calculus
posted by yayah .
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?

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 
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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