A 20-foot ladder is leaning against a vertical wall. If the bottom of the ladder

Question

A 20-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 5 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 12 feet from the wall? A man standing 12 feet from the base of a lamppost casts a shadow 7 feet long. If the man is 6 feet tall and walks away from the lamppost at a speed of 500 feet per minute, at what rate, in feet per minute, will his shadow lengthen?

Explanation / Answer

I try to help you with the second problem. This was a long one, so I hope it didn't take too long. 1) Let A(Foot of the lamp post), C(Foot of the man) and E(end of the shadow point) are the three points of the floor. 2) AB be the lamp of height 'h' ft and CD the height of man = 6 ft; AC = x ft and CE = y ft 3) The two triangles AEB and CED are similar; so applying the proportional parts ratio, AB/AE = CD/CE that is AB/(x+y) = 6/y [Here being given x = 12ft and y = 7ft], we have AB/19 = 6/7; ==> AB = 114/7 ft 4) Hence, 114/(2x+2y) = 6/y ==> 114y = 12x + 12y ==> 8.5y = 1x 5) Differentiating both sides with respect to time, 8.5(dy/dt) = 1(dx/dt) So dy/dt = (1/8.5)*(dx/dt) = (1/8.5)*(500) [given dx/dt = 500ft/min] So shadow changes at the rate of 58.8 ft/min. The change is in the same direction of the man walks; if he walks away from the lamp post, then the length of shadow increase; else if he walks towards the post, then the length of the shadow shortens. [This is because, dy/dt has same sign as that of dx/dt] Hope it helped! :)

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