One end of a ladder 60ft. long is leaning against a perpendicular wall standing on a horizontal plane. Supposing the foot of the ladder to be pulled away from the wall at the rate of 3 ft. per minute;

Question

(a) how fast is the top of the ladder descending when the foot is 14 ft. from the wall? (b) when will the top and bottom of the ladder move at the same rate? (c) when is the top of the ladder descending at the rate of 4ft. per minute? 

Answer 1

x: horizontal distance from the base of the wall to the base of the ladder

y: vertical distance from the base of the wall to the top of the ladder

By pythag thm, x² + y² = 60² ; dx/dt = 3 ft/min

2x·dx/dt + 2y·dy/dt = 0. or dy/dt = - x/y·dx/dt

a) when x = 14, y ~ 58.343 and dy/dt ~ - .72 ft/min

b) From either related rates equation above, it is clear that x and y will be changing at equal (and opposite) rates at the instant that x = y (in other words when the ladder forms a right isosceles triangle with the wall and ground). This will happen when x = y = 30√2, which happens when t = 10√2 min.

c) - 4 = - x/y(3) so y = 3/4·x x² + 9/16·x² = 60² ; x = 48 ft so t = 16 min