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

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?

just to get rid of pesky decimals.

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

Thank you! :D

The equation 5.5x = 4.5y was multiplied by 10 to simplify the equation and make it easier to work with. By multiplying both sides of the equation by 10, we eliminate the decimals and simplify the coefficients. This allows us to easily solve for dy/dt, which represents the rate at which the shadow lengthens.