Asked by Billy the Goat
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?
Answers
Answered by
Reiny
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
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
Answered by
Billy
Correct answer is -.169
Answered by
Reiny
you are right, I got my fraction upside down, sorry
dy/dx = -2/√140 = appr -.169
dy/dx = -2/√140 = appr -.169
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