Sliding Ladder Rates

Modeling a Sliding Ladder

A ladder leaning against a wall creates a right triangle: the wall and ground form the perpendicular sides, and the ladder is the hypotenuse. As the ladder slides, the triangle’s dimensions change, but the length of a rigid ladder stays fixed. This relationship lets us connect how quickly the ladder’s foot moves along the ground to how quickly its top moves up or down the wall.

Let \(x(t)\) be the horizontal distance, in feet, from the wall to the ladder’s foot at time \(t\). Let \(y(t)\) be the vertical height, in feet, of the ladder’s top above the ground. Let \(L\) be the ladder’s fixed length. While the ladder touches both the wall and the ground, \(x\) and \(y\) are positive.

The rates \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) describe the instantaneous changes in the foot’s distance from the wall and the top’s height. Both rates have units of distance per time, such as feet per second. Their signs describe direction: an increasing \(x\) means the foot moves away from the wall, while a decreasing \(y\) means the top moves downward.

Deriving the Rate Relationship

The Pythagorean theorem gives a relation that holds throughout the motion, not just at one instant. Differentiate this relation with respect to time. Because \(x\) and \(y\) depend on \(t\), the chain rule is required; because \(L\) is constant, its derivative is zero.

Dividing by \(2\) and solving for either rate gives the two useful forms below. These formulas apply during the motion when the ladder’s length is fixed and the right-triangle model is valid.

This derivation also explains the direction relationship. With \(x,y>0\), the two rates must have opposite signs unless both are zero. If the foot moves away from the wall, \(\frac{dx}{dt}>0\), then \(\frac{dy}{dt}<0\): the top moves down. If the foot moves toward the wall, \(\frac{dx}{dt}<0\), then \(\frac{dy}{dt}>0\): the top moves up.

Finding the Ladder’s Height Rate

When the foot’s rate is known, use \(\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\). The measurements \(x\) and \(y\) must describe the same instant as the given rate. If the ladder length and one distance are given instead, first find the other distance from the Pythagorean theorem.

Finding the Foot’s Rate

Sometimes the top’s rate is known and the question asks how quickly the foot moves. In that case, solve for \(\frac{dx}{dt}\). Pay close attention to signs: a downward-moving top has a negative height rate, and that should produce a positive foot rate when the ladder slides away from the wall.

Using the Same Model for Motion Toward the Wall

The rate equation works for either direction of motion. A negative horizontal rate represents motion toward the wall. With the ladder still touching both surfaces, that motion makes the top rise, so its rate is positive. The equation determines the sign automatically if the given rate is entered with its direction.

Why the Rate Changes with Position

The multiplier connecting the two rates is a ratio of distances. In particular, \(\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\) shows that the same foot speed can correspond to different vertical speeds at different positions. When \(x/y\) is larger, the height’s rate has greater magnitude for a given horizontal rate. The rates need not remain constant merely because the ladder length is constant.

For example, in the first worked example, the ratio \(x/y\) at the stated instant is \(5/12\). That ratio, together with the instantaneous foot speed, determines the height rate at that instant. If the ladder continues sliding, \(x\) and \(y\) change, so the ratio changes too. A rate found for one position should not be treated as the rate at every position.

Common Pitfall: Substituting Too Early or Losing a Sign

A common mistake is to replace \(x\) and \(y\) with their values at the instant before differentiating. For instance, in the first example, writing \(5^2+12^2=13^2\) and then treating it as an equation for changing quantities loses the relationship between their rates. That equation checks one position; it does not express how the position changes over time.

Another error is to omit the negative sign. For the first example, using \(\frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}\) would give \(\frac{5}{6}\text{ ft/s}\), incorrectly saying the top rises as the foot moves away. The correction is to differentiate the constraint before substituting measurements:

Then enter each rate with its correct sign and check the result in \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\). This check confirms the algebra and catches many direction errors.

A Reliable Setup

The model matters because it turns a geometric constraint into a relationship between instantaneous motions. The Pythagorean theorem describes the ladder’s position; differentiating it describes how that position changes. Keeping those two roles distinct—first the constraint, then its time derivative—is the essential step in solving sliding-ladder rate problems.