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 $x^2+2y^2=k,$ where $c$ and $k$ are constants (see figure). Find $dy/dx$ for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. $xy=a;x^2-y^2=b,$ where $a$ and $b$ are constants

Short answer

Solution: Yes, the given families of curves form orthogonal trajectories since their derivatives, found through implicit differentiation, have a product equal to -1.

Step-by-step solution

Differentiate the equation $xy=a$ with respect to $x$

Using implicit differentiation, we can differentiate each term of the equation $xy=a$ with respect to $x$. First term: Differentiating $x\cdot y$ with respect to $x$ using product rule (let $u=x$ and $v=y$), where $u^{\prime }=\frac{d(x)}{dx}$ and $v^{\prime }=\frac{d(y)}{dx}$. $\frac{d(uv)}{dx}=u^{\prime }v+u{v}^{\prime }=\frac{d(x)}{dx}\cdot y+x\cdot \frac{d(y)}{dx}=1\cdot y+x\cdot \frac{d(y)}{dx}$ Second term: Since "a" is a constant, its derivative is zero: $\frac{d(a)}{dx}=0$ Now, combining these results, we get $\frac{d(xy)}{dx}=y+x\frac{dy}{dx}=0$

Differentiate the equation $x^2-y^2=b$ with respect to $x$

Similarly, we differentiate the equation $x^2-y^2=b$ implicitly with respect to $x$: First term: Differentiating $x^2$ with respect to $x$: $\frac{d(x^2)}{dx}=2x$ Second term: Differentiating $-y^2$ with respect to $x$: $-2y\frac{dy}{dx}$ (since $y^{\prime }=\frac{dy}{dx}$) Third term: Since "b" is a constant, its derivative is zero: $\frac{d(b)}{dx}=0$ Now we combine these results, we get $2x-2y\frac{dy}{dx}=0$

Show that the product of the derivatives is equal to -1

We first find $\frac{dy}{dx}$ for both equations separately, and then check if their product is -1. From step 1, for the equation $xy=a$, we have $y+x\frac{dy}{dx}=0$. Rearranging for $\frac{dy}{dx}$, we have: $\frac{dy}{dx}=-\frac{y}{x}$ From step 2, for the equation $x^2-y^2=b$, we have $2x-2y\frac{dy}{dx}=0$. Rearranging for $\frac{dy}{dx}$, we have: $\frac{dy}{dx}=\frac{x}{y}$ Now we check the condition for orthogonality by multiplying the derivatives and checking if the product is equal to -1: $(-\frac{y}{x})(\frac{x}{y})=-1$ To verify if the product gives -1, we can substitute x and y in the product. Suppose we consider $x=3,y=2,a=6$, and $b=5$. As per the given conditions, $xy=a=6$ and $x^2-y^2=b=5$. Thus, the orthogonality condition holds for these cases. The product of both derivatives equates to -1, indicating that the given families of curves form orthogonal trajectories.

Key concepts

Implicit Differentiation

Implicit differentiation is a method used to take the derivative of an equation when it is not solved for one variable in terms of another. It allows us to find the derivative of a dependent variable with respect to an independent variable without having to explicitly solve for that variable. For instance, consider the equation involving two variables, such as the given equation of the family of curves $xy=a$. When we cannot, or choose not to solve for $y$ in terms of $x$ before differentiating, implicit differentiation comes into play. We differentiate both sides of the equation with respect to $x$ while treating $y$ as a function of $x$ that also needs to be differentiated, which often involves the use of the chain rule and the product rule for differentiation.

It's a useful technique, especially for equations that define $y$ implicitly, or where solving for $y$ explicitly is difficult. This technique was used in Step 1 of our exercise to differentiate the product of $x$ and $y$ by considering $y$ as a function of $x$ that has its own derivative, $\frac{dy}{dx}$ which can be found as a result of the differentiation process.

Tangent Lines

A tangent line is a straight line that touches a curve at a single point without crossing over it. The slope of the tangent line at any given point on the curve is the derivative of the function at that point. In the context of our exercise, the concept of tangent lines is directly related to the slopes of the curves defined by the equations $xy=a$ and $x^2-y^2=b$.

The slopes $\frac{dy}{dx}$ found through implicit differentiation for these curves represent the slopes of the tangent lines at any point on the curves. For understanding orthogonal trajectories, the slopes of tangent lines are crucial, as the condition for two curves to be orthogonal trajectories is that their tangent lines at the points of intersection are perpendicular to each other. This brings us to the idea that the slopes of these tangent lines must be negative reciprocals, which is explored in more detail in the next section.

Negative Reciprocals

Two non-vertical lines are perpendicular if and only if the product of their slopes is $-1$. This means the slopes of perpendicular lines are negative reciprocals of each other. For example, if one line has a slope of $m$, the line perpendicular to it will have a slope of $-\frac{1}{m}$. This concept is a key part in understanding orthogonal trajectories.

Within our exercise, we derived the slopes of the tangent lines for the two families of curves and found them to be $-\frac{y}{x}$ and $\frac{x}{y}$ which are negative reciprocals of each other. Therefore, when the product of these derivatives is taken, as seen in Step 3 of the solution, it results in $-1$, satisfying the condition for orthogonality. This is a mathematical guarantee that the curves are indeed orthogonal trajectories at every point of intersection.

Product Rule Differentiation

The product rule is a fundamental rule in calculus for finding the derivative of a product of two functions. It states that if you have a function that can be expressed as the product of two other functions, $u(x)$ and $v(x)$, then the derivative of this product is given by $u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$, where $u^{\prime }(x)$ and $v^{\prime }(x)$ are the derivatives of $u$ and $v$ with respect to $x$, respectively.

This rule is particularly useful when you are dealing with products of variables as functions, such as $x$ and $y$ in the equation $xy=a$. During the first step of solving our example exercise, we applied the product rule to differentiate the equation $xy=a$ implicitly. We treated $x$ and $y$ as $u$ and $v$, finding the derivative with respect to $x$ while considering $y$ as an implicitly defined function (hence also differentiating $y$ with respect to $x$ within the process). The correct application of the product rule is crucial in such problems to find the correct slope of the tangent lines and establish the relationship between orthogonal trajectories.