The top of a 13 foot ladder is sliding down a vertical wall at a constant rate of 4 feet per minute. When the top of the ladder is 5 feet from the ground, what is the rate of change of the distance between the bottom of the ladder and the wall

Answer 1

In just about any introductory Calculus book this type of question appears as a lead-in to the topic of "rates of change"

Make a diagram, label the height y and the the base length x
then x^2 + y^2 = 13^2
differentiate with respect to t (time)
2x dx/dt + 2y dy/dt = 0

when y = 5 , dy/dx = -4 (negative to show it is getting smaller)
and x^2 + 5^2 = 13^2
x = 12

2x dx/dt = - 2y dy/dt
dx/dt = - (y/x) dy/dt = -5/12 (-4) = 5/3 ft/min

(notice dx/dt is positive, showing the distance to be increasing)