calculus
posted by Anonymous .
A searchlight rotates at a rate of 4 revolutions per minute. The beam hits a wall located 9 miles away and produces a dot of light that moves horizontally along the wall. How fast (in miles per hour) is this dot moving when the angle between the beam and the line through the searchlight perpendicular to the wall is 5? Note that ddt=4(2)=8

let x be the angle
let d be the distance of the dot from the point on the wall closest to the light
tan(x) = d/9
sec^2(x) dx/dt = 1/9 dd/dt
dd/dt = 1/9 sec^2(x) dx/dt
dx/dt = 4*2pi/min
I assume x = 5°, since 5 radians would point away from the wall.
dd/dt = 9 * 1.00765 * 25.13274 = 227.843 mi/min * 60min/hr = 13670.6 mi/hr
Check for sanity. If the wall were a circle 9 miles in radius, the circumference of the wall would be 9*2pi = 56.5 miles. That distance would be covered 4 times per minute, making the dot travel at a constant speed of 13572 mi/hr.
That would be the slowest speed observed when traveling along a straight wall, at the instant when the dot is closes to the light. Since 5° is a small angle, we'd expect the speed to be close to that figure, and it is, but slightly faster.