Question

Related rates problem (question in description)

Answer 1

let AB represents lamppost where point B is on the ground . Let the person is at C, y ft from the base of the pole and let DC = x is the length of the shadow of person. Draw CE perpendicular from C to AD. CE = height of person.
Draw the figure.

so

AB = 15

BC = y and

CD = x

CE = 6 ft

dy/dt = 5 ft/s

x + y = tip of shadow and x = length of shadow

from similar triangles ABD and CDE

AB/CE = BD/CD

15/6 = (x+y)/x

15x = 6(x+y)

x+y = (15/6)x = (5/2)x

y = 5/2 x - x = 3x/2

differentiating with respect to time

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

but dy/dt = 5

5 = 3/2 dx/dt

dx/dt = 10/3

so length of shadow is increasing at the rate of 10/3 ft /s

now rate of change of tip of shadow = dx/dt + dy/dt

= 10/3 + 5

= 25/3

so tip of shadow is changing at the rate of 25/3 ft/s

Answer 2

Thank you so much! I always have had difficulty with similar triangles.