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A ladder 10ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 2ft/sec , how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6ft from the wall?
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away from the wall at a speed of 2ft/sec=dx/dt x^2 +h^2 =100 (2x )dx/dt + 2h dh/dt= 0 2h dh/dt=-2x dx/dt dh/dt= -x/h dx/dt dh/dt= -6/8 * 2 dh/dt= -3/2
I follow you for the first 3 steps, but my prof shows: 2x+2y =0 2(6)(2) + 2(8) = 0 16 dy/dt = -24 -1.5 ft sec
-3/2=-1.5
Im so confused as to where he got the 2 & 8
in my second step
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