If length of tangent at any point on the curve y = f(x) intercepted between the point and the x-axis

Q: Wait, why is the formula for the length of the tangent $L = |y \sqrt{1 + (dx/dy)^2}|$? I thought it was $y \sqrt{1 + (dy/dx)^2}$?

That is a fantastic catch! You are thinking of the length of the tangent segment intercepted on the *y-axis*. In this problem, the tangent is intercepted on the *x-axis*. If you draw the triangle, the vertical side is $y$, and the slope of the tangent is $dy/dx$. The horizontal side of the triangle is $y / (dy/dx)$. Using the Pythagorean theorem on the tangent segment, you get $L = \sqrt{y^2 + (y / (dy/dx))^2}$, which simplifies to $|y \sqrt{1 + (dx/dy)^2}|$. Always check which axis the intercept is on!

Q: Why did we choose the substitution $y = \sin \theta$ instead of $x = \sin \theta$?

Great question. We look at the structure of the integral. We had $\int \frac{\sqrt{1 - y^2}}{y} dy$. The term $\sqrt{1 - y^2}$ screams for a trigonometric substitution. If we set $y = \sin \theta$, then $dy = \cos \theta d\theta$, and the radical becomes $\sqrt{1 - \sin^2 \theta} = \cos \theta$. This substitution transforms the entire integrand into a function of $\theta$ that is much easier to manage. If we had used $x$, we would have been stuck with $dx$ and no clear path to simplify the $y$ terms.

Q: What is the physical significance of the constant $C$ in this equation?

In the context of the Tractrix, the constant $C$ represents the horizontal translation of the curve. Since the differential equation only defines the *shape* of the curve, the constant of integration allows the curve to exist anywhere along the x-axis. It is the 'freedom' of the curve to slide left or right while maintaining its unique geometric property.

Question

If length of tangent at any point on the curve $y = f(x)$ intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

Sigma Percentile · Accepted Answer

### The Geometry of the Tractrix: A Journey into Constant Tangent Lengths

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a differential equation; we are uncovering the secret geometry of a curve known as the **Tractrix**.

Imagine you are standing on a flat plane, holding a string of length $1$ attached to an object. As you walk in a straight line, the object follows you, always being pulled by the taut string. The path that object traces is exactly what we are about to derive.

### Phase 1: Translating Geometry into Calculus

We start with a curve $y = f(x)$. Pick any point $P(x, y)$ on this curve. If we draw a tangent line at $P$, it will eventually strike the x-axis at a point $T$.

The problem tells us that the distance $PT$ is always $1$. This is a powerful constraint, meaning that no matter where you are on the curve, the 'reach' of the tangent to the x-axis is constant.

To turn this into math, we look at the right-angled triangle formed by the point $P$, its projection on the x-axis, and the intercept $T$. The vertical side of this triangle is $y$. The slope of the tangent is $\frac{dy}{dx}$, which means the angle $\phi$ the tangent makes with the x-axis satisfies $	an \phi = \frac{dy}{dx}$.

The horizontal distance from the projection of $P$ to $T$ is $y \cot \phi = y \frac{dx}{dy}$. By the Pythagorean theorem, the length of the tangent $L$ is $\sqrt{y^2 + (y \frac{dx}{dy})^2}$. Factoring out $y$, we get the golden formula:

$$L = \left| y \sqrt{1 + \left( \frac{dx}{dy} ight)^2} ight|$$

### Phase 2: The Differential Equation

We are given $L = 1$. So, we write:

$$\left| y \sqrt{1 + \left( \frac{dx}{dy} ight)^2} ight| = 1$$

To make this manageable, we square both sides:

$$y^2 \left[ 1 + \left( \frac{dx}{dy} ight)^2 ight] = 1$$

Now, we isolate the derivative. Dividing by $y^2$, we get $1 + (\frac{dx}{dy})^2 = \frac{1}{y^2}$. Subtracting $1$ gives us $(\frac{dx}{dy})^2 = \frac{1 - y^2}{y^2}$. Taking the square root, we arrive at our differential equation:

$$\frac{dx}{dy} = \pm \frac{\sqrt{1 - y^2}}{y}$$

This is a first-order separable differential equation. We can separate the variables by moving all $y$ terms to one side and $x$ to the other:

$$dx = \pm \frac{\sqrt{1 - y^2}}{y} dy$$

### Phase 3: The Art of Integration

Integrating the left side is trivial: $\int dx = x$. The right side, however, requires a bit of finesse. We have $\int \frac{\sqrt{1 - y^2}}{y} dy$.

Whenever you see $\sqrt{1 - y^2}$, your intuition should immediately jump to trigonometric substitution. Let $y = \sin 	heta$. Then $dy = \cos 	heta d	heta$. Substituting these into our integral, we get:

$$\pm \int \frac{\sqrt{1 - \sin^2 	heta}}{\sin 	heta} (\cos 	heta d	heta) = \pm \int \frac{\cos^2 	heta}{\sin 	heta} d	heta$$

Since $\cos^2 	heta = 1 - \sin^2 	heta$, we split the fraction:

$$\pm \int \left( \frac{1}{\sin 	heta} - \frac{\sin^2 	heta}{\sin 	heta} ight) d	heta = \pm \int (\csc 	heta - \sin 	heta) d	heta$$

We know these integrals! The integral of $\csc 	heta$ is $\ln |\csc 	heta - \cot 	heta|$, and the integral of $-\sin 	heta$ is $\cos 	heta$. So, we have:

$$x = \pm (\ln |\csc 	heta - \cot 	heta| + \cos 	heta) + C$$

### Phase 4: The Final Reveal

We must return to our original variables $x$ and $y$. Since $y = \sin 	heta$, we have $\csc 	heta = \frac{1}{y}$. Using a right triangle, the adjacent side is $\sqrt{1 - y^2}$, so $\cot 	heta = \frac{\sqrt{1 - y^2}}{y}$ and $\cos 	heta = \sqrt{1 - y^2}$.

Substituting these back, we get:

$$x = \pm \left( \ln \left| \frac{1 - \sqrt{1 - y^2}}{y} ight| + \sqrt{1 - y^2} ight) + C$$

Rearranging, we reach the final equation:

$$\ln \left| \frac{1 - \sqrt{1 - y^2}}{y} ight| + \sqrt{1 - y^2} = \pm x + c$$

This is the **Tractrix**. It is a curve that defies the simplicity of polynomials, born from the elegant constraint of a constant tangent length. You have successfully navigated the calculus, the trigonometry, and the geometry.

Animated Solution for Mathematics - Differential Equations: If length of tangent at any point on the curve $y = f (x)$ intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

Visualized Solution

Visualizing the Geometric Constraint

Formula for Length of Tangent $L$

Applying the Given Constraint $L = 1$

Squaring Both Sides

Isolating the Derivative $(\frac{d x}{d y})^{2}$

Taking the Square Root

Separating the Variables

Integration via Substitution $y = sin θ$

Substituting into the Integral

Simplifying the Integrand

Using Trigonometric Identities

Performing the Integration

Back-Substitution Preparation

Final Substitution

The Equation of the Tractrix

The Sigma Insight: Variable Separable Method

The Geometry of the Tractrix

Phase 1

Phase 2

Phase 3

Phase 4

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