To find the orthogonal trajectories of the family of curves $xy = k$, follow these steps:

Step 1: Rewrite the equation in explicit form

Given that $xy = k$, we can rewrite it as: $y = \frac{k}{x}$ This represents the family of curves for different values of $k$.

Step 2: Differentiate implicitly with respect to $x$

Now, differentiate both sides of $xy = k$ with respect to $x$ implicitly: $\frac{d}{dx}(xy) = \frac{d}{dx}(k)$ Using the product rule on the left-hand side: $x \frac{dy}{dx} + y = 0$ Thus, we obtain the differential equation: $x \frac{dy}{dx} = -y$ or $\frac{dy}{dx} = -\frac{y}{x}$ This is the slope of the tangent lines to the curves $xy = k$.

Step 3: Find the orthogonal trajectories

The slope of the orthogonal trajectories will be the negative reciprocal of the slope of the original family of curves. Thus, if the slope of the curves is $-\frac{y}{x}$, the slope of the orthogonal trajectories is: $\frac{dx}{dy} = \frac{x}{y}$ This is a separable differential equation. Rearrange it as: $\frac{dx}{x} = \frac{dy}{y}$

Step 4: Integrate both sides

Integrate both sides to find the orthogonal trajectories: $\int \frac{dx}{x} = \int \frac{dy}{y}$ This gives: $\ln |x| = \ln |y| + C$ Exponentiating both sides: $|x| = e^C |y|$ Let $e^C = C'$ (since $C'$ is just another constant), we get: $x = C' y$ or equivalently: $x = Cy$ where $C$ is a new constant.

Conclusion:

The family of orthogonal trajectories to the curves $xy = k$ is given by the family of straight lines: $x = Cy$

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Related questions:

1. What is the significance of orthogonal trajectories in physics?
2. How can we use orthogonal trajectories in fluid dynamics?
3. Can we apply this technique to polar coordinates?
4. What would the solution look like if the family of curves was a different form, such as $x^2 + y^2 = r^2$?
5. How do we find the orthogonal trajectories for a more complex implicit equation?

Tip: Orthogonal trajectories can be used in fields like electromagnetism where electric and magnetic field lines are perpendicular to each other.