Shadow Lengths from Similar Triangles

The Geometry of a Moving Shadow

A person walking away from a lamp changes both their distance from the lamp and the length of their shadow. The tip of the shadow moves too, and it does not generally move at the same speed as the person. Similar triangles give a precise relationship among these quantities, which we can then differentiate with respect to time.

We will use a simple model: the lamp is a fixed point at the top of a vertical pole, the ground is level, and the person walks along a straight horizontal path away from the point directly below the lamp. The person’s height and the lamp’s height are constant. We assume the lamp is taller than the person, so its light ray can pass over the person and meet the ground to form a shadow.

Set Up the Variables

Let \(H\) be the fixed height of the lamp and \(h\) the fixed height of the person, with \(H>h>0\). At time \(t\), let \(x(t)\) be the horizontal distance from the lamp’s base to the person, and let \(s(t)\) be the length of the person’s shadow. Let \(y(t)\) denote the distance from the lamp’s base to the shadow tip. Since the person stands between the lamp and the tip, \(y=x+s\).

The corresponding sides must be matched carefully: the full base of the large triangle runs from the pole to the shadow tip, so its length is \(x+s\), not \(x\). The small triangle’s base is the shadow itself, of length \(s\).

Derive the Constraint

Start with the ratio of corresponding sides. Cross-multiplication gives a relationship that can be solved either for the shadow length or for the tip’s distance.

Because \(H-h>0\), division by \(H-h\) is valid. Thus the shadow length is proportional to the person’s distance from the lamp. Adding \(x\) to both sides of the expression for \(s\) gives the tip’s distance.

These equations describe the geometry at every instant. Their coefficients are constant because the lamp and person heights do not change. That makes the rates especially direct to find.

Find the Rates

Differentiate the constraint with respect to time, treating \(x\) and \(s\) as functions of \(t\). Since \(H\) and \(h\) are constants, the result relates the person’s signed walking rate to the rate at which the shadow length changes.

Solving for the shadow-length rate gives one useful formula. Differentiating \(y=x+s\), or differentiating its expression in terms of \(x\), gives the tip’s rate.

The formulas also explain why the shadow tip can move faster than the person. The factor \(H/(H-h)\) is greater than \(1\), because \(H>H-h>0\). Therefore, when the person walks away from the pole at a positive rate, the tip moves away faster. The shadow itself lengthens at the positive rate \(h/(H-h)\) times the person’s walking rate.

Why the Rate Relationship Holds

The geometry fixes the ratio \(s/x\) at \(h/(H-h)\). In other words, as the person’s distance changes, the shadow length must change in the same fixed proportion. Differentiating that proportional relationship preserves the same factor in the rates.

For a direct justification, let \(c=h/(H-h)\), which is constant under the model’s assumptions. Then \(s=cx\). If \(x\) is differentiable in time, differentiating both sides gives \(ds/dt=c\,dx/dt\). Also, \(y=x+s=(1+c)x\), so \(dy/dt=(1+c)\,dx/dt\). Since \(1+c=H/(H-h)\), these are exactly the rate formulas above. Thus the rates follow from the similar-triangle constraint and the derivative of a constant multiple.

Worked Examples

A Common Geometry Mistake

A frequent error is to use \(x\) as the base of the large triangle. That base extends all the way from the lamp’s base to the shadow tip, so it is \(x+s\). For example, with \(H=6\) meters, \(h=1.5\) meters, and \(x=10\) meters, the incorrect proportion \(H/x=h/s\) would give

This is not the shadow length for the stated geometry: the person-to-tip distance is \(s\), but the lamp-to-tip distance is \(x+s\), not \(x\). The correct proportion gives

Another error is to report the person’s walking rate as the tip’s rate. Since \(y=x+s\), the tip’s rate includes both the person’s motion and the change in shadow length: \(dy/dt=dx/dt+ds/dt\). Keep the two rates distinct.

Conditions and Interpretation

The model depends on fixed heights, level ground, and straight-line motion along the ground directly away from or toward the lamp. The person must be shorter than the lamp, so that \(H-h>0\). If \(H=h\), the derived denominator is zero and this ground-shadow configuration does not give a finite shadow length for a person at positive distance. If \(H<h\), the assumed ray geometry does not describe a shadow tip on the ground beyond the person.

Finally, rates are signed. If the person is walking away, \(dx/dt>0\), and both \(ds/dt\) and \(dy/dt\) are positive. If the person walks toward the lamp, \(dx/dt<0\), so these rates are negative. When a question asks for a speed rather than a signed rate, report the nonnegative magnitude and state the direction separately.