How to Find the Area in a Ladder Related-Rates Problem

In the standard ladder related-rates problem, a ladder leans against a vertical wall while its foot slides along the floor. The ladder, wall, and floor form a right triangle. If the question asks for the area of that triangular region, use its horizontal leg and vertical leg—not the ladder’s length—as the base and height.

The main steps are to write the triangle’s area, relate the two changing distances using the fixed ladder length, and differentiate with respect to time. Keep the position and rate variables distinct: a position such as $x=6$ is substituted after differentiation, not before.

Set up the area and length equations

Let $x(t)$ be the distance from the wall to the foot of the ladder, and let $y(t)$ be the height of the top of the ladder above the floor. If the ladder has fixed length $L$, the Pythagorean theorem gives

$$x^2+y^2=L^2.$$

The area $A$ of the right triangle is

$$A=\frac{1}{2}xy.$$

Both $x$ and $y$ usually change over time, so the area changes too. Differentiate the area equation with respect to time using the product rule:

$$\frac{dA}{dt}=\frac{1}{2}\left(x\frac{dy}{dt}+y\frac{dx}{dt}\right).$$

This is the key area formula for the problem. The units of $dA/dt$ are square units per unit of time, such as square feet per second.

Use the fixed-length constraint to find a missing rate

Differentiate $x^2+y^2=L^2$ with respect to time. Since $L$ is constant,

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

Solving for the vertical rate gives

$$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}.$$

The minus sign reflects the geometry: as the foot moves away from the wall, the top slides downward. If the problem gives the top’s rate instead, you can solve the same equation for $dx/dt$.

Worked example

A 10-foot ladder leans against a wall. Its foot moves away from the wall at $2$ feet per second. How fast is the triangular area changing when the foot is $6$ feet from the wall?

First find the height at that instant. The fixed-length equation gives

$$6^2+y^2=10^2,$$

so $y^2=64$ and $y=8$ feet. The positive root is appropriate because the top of the ladder is above the floor.

Next find the rate at which the top moves. Here $dx/dt=2$ feet per second, so

$$\frac{dy}{dt}=-\frac{6}{8}(2)=-\frac{3}{2}\text{ feet per second}.$$

Now substitute the positions and rates into the differentiated area formula:

$$\begin{aligned} \frac{dA}{dt} &=\frac{1}{2}\left(6\left(-\frac{3}{2}\right)+8(2)\right)\\ &=\frac{1}{2}(-9+16)\\ &=\frac{7}{2}\text{ square feet per second}. \end{aligned}$$

The triangular area is increasing at $3.5$ square feet per second at that instant. Although the triangle is getting shorter, its base is growing quickly enough that its total area is increasing.

A useful equivalent formula

You can combine the two rate equations to express the area rate using just the horizontal rate:

$$\begin{aligned} \frac{dA}{dt} &=\frac{1}{2}\left(x\left(-\frac{x}{y}\frac{dx}{dt}\right)+y\frac{dx}{dt}\right)\\ &=\frac{y^2-x^2}{2y}\frac{dx}{dt}. \end{aligned}$$

This version can be convenient when the foot’s speed is given. It also shows that the area is momentarily constant when $x=y$, provided the ladder is moving. At that instant, the base’s increase and the height’s decrease balance in the area formula.

Common mistakes

• Using the ladder length as a triangle side in the area formula. The ladder is the hypotenuse, not the base or height. The area is $xy/2$, not $Lx/2$ or $Ly/2$.
• Treating the rates as positive magnitudes. If upward is the positive direction for $y$, a top sliding down has $dy/dt<0$. Preserve that sign in the calculation.
• Differentiating after substituting the instant’s measurements. Keep $x$ and $y$ as variables until after differentiating. Substituting fixed numbers too early erases their time dependence.
• Confusing area with area rate. Area is measured in square feet; its rate is measured in square feet per second.
• Assuming the area must decrease because the top descends. The base also changes. Evaluate the full expression for $dA/dt$ to determine whether the area grows or shrinks.

If a problem asks for the area of a different shape—for example, a rectangle beside the ladder—identify that shape and its dimensions first. The formula $A=xy/2$ applies specifically to the right triangle formed by the wall, floor, and ladder.