Rate of change — ladder sliding down wall related rates problem

Question

A ladder 10 m long is leaning against a vertical wall. The bottom of the ladder is sliding away from the wall at a rate of 2 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 6 m from the wall?

(CBSE 2023)

Solution — Step by Step

Let $x$ = distance of the bottom of the ladder from the wall, and $y$ = height of the top of the ladder on the wall.

The ladder length is constant at 10 m. By the Pythagorean theorem:

x^2 + y^2 = 100

Given: $\frac{dx}{dt} = 2$ m/s (bottom moving away from wall)

Find: $\frac{dy}{dt}$ when $x = 6$ m.

6^2 + y^2 = 100

y^2 = 64

y = 8 \text{ m}

Differentiate $x^2 + y^2 = 100$ implicitly with respect to $t$ :

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

2(6)(2) + 2(8)\frac{dy}{dt} = 0

24 + 16\frac{dy}{dt} = 0

\frac{dy}{dt} = -\frac{24}{16} = -\frac{3}{2}

\boxed{\frac{dy}{dt} = -1.5 \text{ m/s}}

The negative sign means $y$ is decreasing — the top of the ladder is sliding down at 1.5 m/s.

Why This Works

This is a related rates problem. Two quantities ( $x$ and $y$ ) are changing with time, and they’re connected by a geometric equation ( $x^2 + y^2 = 100$ ). Differentiating this equation with respect to time links their rates of change.

The constant right-hand side (100) differentiates to zero — this is crucial and reflects the fact that the ladder doesn’t stretch or shrink. The negative sign in the answer makes physical sense: as the bottom moves out, the top must move down.

Alternative Method — Direct formula

From $x^2 + y^2 = L^2$ , the general related rates formula is:

\frac{dy}{dt} = -\frac{x}{y} \cdot \frac{dx}{dt}

Substituting directly: $\frac{dy}{dt} = -\frac{6}{8} \times 2 = -\frac{3}{2}$ m/s.

For CBSE, always state the negative sign and interpret it: “The negative sign indicates that $y$ is decreasing, i.e., the top of the ladder is sliding down.” Examiners award marks for interpretation, not just the numerical value. In JEE, the answer is typically asked as a magnitude, so write $1.5$ m/s.

Common Mistake

Students often forget that the right side of $x^2 + y^2 = 100$ is a constant and differentiates to zero. Some mistakenly differentiate it as $2L \frac{dL}{dt}$ — but the ladder length isn’t changing, so $\frac{dL}{dt} = 0$ . If the question said the ladder was extending or retracting, only then would the right side have a non-zero derivative.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct formula

Common Mistake

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