Similar Triangles and Moving Shadows

Why Similar Triangles Model a Moving Shadow

A person walking away from a lamppost changes the length of the shadow cast on level ground. The person’s distance from the pole, the shadow length, and the distance from the pole to the shadow tip all change together. Similar triangles turn that geometry into an equation, and differentiating the equation relates the corresponding rates.

The key is to label the distances carefully. The shadow itself is not the same distance as the shadow tip’s position measured from the pole. That distinction determines whether a calculated rate describes the shadow length or the tip’s motion.

Set Up the Geometry

Let \(H\) be the height of the lamppost and \(h\) the height of the person. Assume both heights remain constant, the ground is level, and the lamppost is taller than the person, so \(H>h>0\).

At time \(t\), let \(x(t)\) be the person’s horizontal distance from the base of the lamppost, and let \(s(t)\) be the length of the person’s shadow. The shadow tip is therefore \(x(t)+s(t)\) units from the pole. The lamppost, the person, and the shadow tip lie along the same straight ground line.

The large right triangle has vertical side \(H\) and horizontal side \(x+s\). The smaller right triangle, formed by the person and the person’s shadow, has vertical side \(h\) and horizontal side \(s\). The triangles share an acute angle along the light ray, so they are similar by the angle-angle criterion.

Cross-multiplying gives a useful equivalent form. Because \(H>h\), it can also be solved for the shadow length:

This relationship says that, for fixed heights, the shadow length is proportional to the person’s distance from the pole. It describes the geometry at every instant; it is not yet a statement about speed.

Differentiate the Constraint

To relate rates, differentiate the equation \((H-h)s=hx\) with respect to time. Since the heights \(H\) and \(h\) are constant, the result is \((H-h)\frac{ds}{dt}=h\frac{dx}{dt}\). Thus the shadow-length rate is a constant multiple of the person’s distance rate.

The tip’s distance from the pole is \(x+s\). Differentiating this position gives its signed velocity along the ground. Substituting the shadow-length rate produces a second, especially useful relationship:

Why the Rate Relationship Works

The derivation follows from two facts. First, similar triangles give a fixed proportion between the horizontal lengths: \((H-h)s=hx\). Second, the heights are constant, so differentiating this constraint leaves only the rates of the changing horizontal distances.

The result also has a direct interpretation. If \(H>h\), then both multipliers \(\frac{h}{H-h}\) and \(\frac{H}{H-h}\) are positive. So if the person walks away from the pole, with \(\frac{dx}{dt}>0\), the shadow length grows and the tip moves away too. The tip moves faster than the person because its rate multiplier is greater than \(1\): \(\frac{H}{H-h}>1\).

If the person walks toward the pole, then \(\frac{dx}{dt}<0\). The shadow length is shrinking, so \(\frac{ds}{dt}<0\), and the tip’s signed velocity is also negative. A negative signed velocity here means motion toward the pole; it does not mean the tip’s speed is negative.

Worked Examples

A Common Mistake: Using the Wrong Horizontal Side

A frequent error is to use \(x\), the person’s distance from the pole, as the large triangle’s horizontal side. The large triangle extends all the way from the pole to the shadow tip, so that side has length \(x+s\).

For example, with \(H=6\), \(h=2\), and \(x=8\), the incorrect equation \(\frac{H}{x}=\frac{h}{s}\) would give \(\frac{6}{8}=\frac{2}{s}\), or \(s=\frac{8}{3}\) meters. But substituting this into the correct geometric constraint gives \(\frac{6}{8+\frac83}=\frac{9}{16}\), while \(\frac{2}{\frac83}=\frac34\). These ratios are not equal, so the proposed shadow length fails the similarity check.

The correct large-triangle base is \(x+s\). Using it gives \(\frac{6}{8+s}=\frac{2}{s}\), so \(6s=16+2s\), \(4s=16\), and \(s=4\) meters. Substitution confirms \(\frac{6}{8+4}=\frac12=\frac{2}{4}\).

Another related distinction is that the tip’s position is \(x+s\), not \(s\). Its rate is therefore \(\frac{dx}{dt}+\frac{ds}{dt}\), not merely \(\frac{ds}{dt}\). Always identify which changing length the question asks about.

Using the Model Reliably

In a related-rates problem, the geometry supplies a constraint that connects changing quantities. For this model, the constraint comes from the two similar triangles. After differentiating it with respect to time, use the given rate with its sign and units. Do not confuse the shadow length with the tip’s distance from the pole.

The model relies on the stated geometry: fixed heights, level ground, and a lamppost taller than the person. If the person’s height or the lamp’s height changes, the corresponding quantities cannot be treated as constants during differentiation. For the fixed-height situation, the similar-triangle constraint provides both the lengths and the rates needed to describe the moving shadow.