Posted by Preston on Monday, November 1, 2010 at 5:17pm.
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
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