Searchlight—wide beam A revolving searchlight, which is 100 m from the nearest point on a straight highway, casts a horizontal beam along a highway (see figure). The beam leaves the spotlight at an an?gle of ? ? /16 rad and revolves at a r?ate o? /6 rad? /s. Le? be the width of the beam as it sweeps along the highw ? ay and ? be the angle that the center of the beam makes with the perpendicular to the highway. What is the rate of c? hange o?? ?when ?? = ?? /3? Neglect the height of the lighthouse.

Solution 46E Step 1 To find What is the rate of change of wh en /3 Step 2 Given that A revolving searchlight, which is 100 m from the nearest point on a straight highway, casts a horizontal beam along a highway (see figure). The beam leaves the spotlight at an angle o /16 rad and revo lves at a rat f /6 rad/s. L be the width of the beam as it sweeps along the highway and be the angle that the center of the beam makes with the perpendicular to the highway. Step3 Step4 Here s is the spotlight, the heavy horizontal line is the highway, the line sc marks the centre of the spotlight’s beam, and the lines sa and sd mark the edges of the beam. We have angle bsd=/16, so bsc=csd=/32. The angle is the angle that the centre of the beam makes with the perpendicular sa from s to the highway, so =asc. Finally, w is the width of the beam measured along the highway, so w=|bd|. Step 5 ( - (/32)) radians with the perpendicular and the 'leading' edge of the beam will make an angle of( - (/32)) radians with the perpendicular. So the width w of the beam on the highway will be w = 100*tan( + (/32))- 100*tan( - (/32)). Now differentiate w with respect to time t , we get, dw/dt = 100*sec^2( + (/32))*d/dt - 100*sec^2( - (/32))*d/dt. Now with = /3 and d/dt = /6 rad/s we have dw/dt = 100*(/6)*[sec^2((/3) + (/32)) - sec^2((/3) - (/32))] = 100*(0.00696-0.01014) = 100*(-0.00318) = -0.318 dw/dt-0.318 m/s Therefore, 0.318 m/s is the rate of change of w