a 12 foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 1 feet/second, how fast is the top of the ladder moving down when the foor of the ladder is 2 feet from the wall?

Answer 1

This is the classic example used by most textbooks in the introduction to Related Rates in Calculus.

let the foot of the ladder be x ft from the wall, let the ladder reach y ft above the ground

x^2 + y^2 = 12^2
2x dx/dt + 2y dy/dt = 0
or
x dx/dt + y dy/dt = 0

given: dx/dt = 1 ft/s
find: dy/dt when x = 2

when x = 2
4 + y^2 = 144
y^2 = 140
y = √140

2(1) + √140 dy/dt = 0
dy/dt = -√140/2 ft/s
= ....

notice the dy/dt is negative, indicating that y is decreasing

Answer 2

Correct answer is -.169

Answer 3

you are right, I got my fraction upside down, sorry

dy/dx = -2/√140 = appr -.169