Problem 39 - Calculus - AP Edition

Question

Two curves are said to be orthogonal if their tangent lines are perpendicular at each point of intersection, and two families of curves are said to be orthogonal trajectories of one another if each member of one family is orthogonal to each member of the other family. This terminology is used in these exercises. The accompanying figure on the next page shows some typical members of the families of circles \(x^{2}+(y-c)^{2}=c^{2}\) (black curves) and \((x-k)^{2}+y^{2}=k^{2}\) (gray curves). Show that these families are orthogonal trajectories of one another. [Hint: For the tangent lines to be perpendicular at a point of intersection, the slopes of those tangent lines must be negative reciprocals of one another.]

Step-by-Step Solution

Answer

The given families of curves are orthogonal trajectories because their tangent slopes at intersections are negative reciprocals.

Step 1: Differentiate the First Family of Curves

Consider the family of curves given by the equation \(x^2 + (y-c)^2 = c^2\). To find the slope of the tangent line at a point, we first differentiate this equation implicitly with respect to \(x\).Differentiate the equation: \[ \frac{d}{dx}(x^2 + (y-c)^2 = c^2) \]This becomes: \[ 2x + 2(y-c)\frac{dy}{dx} = 0 \]Solving for \( \frac{dy}{dx} \), we get:\[ \frac{dy}{dx} = -\frac{x}{y-c} \]

Step 2: Differentiate the Second Family of Curves

Now consider the family of curves given by the equation \((x-k)^2 + y^2 = k^2\). Again, we find the slope of the tangent line by differentiating implicitly with respect to \(x\).Differentiate the equation: \[ \frac{d}{dx}((x-k)^2 + y^2 = k^2) \]This becomes: \[ 2(x-k) + 2y\frac{dy}{dx} = 0 \]Solving for \( \frac{dy}{dx} \), we get:\[ \frac{dy}{dx} = -\frac{x-k}{y} \]

Step 3: Check Slopes for Orthogonality

For the two families of curves to be orthogonal trajectories, the product of their slopes must be \(-1\) at points of intersection.Multiply the slopes obtained:\[ \left(-\frac{x}{y-c}\right) \times \left(-\frac{x-k}{y}\right) \]This becomes:\[ \frac{x(x-k)}{(y-c)y} \]Now identify any point of intersection. For simplicity, consider \(x = 0\), then the coordinates at intersection must satisfy both original equations. Substituting, we confirm:\[ x^{2}+(y-c)^{2}=c^{2} \rightarrow (y-c) = \pm c \]\[ (x-k)^{2}+y^{2}=k^{2} \rightarrow y = \pm k \]At these simpler points, \(y-c\) and \(y\) are negatives of each other, hence:\[ \frac{x(x-k)}{(y-c)y} = -1 \]Therefore, the families of curves are orthogonal trajectories.

Key Concepts

Perpendicular Tangent Lines

Understanding perpendicular tangent lines is key when it comes to orthogonal trajectories. At each point where two curves intersect, their tangent lines need to be perpendicular. This means that their orientations differ by 90 degrees.
To find out if two tangent lines are perpendicular, we look at their slopes. In simpler terms, the slope of a line indicates how steep it is. For two lines to be perpendicular, the slopes must multiply together to give -1.

• Think of a pair of scissors: as you open them fully, the blades are perpendicular to each other. Similarly, two lines intersecting at a right angle are perpendicular.
• When two tangent lines at the intersection point of two curves are perpendicular, it implies that the tangent at that point behaves just like intersecting blades of scissors.

In this exercise, consider two curves described by specific equations. For the tangent lines to be perpendicular at their intersections, the slopes calculated by differentiating these equations should be negative reciprocals.

Implicit Differentiation

Differentiation is a technique used to find the slope of a tangent line to a curve at any point. Implicit differentiation is a specialized method used when dealing with equations that are not easily solved for one variable in terms of another.
In our exercise, implicit differentiation helps us find the slopes of the tangent lines for both families of curves. Here’s how it works:

• Instead of solving the equation for one variable, you differentiate the whole equation. This treats one variable as an implicit function of the other.
• When differentiating, each term involving "y" requires the chain rule because "y" is considered a function of "x". Thus, \( \frac{dy}{dx} \) represents the slope.

For example, when differentiating the curve \( x^2 + (y-c)^2 = c^2 \), we find that the slope of the tangent line is \( -\frac{x}{y-c} \). The same process applies to the curve \((x-k)^2 + y^2 = k^2\), leading to a different slope. Using implicit differentiation allows us to find these slopes without needing to explicitly solve for \(y\).

Slopes as Negative Reciprocals

Why do the slopes of orthogonal trajectories appear as negative reciprocals? Understanding this concept is crucial for determining orthogonality between curves.
A slope is a ratio that describes how a line rises or falls. Two lines are orthogonal when the product of their slopes is -1. This relationship comes from the fact that for any slope \( m \), its negative reciprocal \( -\frac{1}{m} \) will result in their product being -1.

• If one line’s slope is fairly steep (like going uphill), the perpendicular line (orthogonal) must be oriented such that it cuts through the base of the incline (like going downhill).
• In our context, after calculating the slopes of tangent lines to each curve, you determine whether they are negative reciprocals to confirm orthogonality.

For the given curves, when solving for the product of \( -\frac{x}{y-c} \) and \( -\frac{x-k}{y} \), the resulting expression equals -1 at intersections, validating the orthogonality of these trajectory families. This approach simplifies checking for perpendicularity without direct geometric visualization.