The plane flying horizontally at an altitude of 2 km at a speed of 800 km/h passes over a radar station. After a time t, the distance between the plane and the radar station is `D = sqrt(2^2 + (800*t)^2)` = `sqrt(4 + 640000*t^2)`
The rate at which D is changing with time is `(dD)/(dt) = (1/2)*1280000*t*(1/sqrt(4 + 640000*t^2))`
=> `640000*t/sqrt(4 + 640000*t^2)`
If D = 3, `sqrt(4 + 640000*t^2) = 3`
=> `4 + 640000*t^2 = 9`
=> `640000*t^2 = 5`
=> `t = sqrt 5/800`
At `t = sqrt 5/800` , `(dD)/(dt) = (640000*(sqrt 5/800))/(sqrt(4 + (640000*5/640000)))`
=> `(dD)/(dt) = (800*(sqrt 5))/(sqrt9)`
=> `800*sqrt 5/3`
The rate at which the distance of the plane from the radar station is increasing is `(800*sqrt5)/3`
See eNotes Ad-Free
Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.
Already a member? Log in here.