A drone is descending vertically at a constant speed of 5 m/s while its camera films a square structure on the ground. The side length of the structure is 100 m and the focal length of the camera lens is 10 cm . At the instant the drone is at a height of 1 km above the ground, the rate at which the side length of the structure's image on the camera sensor is increasing is :

Step-by-step solution

Let the side length of the structure be x , the focal length be f , and the instantaneous height of the drone be H . For a distant object, the image distance v f . The linear magnification m is given by: m = y x = f H where y is the side length of the image on the sensor. Thus, y = x f H Differentiating both sides with respect to time t : dy dt = x f ( - 1 H^2 ) dH dt Given: x = 100 m f = 10 cm = 0.1 m H = 1 km = 1000 m The drone is descending, so the rate of change of height is dH dt = -5 m/s . Substituting the va