Question

As Collin walks away from a 264 cm lamppost, the tip of his shadow moves twice as fast as he does? What is Collin's height?

How should I do this?


The moment that a rectangle is 5 ft long and 4 ft wide, its length is increasing at 1 ft/min and its width is decreasing at 0.5 ft/min . How is the area changing at that moment? A= l*w , Da/dt = w*dw/dt +l*dl/dt =4*1(-0.5 ft/min)+ 5(1 ft/min) = 3 ft/min Da/dt = 3 ft/min is this correct?

Answer 1

Let
h = Colin's height
x = Colin's distance from post
s = distance of tip of shadow from post

(s-x)/h = s/264
264s -264x = hs
(264-h)s = 264x
s = 264/(264-h) x
ds/dt = 264/(264-h) dx/dt
since ds/dt = 2 dx/rt,
2 = 264/(264-h)
h = 132