Spotlight Beam and Shadow Speed

How a Moving Person Moves a Shadow Tip

A fixed spotlight shines across level ground while a person walks along that ground. The person’s shadow changes length, and the tip of the shadow moves too. The tip does not generally move at the same speed as the person: its speed depends on the heights of the spotlight and the person.

Similar triangles provide the geometric relationship between the person’s position and the shadow tip. Differentiating that relationship with respect to time then gives the shadow tip’s speed. Careful variable definitions matter: the shadow’s length and the shadow tip’s position are different quantities.

Set Up the Geometry

Place the spotlight at a fixed point on the ground, and measure horizontal distances from its base in the direction the beam shines. Let \(H\) be the height of the spotlight and \(h\) the person’s height, with \(H>h>0\). Let \(x\) be the person’s distance from the spotlight’s base, and let \(s\) be the position of the shadow tip, measured from that same base.

Assume the person is between the spotlight and the shadow tip, so \(0<x<s\). The length of the person’s shadow is therefore \(s-x\), not \(s\). Treat \(H\) and \(h\) as constant, while \(x\) and \(s\) may change with time.

The beam from the top of the spotlight to the shadow tip forms a large right triangle. The beam from the top of the person’s head to the shadow tip forms a smaller right triangle. These triangles are similar: both have a right angle where they meet the ground, and both share the angle made by the light beam and the ground. Their corresponding vertical and horizontal sides are \(H,s\) for the large triangle and \(h,s-x\) for the small one.

This equation relates the heights and horizontal distances. To find a speed, first rearrange it into a convenient constraint between \(x\) and \(s\).

Derive the Shadow-Tip Rate

Cross-multiplying the similar-triangle equation gives \(H(s-x)=hs\). Expanding and collecting the terms involving \(s\) gives the position relationship below.

Because \(H\) and \(h\) are fixed, differentiate both sides with respect to time. The quantities \(x\) and \(s\) depend on time, so their derivatives are their signed velocities along the ground.

Since \(H>h\), division by \(H-h\) is valid. The resulting formula gives the shadow tip’s velocity in terms of the person’s velocity.

The multiplier \(H/(H-h)\) is greater than \(1\), because \(H>H-h>0\). Thus, when the person walks away from the spotlight, the shadow tip moves away faster than the person. When the person walks toward the spotlight, both signed velocities are negative, and the tip moves toward the spotlight faster in magnitude.

Why the Rate Formula Works

The derivation follows from the similar-triangle constraint, rather than from an assumption about how fast the shadow moves. Here is the complete chain of reasoning:

This is an example of differentiating a constraint with respect to time. The geometry supplies the constraint; differentiation converts it into a relationship between instantaneous rates. The rate equation applies while the stated arrangement holds and the heights remain fixed. If the person’s height or the spotlight’s height changes, extra derivative terms would be needed.

Worked Examples

A Common Mistake: Confusing Shadow Length with Tip Position

The shadow length is \(s-x\), while the tip’s position from the spotlight is \(s\). They do not have the same rate. For the first example, the rearranged geometric constraint also gives the shadow length directly:

Using \(H=6\), \(h=2\), and \(dx/dt=1.5\), a tempting but incorrect calculation is to treat the rate of shadow-length change as the tip’s velocity:

This \(0.75\) metres per second is the rate at which the shadow length increases, not the rate at which the tip moves from the spotlight. Since the tip position is the person’s position plus the shadow length, differentiate \(s=x+(s-x)\):

This agrees with the rate formula. The correction is to identify the requested quantity precisely before differentiating or interpreting a rate.

Interpreting and Checking the Result

A related-rates answer should make sense in the geometry, in its sign, and in its units. Here, \(H/(H-h)\) is a positive dimensionless factor greater than \(1\). Therefore the tip and person move in the same direction, and the tip’s speed is greater than the person’s speed. A negative velocity means motion toward the spotlight, not a negative physical speed.

The equation also passes a dimensional check. Both \(H\) and \(H-h\) are lengths, so their ratio has no units. Multiplying that ratio by \(dx/dt\), measured in distance per time, gives \(ds/dt\) in distance per time. If a calculated rate has units of length, or if it points in the wrong direction, revisit the definitions and the sign convention.