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The Lighthouse Problem
Louis A. Talman, Ph.D.
Department of Mathematical & Computer Sciences
Metropolitan State University of Denver
March 18, 2015
Problem: The axis of a light in a lighthouse is tilted. When the light, which is mounted at a right
angle to the axis of the lighthouse and rotates about that axis, points east, its beam is inclined upward
at 8◦ . When it points north, the beam is inclined upward at 5◦ . What is its maximum angle of elevation?
Solution: The rotating light beam sweeps out a plane, whose normal is the axis of the lighthouse.
We take center of the light’s rotation to be the origin of a three-dimensional coordinate system
whose x- and y-axes are both horizontal, and whose positive directions point, respectively, east
and north. The positive direction of the z-axis is directly upward. Then the equation of the light
beam’s plane has the form z = f (x, y) = Ax + By, where A and B are constants.
The light beam is inclined upward at eight degrees when it points eastward, so the trace of the light
beam’s plane on the xz-plane has slope tan 8◦ . But the slope of the tangent to a curve z = f (x, y)
at a point (x0 , y0 ) in the direction of a unit vector u is the directional derivative Du f (x0 , y0 ) =
∇f (x0 , y0 )·u In our case, ∇f = hA, Bi and u = i, so that tan 8◦ = Di f (0, 0) = hA, Bi·h1, 0i = A. An
altogether similar calculation leads us to the conclusion that tan 5◦ = Dj f (0, 0) = hA, Bi·h0, 1i = B.
Hence
∇z = htan 8◦ , tan 5◦ i.
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This gradient vector is a plane vector that points in the direction of most rapid increase
of the plane of the light beam, so slope (i.e. the directional derivative) in this direction
is maximal.
We put u = ∇z/k∇zk and we find this directional derivative as
Du z(0, 0) = ∇z · u = ∇z ·
∇z · ∇z
k∇zk2
∇z
=
=
= k∇zk.
k∇zk
k∇zk
k∇zk
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√
The maximal slope is therefore tan2 8◦ + tan2 5◦ , and this slope is the tangent of the maximum
angle of elevation. The solution is therefore
q
2
2
◦
◦
arctan
(tan 8 ) + (tan 5 ) ∼ 9.399923432◦ .
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