Posted by **yayah** on Saturday, December 15, 2012 at 11:12am.

A ladder 25 m long leans against a vertical all. If the lower end is being moved away from the wall at the rate 6m/sec, how fast is the height of the top decreasing when the lower end is 7m from the wall?

- calculus -
**Steve**, Saturday, December 15, 2012 at 11:18am
at the moment in question, the ladder top is 24m up the wall.

x^2+y^2 = 25^2

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

2(7)(6) + 2(24) dy/dt = 0

dy/dt = -4/7 m/s

- calculus -
**Elsi**, Saturday, December 15, 2012 at 1:45pm
start by drawing a diagram. you should notice that you have a triangle with the ladder as the hypoteneuse. you are given that the lower end is moving away from the wall at 6m/sec, so your dx/dt =6.

To find y when x=7:

x^2 +y^2 = 25^2

7^2 +y^2 = 625

49 + y^2 = 625

y^2 = 576

y = 24

Then, since we know that x is changing:

x^2 + y^2 = 25^2

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

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

plug in values that you know:

7(6) + 24(dy/dt) = 0

24(dy/dt) = -42

dy/dt = -1.75 m/sec

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