy/dx\) is a stepping stone towards understanding the behavior of the curve, and more specifically, in finding orthogonal trajectories, which are curves that intersect the original curves at right angles.

Differential Equations

Differential equations form the backbone of many phenomena in the physical sciences. They are equations that relate functions of one or more variables to their derivatives. In our context, differential equations come into play when we seek to find the orthogonal trajectories of a family of curves.

For instance, when dealing with the curve given by x^2 - y^2 = c^2, we first find the derivative using implicit differentiation, which produces a first-order differential equation. To discover the orthogonal trajectories, we set up another differential equation where the slope is the negative reciprocal of the original slope. This change reflects the right-angle intersection characteristic of orthogonal trajectories.

In solving these new differential equations, we often use separation of variables or integration, which allows us to find a new function that describes the orthogonal trajectories. The resulting equations, after integration and simplification, give us the family of curves that are orthogonal to the original family.

Integration

Integration is the process of finding the antiderivative or integral of a function, essentially the reverse process of differentiation. It is a fundamental tool in finding solutions to differential equations and is pivotal in our quest for the orthogonal trajectories.

The act of integrating both sides of the differential equation is akin to 'summing up' the slope information (given by \(dy/dx\)) to get the equation for the curve itself. In the case of our example, integrating \(-y/x\) with respect to x gives us one part of the equation for the orthogonal trajectories. We often encounter two types of integration: indefinite and definite. Indefinite integration, where we find a general form of the antiderivative, is what we typically use in the given context.

In solving for orthogonal trajectories, integration provides the link between the slope of a curve and the curve itself. By integrating, we accumulate the small slope elements \(dy/dx\) over x to construct the overall shape of the family of orthogonal curves. It's a crucial step in moving from the behavior of the curve at infinitesimal points (through differentiation) to understanding the whole picture — the geometric path that the curve traces.