Sliding Ladder Constraints

Introduction

A ladder leaning against a vertical wall forms a right triangle with the wall and the floor. As the ladder slides, its foot moves horizontally and its top moves vertically. These motions are linked: the ladder's length stays fixed, so the two endpoint rates cannot vary independently.

The useful starting point is not a rate formula but the geometric constraint. Once the distances are defined and the fixed-length relation is written down, differentiating that relation with respect to time gives the link between the rates. The position of the ladder matters: the same horizontal speed can correspond to different vertical speeds at different angles.

Set Up the Ladder Model

Assume the floor and wall meet at a right angle, and the ladder is rigid, straight, and of fixed length \(L\). Let \(x(t)\) be the horizontal distance from the wall to the foot of the ladder at time \(t\), and let \(y(t)\) be the vertical height of its top above the floor. We consider positions where the ladder touches both the floor and wall, so \(x(t)>0\) and \(y(t)>0\).

The ladder is the hypotenuse of a right triangle whose legs have lengths \(x(t)\) and \(y(t)\). The Pythagorean theorem therefore gives the constraint

Here \(L\) is constant, while \(x\) and \(y\) change over time. In related-rates notation, their instantaneous rates are \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). A positive value of \(\frac{dx}{dt}\) means the foot moves away from the wall; a negative value means it moves toward the wall. A positive value of \(\frac{dy}{dt}\) means the top rises; a negative value means it descends.

Derive the Rate Relation

Differentiate the constraint with respect to time. Since \(x\) and \(y\) are functions of \(t\), the chain rule applies to both squared terms. Since \(L\) is fixed, the derivative of \(L^2\) is zero.

Dividing by \(2\) gives the rate equation. Because \(y>0\) for the positions considered, we can also solve it for the top's rate. Because \(x>0\), we can solve it for the foot's rate as well.

This derivation explains why the rate relation works: it follows by differentiating the geometric equation that must hold at every instant. It is not a separate physical law or an approximation. The relation is instantaneous, so when a problem supplies a particular position and rate, substitute those values into the differentiated equation.

A Reliable Calculation Sequence

Worked Examples

Signs, Units, and Interpretation

Rates should be signed before they are substituted. Words such as “away,” “toward,” “rising,” and “descending” determine whether a rate is positive or negative under the definitions of \(x\) and \(y\). For example, “the foot slides toward the wall at \(2\) ft/s” means \(\frac{dx}{dt}=-2\text{ ft/s}\), not \(+2\text{ ft/s}\). Likewise, a downward top rate is negative.

The equation \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\) is dimensionally consistent. If lengths are measured in feet and time in seconds, then \(x\frac{dx}{dt}\) and \(y\frac{dy}{dt}\) both have units of square feet per second. Solving for \(\frac{dy}{dt}\) gives a length ratio \(x/y\), which has no units, multiplied by \(\frac{dx}{dt}\), a length per time. The resulting vertical rate therefore has the required units of feet per second.

The factor \(x/y\) also shows why position affects the rate. When \(y\) is small compared with \(x\), the top's rate can be large in magnitude relative to the foot's rate. When \(x\) is small compared with \(y\), the top's rate can be smaller. The formula applies at positions with \(x>0\) and \(y>0\); at a limiting position where one endpoint distance is zero, the solved formula that divides by that distance cannot be used as written.

A Common Sign Error

Return to the first example, where \(x=5\), \(y=12\), and \(\frac{dx}{dt}=2\text{ ft/s}\). A common incorrect step is to write \(\frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}\), dropping the minus sign. That gives

This result says the top rises while the foot moves away. Test it in the differentiated constraint:

So the proposed rates do not preserve the fixed-length constraint. The correct rearrangement is \(y\frac{dy}{dt}=-x\frac{dx}{dt}\), which gives \(\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{5}{6}\text{ ft/s}\). Substitution gives \(5(2)+12(-5/6)=0\), as required.

Why the Constraint Is the Essential Step

The ladder problem is a useful model of related rates because it separates geometry from motion. The right triangle gives a relation between positions, and differentiation turns that relation into a connection between rates. The rates are not inferred from a drawing alone: the equation specifies their precise relationship at the instant in question.

The method also clarifies what information is needed. A ladder length and one endpoint position determine the other endpoint's position by the triangle equation. To determine an endpoint's rate, one additionally needs a rate for the other endpoint at that instant. After these quantities are identified, the rate equation can be solved directly. No assumption that either endpoint moves at constant speed is required; the relation holds instant by instant as long as the ladder length remains fixed and the endpoints slide along the wall and floor.

The main practical safeguards are to keep \(L\) constant, differentiate before substituting the instant's values, assign signs from the stated directions, and verify the result in the differentiated constraint. These checks catch the most common mistakes without changing the geometric model.