Finding the ladder's sliding speed using related rates

Key concept: The 3 m ladder gives a right triangle whose side lengths change together, so related rates and the Pythagorean theorem are the tools to use.

Step by step

Set up the moving right triangle

The ladder, the wall, and the ground form a right triangle with fixed hypotenuse $3$ m. Let $x$ be the distance from the wall to the bottom of the ladder, and let $y$ be the height of the top of the ladder above the ground. Then

$x^2+y^2=3^2=9$

The bottom moves away at $\frac{dx}{dt}=0.5\text{ m/s}$. We want the downward speed of the top when $y=2\text{ m}$.

Differentiate the geometric relationship

Because both $x$ and $y$ depend on time, differentiate $x^2+y^2=9$ with respect to $t$:

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

Divide by 2:

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

Now find $x$ when $y=2$:

$x^2+2^2=9$

$x^2=5$

$x=\sqrt{5}$

Substitute $x=\sqrt{5}$, $y=2$, and $\frac{dx}{dt}=0.5$:

$\sqrt{5}(0.5)+2\frac{dy}{dt}=0$

$2\frac{dy}{dt}=-0.5\sqrt{5}$

$\frac{dy}{dt}=-\frac{\sqrt{5}}{4}\text{ m/s}$

Interpret the negative sign

The negative sign shows that the top is sliding downward. The speed, as a positive magnitude, is

$\frac{\sqrt{5}}{4}\text{ m/s} \approx 0.56\text{ m/s}$

So the top of the ladder is descending at about $0.56\text{ m/s}$.

Useful checking habit

The answer should be slower than the bottom's horizontal speed because the vertical motion is constrained by the ladder length. If your result is larger than $0.5\text{ m/s}$, that is a warning sign that the geometry or substitution went wrong. The fixed ladder length must stay in the equation until the instant values are inserted.

Pitfall alert

The most common place this ladder problem breaks down is when the 3 m length is treated as if it were changing. The ladder is rigid, so $x^2+y^2=9$ stays constant, and only $x$ and $y$ vary with time. Another trap is substituting $y=2$ too early and forgetting to compute the matching $x$ value from the Pythagorean theorem. Without $x=\sqrt{5}$, the rate equation cannot be solved correctly. Students also sometimes report $\frac{dy}{dt}$ as positive because they are thinking about speed rather than direction. The derivative gives a signed rate, and here the sign must be negative because the top moves downward. A final check is units: the bottom's speed is in m/s, so the top's rate must also be in m/s.

Try different conditions

If the ladder were 5 m long and the bottom rolled away at 0.5 m/s, the same method would apply with $x^2+y^2=25$. When the top is 4 m above the ground, we would find $x=3$ because $3^2+4^2=25$. Differentiating gives $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$, so substituting $x=3$, $y=4$, and $\frac{dx}{dt}=0.5$ gives $3(0.5)+4\frac{dy}{dt}=0$. Then $\frac{dy}{dt}=-0.375\text{ m/s}$. Changing the ladder length changes the geometry, but the same related-rates structure remains. That makes this a good test of whether you can recompute the instant triangle correctly instead of memorizing one number.

Further reading

ladder related rates, Pythagorean theorem rates, sliding ladder problem