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October 20, 2014

October 20, 2014

Posted by **Julia** on Thursday, May 6, 2010 at 9:53pm.

- Calculus -
**bobpursley**, Thursday, May 6, 2010 at 10:08pmLet y be the wall height

let x be the horizontal distance

x^2+y^2=100

2x dx/dt+2y dy/dt=0

dy/dt=-x dx/dt

so calculate the x,y positions after 2 seconds, I have no idea the starting position.

- Calculus -
**Reiny**, Thursday, May 6, 2010 at 10:18pmThis problem is the most popular lead-in question to "rate of change" questions for almost any Calculus text I have seen

Let the ladder be x ft from the wall, and y m up the wall

we know x^2 + y^2 = 100

2x(dx/dt) + 2y(dy/dt) = 0

given: dx/dt = 1 ft/s

find dy/dt when t = 2 or

in other words, when x = 2 ft since we know in 2 s it moved 2 ft.

also when x = 2

2^2 + y^2 = 100

y = √96

2(2)(1) + 2√96(dy/dt) = 0

dy/dt = -4/(2√96

= -.204 ft/s

at that moment the top of the ladder is dropping (note the - sign) at .204 ft/s

Just realized I read the question as the foot of the ladder moving away from the wall, so just change

dx/dt to -1,

the result will change to dy/dt = +.204

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