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

A 10 foot ladder leans against a 20 foot wall. Someone begins pushing the base of the ladder toward the wall at the rate of one foot per second. How quickly is the top of the ladder moving up the wall after 2 seconds?

- Calculus -
**bobpursley**, Thursday, May 6, 2010 at 10:08pm
Let 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 10:18pm
This 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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