4.6 worksheet 2

Math 1300 4.6 Rates and Related Rates (part 2) Name: Solutions5. An FBI agent with a powerful spyglass is located in a boat anchored 0.4 km offshore. Agangster under surveillance is walking along the shore. Assuming the shoreline is straightand that the gangster is walking at the rate of 2 km/hr, how fast must the FBI agent rotatethe spyglass to track the gangster when the gangster is 1 km from the point on the shorenearest to the boat? (In other words, find dθ/dt.)xgangster →0.4 kmθFBItan θ = x0.4sec 2 θ dθdt = 1 dx0.4 dt = 2.5dx dtdxdt = 2 km/hrWhen x = 1, we have tan θ = 1/0.4 and θ = arctan(1/0.4) = arctan(2.5)Useful trig identity: 1 + tan 2 ∗ = sec 2 ∗ for any value of ∗.dθdt=2.5dxdtsec 2 θ ==2.5(2 km/hr)sec 2 (arctan(2.5))51 + tan 2 (arctan(2.5)) = 51 + (2.5) 2 = 57.25 = 2029 radians/hr.