The ends of a rod AB = 10 cm slide along the coordinate

The ends of a rod $AB = 10\ \text{cm}$ slide along the coordinate axes, $A$ along $Ox$ and $B$ along $Oy$. The point $A$ moves away from the origin at $\dfrac{dx}{dt} = 2\ \text{cm/s}$. Find the speed of $B$ at the moment when $OA = 6\ \text{cm}$.

Step 1: The constraint

With $x = OA$ and $y = OB$, the constant length gives

\[ x^{2}+y^{2} = 100. \tag{1} \]

Both $x$ and $y$ vary with time, but their combination does not — that is what makes the rates related.

Step 2: Differentiate with respect to time

\[ 2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0 \;\Longrightarrow\; x\frac{dx}{dt}+y\frac{dy}{dt} = 0. \tag{2} \]

Step 3: Locate the moment

When $x = 6$, equation (1) gives $y^{2} = 64$, so $y = 8\ \text{cm}$ — the familiar $6$–$8$–$10$ triangle.

Step 4: Solve for the speed of $B$

\[ 6\cdot2+8\frac{dy}{dt} = 0 \;\Longrightarrow\; \frac{dy}{dt} = -\frac{12}{8} = -1.5\ \frac{\text{cm}}{\text{s}}. \]

The minus sign says $B$ is descending towards the origin while $A$ recedes from it.

Step 5: Verification

\[\text{From (2): } \frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} = -\frac{6}{8}\cdot2 = -1.5 \;\;\checkmark\]

\[\text{At } x = 8,\ y = 6:\ \frac{dy}{dt} = -\frac{8}{6}\cdot2 \approx -2.67\ \frac{\text{cm}}{\text{s}} \ \text{(faster, as } B \text{ nears the origin)}\]

\[\text{The two speeds are equal only at } x = y = 5\sqrt2 \approx 7.07\ \text{cm} \;\;\checkmark\]