Question

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=m x ; x^{2}+y^{2}=a^{2},\) where \(m\) and \(a\) are constants

Short answer

Answer: Yes, the families of curves \(y=mx\) and \(x^2+y^2=a^2\) form orthogonal trajectories.

Step-by-step solution

Find the derivative of y=mx

We will find the derivative of the first equation with respect to x: \(\frac{dy}{dx} = m\)

Find the derivative of x^2 + y^2 = a^2

We will find the derivative of the second equation with respect to x using implicit differentiation: \(\frac{d}{dx}(x^2+y^2) = \frac{d}{dx}(a^2)\) \(2x + 2y\frac{dy}{dx} = 0\) Now, solving for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = -\frac{x}{y}\)

Check if the slopes are negative reciprocals

The product of the slopes is: \(m \times -\frac{x}{y} = -\frac{mx}{y}\) In order to have orthogonal trajectories, we need this product to be -1. However, we need more information to determine if this product is -1. We have to find the intersection points between the two curves.

Find the intersection points

Set the two curves equal to each other to find the intersection points: \(mx = x^2 + y^2 = a^2\) Substitute \(y = mx\) into the second equation: \(x^2 + (mx)^2 = a^2\) \(x^2(1 + m^2) = a^2\) Now, solving for x, we get: \(x = \pm\frac{a}{\sqrt{1+m^2}}\) Since \(y = mx\), we have: \(y = m\left(\pm\frac{a}{\sqrt{1+m^2}}\right)\) \(y = \pm\frac{ma}{\sqrt{1+m^2}}\) The intersection points are \(\left(\pm\frac{a}{\sqrt{1+m^2}}, \pm\frac{ma}{\sqrt{1+m^2}}\right)\).

Check if the product of slopes is -1 at the intersection points

Substitute the intersection points into the product of slopes: \(\left(-\frac{mx}{y}\right)\left(m\right) = -\frac{mx^2}{(mx)^2 + y^2}\) Substitute \(x^2 + (mx)^2 = a^2\) into this product: \(-\frac{mx^2}{a^2} = -\frac{mx^2}{x^2 + (mx)^2}\) Now plug in the intersection points \(\left(\pm\frac{a}{\sqrt{1+m^2}}, \pm\frac{ma}{\sqrt{1+m^2}}\right)\) into the above equation: \(-\frac{m\left(\frac{a}{\sqrt{1+m^2}}\right)^2}{a^2} = -\frac{m\left(\frac{a^2}{1+m^2}\right)}{a^2}= -1\) Thus, the product of slopes at the intersection points is -1, which confirms that the given families of curves form orthogonal trajectories.

Key concepts

Implicit Differentiation

Implicit differentiation is a handy technique to find derivatives of equations that may not be easily solved for one variable in terms of another. In cases where you have both variables, like in the equation \(x^2 + y^2 = a^2\), solving directly for \(y\) isn't straightforward. Here's where implicit differentiation shines, allowing us to differentiate both sides of an equation with respect to \(x\), even if we are not isolating \(y\) first.

When applying implicit differentiation, we treat \(y\) as a function of \(x\), using the chain rule when differentiating terms involving \(y\). So, when we found the derivative of \(x^2 + y^2 = a^2\),

• We derivative \(x^2\) as usual with respect to \(x\), getting \(2x\).
• The term \(y^2\) involves \(y\), so using the chain rule gives \(2y \cdot \frac{dy}{dx}\).

Setting the derivative of the constant \(a^2\) to zero, we obtained the equation: \(2x + 2y\frac{dy}{dx} = 0\).
Solving for \(\frac{dy}{dx}\) provides us with the slope of the tangent to the curve, in terms of both \(x\) and \(y\) variables.

Perpendicular Slopes

Perpendicular slopes are at the core of understanding orthogonal trajectories in geometry. Simply put, two lines are perpendicular when the product of their slopes is -1. This is due to the definition of perpendicular lines, where the angle between them is 90 degrees, reflected mathematically by this specific slope relationship.

• If one line has a slope of \(m\), the line perpendicular to it must have a slope of \(-\frac{1}{m}\).
• This concept extends to curves, where the tangent lines at their intersection points need to show this perpendicularity.

In the original problem, the task was to determine the slopes of the tangent lines for each curve, \(y = mx\) and \(x^2 + y^2 = a^2\).
Using implicit differentiation, we found the derivative, \(\frac{dy}{dx} = -\frac{x}{y}\) for the circle. These derivatives represent the slopes of tangents at any point. Their relationship ensures that when calculated at intersections, they should yield a product of -1 to show perpendicularity. This confirms the curves are orthogonal trajectories.

Derivative Calculation

Derivative calculation is crucial in the process of finding tangent slopes, which is significant when analyzing curve behavior, especially for orthogonal sets. Calculating derivatives requires applying various differentiation rules, such as power rule and chain rule while keeping in mind the function type being differentiated.

For straightforward functions like \(y = mx\), the derivative calculation is uncomplicated, as applying the power rule directly gives \(\frac{dy}{dx} = m\), representing a constant slope. For more complex functions like \(x^2 + y^2 = a^2\), implicit differentiation comes into play, involving a sequence of steps to unravel the relationships between differentials.

• The direct derivative \(2x\) is computed for \(x^2\).
• For \(y^2\), applying the chain rule yields \(2y \cdot \frac{dy}{dx}\).

Solving these gives the derivative \(\frac{dy}{dx} = -\frac{x}{y}\), which aligns well with understanding the slope relationship of intersecting curves.
In summarizing, derivative calculations determine slopes and allow further geometric interpretations pertinent to orthogonal trajectories, where constant or implicit differentiation methods are applied as necessary to describe various curves accurately.