Calc.
posted by Preston .
A ladder 25ft long is leaning against the wall of a house. The ladder is pulled away from the wall at a rate of 2 ft per second. How fast is the top of the ladder moving down the wall when its base is 7ft from the wall?

This is the classic problem that almost every text in Calculus uses to introduce "rate of change"
Make a diagram
Let the ladder by y ft above the ground, and
x ft away from the wall
so you have a right angled triangle
x^2 + y^2 = 25
2x dx/dt + 2y dy/dt = 0 (#1)
given : dx/dt = 2 ft/s
find : dy/dt when x = 7
when x = 7
49 + y^2 = 625
y = √576 = 24
back in #1
2(7)(2) + 2(24)dy/dt = 0
dy/dt = 28/48 = 7/12 ft/s
the negative indicates that the value of y is decreasing, or in other words, the ladder is dropping along the wall