Let Γ denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let

Q: Wait, why do we choose the negative sign for $y'$?

Great observation! The differential equation $x^2 + (xy')^2 = 1$ simplifies to $(y')^2 = \frac{1 - x^2}{x^2}$, which means $y' = \pm \frac{\sqrt{1 - x^2}}{x}$. We are given that the curve passes through $(1, 0)$ and lies in the first quadrant. As $x$ increases from $0$ to $1$, the curve must descend to reach the x-axis at $x=1$. A descending curve has a negative slope, so we must select the negative root.

Q: Why use the substitution $x = \sin \theta$?

This is a classic trigonometric substitution. When you see expressions like $\sqrt{1 - x^2}$, substituting $x = \sin \theta$ is often the most elegant path. It transforms the algebraic expression into $\sqrt{1 - \sin^2 \theta} = \cos \theta$, which simplifies the integral significantly, turning a difficult algebraic integral into a manageable trigonometric one.

Q: How do we handle the absolute value in the natural log during integration?

The term $\ln|\csc \theta - \cot \theta|$ arises from the integral of $\csc \theta$. Since our curve is in the first quadrant and $x$ is between $0$ and $1$, the argument inside the absolute value is positive. Therefore, we can safely remove the absolute value bars, which helps in simplifying the expression to the final form.

Question

Let $\Gamma$ denote a curve $y = y(x)$ which is in the first quadrant and let the point $(1, 0)$ lie on it. Let the tangent to $\Gamma$ at a point $P$ intersect the y-axis at $Y_P$. If $PY_P$ has length 1 for each point $P$ on $\Gamma$, then which of the following options is/are correct?

Options:

(A) $y = \log_e \left( \frac{1 + \sqrt{1 - x^2}}{x} \right) - \sqrt{1 - x^2}$

(B) $xy' - \sqrt{1 - x^2} = 0$

(C) $y = -\log_e \left( \frac{1 + \sqrt{1 - x^2}}{x} \right) + \sqrt{1 - x^2}$

(D) $xy' + \sqrt{1 - x^2} = 0$

Sigma Percentile · Accepted Answer

### The Geometry of the Tractrix: A Journey into Curves

Imagine you are standing in the first quadrant of the Cartesian plane. You have a curve, $\Gamma$, that is not just any curve; it is a special one that holds a secret. It passes through the point $(1, 0)$, and it is defined by a beautiful, constant property: the length of the tangent segment from any point $P$ on the curve to the $y$-axis is always exactly $1$.

This is the famous **Tractrix**, often described as the path a dog takes when its owner walks in a straight line while the dog is on a leash. Let us unravel the mathematics behind this elegant shape.

### Phase 1: The Tangent Constraint

To begin, let us place an arbitrary point $P(x, y)$ on our curve $\Gamma$. We draw the tangent line at this point. The equation of this tangent line, using the point-slope form, is:

$$Y - y = y'(X - x)$$

where $y'$ is the derivative $\frac{dy}{dx}$. This line intersects the $y$-axis at a point $Y_P$. To find the coordinates of $Y_P$, we set $X = 0$, which gives us $Y = y - xy'$.

Thus, the coordinates of $Y_P$ are $(0, y - xy')$. We are told that the distance $PY_P$ is $1$. Using the distance formula, we have:

$$\sqrt{(x - 0)^2 + (y - (y - xy'))^2} = 1$$

This simplifies to $\sqrt{x^2 + (xy')^2} = 1$. Squaring both sides, we get:

$$x^2(1 + (y')^2) = 1$$

This is the heart of our problem.

### Phase 2: The Differential Equation

Now, we isolate the derivative. We have $(y')^2 = \frac{1 - x^2}{x^2}$. Taking the square root, we get:

$$y' = \pm \frac{\sqrt{1 - x^2}}{x}$$

As we discussed, the curve is in the first quadrant and must hit $(1, 0)$, meaning it is decreasing. Thus, we choose the negative sign:

$$y' = -\frac{\sqrt{1 - x^2}}{x}$$

This gives us the differential equation $xy' + \sqrt{1 - x^2} = 0$, which is one of our correct options.

### Phase 3: The Integration Adventure

To find the equation of the curve, we separate the variables: $dy = -\frac{\sqrt{1 - x^2}}{x} dx$. We integrate both sides:

$$y = -\int \frac{\sqrt{1 - x^2}}{x} dx$$

Let us use the substitution $x = \sin 	heta$, so $dx = \cos 	heta d	heta$. The integral becomes:

$$y = -\int \frac{\cos 	heta}{\sin 	heta} \cos 	heta d	heta = -\int \frac{\cos^2 	heta}{\sin 	heta} d	heta$$

Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$, we get:

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

The integral of $\sin 	heta$ is $-\cos 	heta$, and the integral of $\csc 	heta$ is $\ln|\csc 	heta - \cot 	heta|$. Thus:

$$y = -\cos 	heta - \ln|\csc 	heta - \cot 	heta| + C$$

Substituting back $x = \sin 	heta$, we have $\cos 	heta = \sqrt{1 - x^2}$, $\csc 	heta = \frac{1}{x}$, and $\cot 	heta = \frac{\sqrt{1 - x^2}}{x}$. The equation becomes:

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

Using log properties, this simplifies to:

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

Since the curve passes through $(1, 0)$, we plug in $x=1$ and $y=0$ to find $C = 0$. The final equation is:

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

We have successfully derived the equation of the **Tractrix**! It is a beautiful result, showing how a simple geometric constraint can lead to such a sophisticated logarithmic curve.

Select Answer:

Visualized Solution

Visualizing the Curve

The Tangent Constraint

Equation of Tangent Line

Applying the Distance Formula

Simplifying to Differential Equation

Determining the Correct Sign

Setting up Integration

Executing the Integration

Reverting Substitution and Finding C

The Sigma Insight: Variable Separable Method

The Geometry of the Tractrix

Phase 1

Phase 2

Phase 3

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