An airplane is flying overhead at a constant elevation of 405 ft. A person is viewing the plane from a position 3040 ft from the base of a radio tower. The airplane is flying horizontally away from the person. If the plane is flying at the rate of 640 ft/s, at what rate is the distance between the person and the plane increasing when the plane passes over the radio tower? At the moment the plane passes over the tower, the distance between the person and plane is changing at a rate of ft/s

δL/δt = 634,38 ft/s

Step-by-step explanation:

A right triangle is shaped by  ( y = distance between aircraft and ground which is constant and equal to  405 f ) a person who is at ground level 3040 f away from the tower distance x = 3040 f   and the line between the aircraft and the person. Then we can use Pythagoras theorem and write

L ( distance between aircraft and person )

L²  =  x²  +  y²      or    L²  =  x²  + (405)²

Taken partial derivatives with respect to t we get:

2*L*δL/δt    =  2*x*δx/t + 0

Then      L*δL/δt = x*δx/dt

At the moment of the aircraft passing over the tower

x = 3040 ft       δx/δt = 640 ft/s   and L = √ ( 3040)² + (405)²

So   L  = √9241600  + 164025        L = √9405625   L ≈3066,9 ft

Then:

δL/δt = 3040*640/ 3066,9         units   [ ft * ft/s / ft ]   ft/s

δL/δt = 634,38 ft/s

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

i think 1: 2

step-by-step explanation: