Orthogonal trajectories and curve problems

What you will be able to do

• find the orthogonal trajectories of a family of curves;
• write the tangent, the normal, the sub-tangent and the sub-normal of a curve in terms of $y$ and $\dfrac{dy}{dx}$;
• turn a word statement about a curve into a differential equation, and solve it.

Two families that cross at right angles

On a weather map the lines of equal pressure and the paths of the wind cross at right angles. In mathematics, two families of curves are orthogonal trajectories of each other when each curve of one family cuts each curve of the other at right angles. At a crossing point the two tangents are perpendicular, so the product of their slopes is $-1$.

The method

Start with a family that has one constant, such as $y = Cx^2$.

1. Form its differential equation. Differentiate and remove the constant. For $y = Cx^2$ we get $\dfrac{dy}{dx} = 2Cx$ and $C = \dfrac{y}{x^2}$, so $\dfrac{dy}{dx} = \dfrac{2y}{x}$.
2. Swap the slope. The slope of the orthogonal curve is the negative reciprocal. Replace $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$, which gives $-\dfrac{dx}{dy} = \dfrac{2y}{x}$, or $\dfrac{dy}{dx} = -\dfrac{x}{2y}$.
3. Solve the new equation. Here $2y\,dy = -x\,dx$, so $y^2 = -\dfrac{x^2}{2} + c$, which is $x^2 + 2y^2 = k$.

The orthogonal trajectories of the parabolas $y = Cx^2$ are the ellipses $x^2 + 2y^2 = k$. Do step 1 first and remove the constant before you swap. If you swap the slope while the constant is still there, you mix the two families.

More examples.

• Circles $x^2 + y^2 = r^2$. Then $2x + 2y\dfrac{dy}{dx} = 0$ gives $\dfrac{dy}{dx} = -\dfrac{x}{y}$. The swap gives $\dfrac{dy}{dx} = \dfrac{y}{x}$, so $y = mx$. The trajectories are straight lines through the origin.
• Hyperbolas $xy = c$. Then $\dfrac{dy}{dx} = -\dfrac{y}{x}$, and the swap gives $\dfrac{dy}{dx} = \dfrac{x}{y}$. So $y\,dy = x\,dx$ and $x^2 - y^2 = k$.

If you swap twice you get the original family back, so the relation is symmetric.

Tangent, normal and the words of curve problems

Curve problems state a property of the tangent or the normal at every point $P(x, y)$. To write them as equations, use these results. The slope of the tangent is $y' = \dfrac{dy}{dx}$ and the slope of the normal is $-\dfrac{1}{y'}$.

• Tangent: $Y - y = y'(X - x)$. It meets the $x$-axis at $X = x - \dfrac{y}{y'}$ and the $y$-axis at $Y = y - xy'$.
• Normal: $Y - y = -\dfrac{1}{y'}(X - x)$. It meets the $x$-axis at $X = x + yy'$.
• Sub-tangent: the length along the $x$-axis from the foot of the perpendicular to the tangent, $\left|\dfrac{y}{y'}\right|$.
• Sub-normal: the length from the foot of the perpendicular to the normal, $|yy'|$.

Example 4. The sub-normal of a curve is $4$ at every point. Then $yy' = 4$, so $y\,dy = 4\,dx$ and $y^2 = 8x + C$. Curves with a constant sub-normal are parabolas.

Example 5. The slope at every point $(x, y)$ is twice $\dfrac{y}{x}$. Then $\dfrac{dy}{dx} = \dfrac{2y}{x}$, so $\dfrac{dy}{y} = \dfrac{2\,dx}{x}$ and $y = Cx^2$. If the curve passes through $(2, 12)$, then $C = 3$.

Example 6. The normal at every point passes through the origin. The normal at $(x, y)$ has the equation $Y - y = -\dfrac{1}{y'}(X - x)$. Putting $(0, 0)$ gives $-y = \dfrac{x}{y'}$, so $yy' = -x$. Then $y\,dy = -x\,dx$ and $x^2 + y^2 = c$. The curves are circles centred at the origin, which fits: the radius is normal to a circle.

Always read the words in this order. Name the property, write it with $x$, $y$ and $y'$, solve, and use any point given to find the constant.

Practise with new numbers

Try it in Python

The program takes crossing points of the parabolas and the ellipses, finds both slopes at each point and multiplies them.

Common mistakes

• Swapping before eliminating the constant. First remove $C$, then replace $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$.
• Forgetting the minus sign. The slopes of perpendicular curves have a product of $-1$, not $+1$.
• Mixing the tangent and the normal. The tangent has slope $y'$ and the normal has slope $-\dfrac{1}{y'}$.
• Confusing the sub-tangent with the sub-normal. The sub-tangent is $\dfrac{y}{y'}$, and the sub-normal is $yy'$.
• Leaving out the point. A condition such as "passes through $(2, 12)$" fixes the constant of integration.

Remember these

• Orthogonal trajectories: form $\dfrac{dy}{dx} = f(x, y)$ without the constant, replace $\dfrac{dy}{dx}$ by $-\dfrac{1}{f}$, then solve.
• At a crossing point the product of the slopes is $-1$.
• Intercepts of the tangent: $x - \dfrac{y}{y'}$ and $y - xy'$. Intercept of the normal on the $x$-axis: $x + yy'$.
• Sub-tangent $\left|\dfrac{y}{y'}\right|$ and sub-normal $|yy'|$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the orthogonal trajectories of the family of lines $y = mx$.

2. Find the orthogonal trajectories of the parabolas $y^2 = 4ax$.

3. The sub-tangent of a curve is $2$ at every point, with $y > 0$. Find the curve through $(0, 5)$.

4. The normal at every point of a curve passes through the origin, and the curve passes through $(3, 4)$. Find the curve.

5. True or false: if a family $F_2$ is orthogonal to a family $F_1$, then $F_1$ is orthogonal to $F_2$.

References

• OpenStax, Calculus Volume 2, Chapter 4, section 4.1 Basics of Differential Equations and section 4.3 Separable Equations.
• NCERT, Mathematics Part II for Class XII, Chapter 9, the sections on forming and solving first-order equations. Orthogonal trajectories are a standard exercise in first-order equations.