Just about every Calculus book I have ever seen uses this question as an introductory example of "rate of change" problem.
At a time of t seconds, let the foot of the ladder be x ft from the wall, and the top of the ladder by y ft above the ground>
then,
x^2 + y^2= 25^2
2x dx/dt + 2y dy/dt = 0
or
dy/dt = -2x dx/dt/2y = -x dx/dt/y
given: dx/dt = 4 ft/sec
find : dy/dt when y = 20
when y = 20
x^2 + 20^2 = 25^2
x^2 = 225
x = 15
dy/dt = -15(4)/20 ft/sec
= -3 ft/sec
(the negative tells me that y is decreasing)
A ladder 25 feet long that was leaning against a vertical wall begins to slip. Its top slides down the wall while its bottom moves along the lever ground at a constant speed of 4 ft/sec. How fast is the top of the ladder moving when it is 20 feet above the ground?
1 answer
1 answer