Posted by hey on Monday, August 10, 2015 at 11:28pm.

Question

A man 5.5 ft tall walks away from a lamp post 10 ft high at the rate of 8 ft/s. (a.) How fast does his shadow lengthen?
(b.) how fast does the tip of the shadow move?

Calculus - Reiny, Monday, August 10, 2015 at 11:43pm

make a sketch
let the distance of the man from the lamp-post be x
let the length of his shadow be y

by ratios:
5.5/y = 10/(x+y)
5.5x + 5.5y = 10y
5.5x = 4.5y
times 10
55x = 45y
11x = 9y
11 dx/dt = 9 dy/dt
11(8) = 9 dy/dt
dy/dt = 88/9 ft/s or 9 7/9 ft/s --> rate at which the shadow is lengthening.

How fast is the shadow moving??
d(x+y)/dt = dx/dt + dy/dt
= 8 + 88/9 = 160/9 ft/s or 17 7/9 ft/s

Why was 5.5x = 4.5y multiplied to 10?

Answer 1

just to get rid of pesky decimals.

Fractions are exact; decimals can often be only an approximation, as in 2/3 = 0.666...

Answer 2

Thank you! :D