Question

a woman 2 meter tall walk at a rate of 6m per minute away from a source of light that is 5 m above the ground. how fast is the length of her shadow increasing when she is 3m away from base of the light

Answer 1

                 

                     B

                     |\

                     |      \          

                     |           \D

                     |           |    \

                     |_____|____\

                    A           C          E

AB - the lamp post

suppose at any time t, the man CD at a distance of x m. from the lamp post & y  m. is the length of his shadow CE.

since the triangles ABE , CDE are similar

AB/CD = AE/CE

=> 5/2 = (x+y) / y

=> x/y = 3/2

=> y = (2/3) x

now differentiating both sides w.r.to 't'-

=> dy/dt = 2/3 (dx / dt)

=> dy/dt = (2/3) * 6             (as dx/dt = 6, given)

=> dy/dt = 4

so, the length of her shadow is increasing at a rate of 4m/min, when she is 3m away from base of the light

I think this answer may help you.