A man walks along a straight path at a speed of 6 ft/s. A searchlight is located on the ground 24 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 10 ft from the point on the path closest to the searchlight? Solution We draw the figure below and let x be the distance from the man to the point on the path closest to the searchlight. We let 0 be the angle between the beam of the searchlight and the perpendicular to the path. 24

Trigonometry (MindTap Course List)
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Chapter3: Radian Measure
Section3.5: Velocities
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Differentiating each side with respect to t, we get
dx
d0
dt
dt
so
do
dx
dt
dt
(6)
24
and
26
When x = 10, the length of the beam is 26, so cos(0)
do
24 \2
dt
26
The searchlight is rotating at a rate of
rad/s.
II
II
II
II
Transcribed Image Text:Differentiating each side with respect to t, we get dx d0 dt dt so do dx dt dt (6) 24 and 26 When x = 10, the length of the beam is 26, so cos(0) do 24 \2 dt 26 The searchlight is rotating at a rate of rad/s. II II II II
A man walks along a straight path at a speed of 6 ft/s. A searchlight is located on the ground 24 ft from the path and is kept focused on the man. At what rate is the
searchlight rotating when the man is 10 ft from the point on the path closest to the searchlight?
Solution
We draw the figure below and let x be the distance from the man to the point on the path closest to the searchlight. We let 0 be the angle between the beam of the searchlight
and the perpendicular to the path.
24
dx
We are given that
dt
when x = 10. The equation that relates x and 0 can be written from the figure.
dt
6 ft/s and are asked to find
tan(0)
X =
tan(0).
Transcribed Image Text:A man walks along a straight path at a speed of 6 ft/s. A searchlight is located on the ground 24 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 10 ft from the point on the path closest to the searchlight? Solution We draw the figure below and let x be the distance from the man to the point on the path closest to the searchlight. We let 0 be the angle between the beam of the searchlight and the perpendicular to the path. 24 dx We are given that dt when x = 10. The equation that relates x and 0 can be written from the figure. dt 6 ft/s and are asked to find tan(0) X = tan(0).
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