A 12-m long ladder leans against a vertical building. If the top of the ladder is sliding down the building at .25m/s, how fast is the bottom of the ladder sliding away from the building when the top of the ladder is 6m above the ground?

Question

anonymous · Answer

looks like a related rate pythagoras problem

anonymous · Answer

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anonymous · Answer

$$x^2+y^2=12^2$$
$$2xx'+2yy'=0$$
$$yy'=-xx'$$ and you are told $x'=.25$ so plug the number in, solve for $y'$

majikdusty · Answer

Thank you very much! For some reason i was using .25 ad dy/dt instead of dx/dt! I appreciate the help! the answer i got was 1/(4sqrt(3))

anonymous · Answer

oh i made a tiny mistake. it should be $x'=-.25$ because the ladder is sliding down the wall, and so it is decreasing, making the rate of change negative