Q.Two men $A$ and $B$ start with velocities $v$ at the same time from the junction of two roads inclined at $45^\circ$ to each other. If they travel by different roads, find the rate at which they are being separated.

Concept: Related Rates — we relate the distance between the two men to the given velocities using geometry, then differentiate with respect to time.

The two roads form an angle of $45^\circ$. Let the distance each man has travelled from the junction after time $t$ be $x = vt$ (since both start at the same time with speed $v$). The distance $s$ between them is the third side of a triangle with two sides $x$ and included angle $45^\circ$.

By the law of cosines:

Differentiate both sides with respect to $t$:

Since $\frac{dx}{dt} = v$ and $s = x\sqrt{2 - \sqrt{2}}$, substitute: …

Why this is a Related Rates problem

When two objects move away from a common point along fixed paths, the distance between them changes over time. We know their individual speeds, but we want the rate of change of the separation distance. That is the essence of related rates: connect the changing quantities through geometry, then differentiate with respect to time.

Here the geometry is simple: two roads meeting at $45^\circ$, both men starting together at the junction, each moving at speed $v$ along his own road. At any time $t$, each has travelled a distance $vt$ from the start. The separation $s(t)$ is the third side of a triangle with two known sides and the included angle.

Step-by-step solution

1. Set up the geometry at time $t$

Let the junction be point $O$. Man $A$ travels along road $OA$, man $B$ along road $OB$, with $\angle AOB = 45^\circ$. Both start at $t=0$ from $O$ with speed $v$.

At time $t$:

• $OA = vt$
• $OB = vt$
• $\angle AOB = 45^\circ$ (constant, because the roads are fixed)

The distance $AB$ between them is the side opposite the $45^\circ$ angle in triangle $OAB$.

2. Apply the law of cosines

For any triangle with sides $a$, $b$ and included angle $\theta$, the third side $c$ satisfies:

Here $a = vt$, $b = vt$, $\theta = 45^\circ$, and $\cos 45^\circ = \frac{\sqrt{2}}{2}$. So:

Notice that $2 - \sqrt{2} > 0$, so the square root is real. The distance $s$ is directly proportional to $t$ — that already hints the rate will be constant.

3. Differentiate to find the rate of separation

We want $\frac{ds}{dt}$. Since $s = v\sqrt{2-\sqrt{2}} \cdot t$, differentiate with respect to $t$:

The speed $v$ and the constant $\sqrt{2-\sqrt{2}}$ are both constants, so the rate is constant — the men separate at a steady pace. …

Method: Related Rates via the Law of Cosines (Fixed-Angle Separation)

When two objects move away from a common starting point along two straight paths that meet at a fixed angle, the distance between them is governed by the Law of Cosines — the Pythagorean theorem only applies when that angle happens to be $90^\circ$.

Steps

Step 1: Set up the triangle.

Let the two paths meet at a constant angle $\theta$. Let $a$ and $b$ be the distances each object has covered from the junction at time $t$, and let $s$ be the distance between them.

Step 2: Write the Law of Cosines.

If both objects move along their own path at the same constant speed $v$, then $a=b=vt$, and the relation simplifies before you even differentiate:

Step 3: Solve for $s$ directly, or differentiate implicitly — both routes agree.

Taking the square root, $s=vt\sqrt{2(1-\cos\theta)}$ is linear in $t$, so …

Common Mistakes

Mistake 1: Assuming the two roads meet at a right angle and using Pythagoras

Why it's wrong: the angle between the roads is $45^\circ$, not $90^\circ$; using $s^2=x^2+x^2$ (implicitly $\cos 90^\circ=0$) instead of the law of cosines gives the wrong coefficient $\sqrt{2}$ instead of the correct $\sqrt{2-\sqrt2}$. Correct approach: for a non-right angle between two known sides, the law of cosines $c^2=a^2+b^2-2ab\cos\theta$ is required.

Mistake 2: Mishandling the surd $\cos 45^\circ=\frac{\sqrt2}{2}$ while simplifying

Why it's wrong: dropping the factor of $\frac12$ or mis-simplifying $2x^2\cdot\frac{\sqrt2}{2}$ to $2\sqrt2\,x^2$ (instead of $\sqrt2\,x^2$) changes the coefficient inside the square root and gives a wrong final rate. Correct approach: substitute $\cos45^\circ=\frac{\sqrt2}{2}$ carefully and simplify step by step before taking the square root. …