Let #x# be the distance of the bottom of the ladder from the building, and let #y# be the distance of the top of the ladder from the ground.
By Pythagorean Theorem,
#x^2+y^2=20^2#
Since the bottom of the ladder slides away at #3# ft/s,
#{dx}/{dt}=3#.
Since the top of the ladder is #8# ft from the ground,
#y=8#.
By plugging #y=8# in the Pythagorean theorem above,
#x^2+(8)^2=20^2 Rightarrow x=sqrt{336}=4sqrt{21}#
(Note: Only the positive #x#-value is considered since it is a distance.)
By implicitly differentiating with respect to #t#,
#d/{dt}(x^2+y^2)=d/{dt}(20^2) Rightarrow 2x{dx}/{dt}+2y{dy}/{dt}=0#
by subtracting #2x{dx}/{dt}#,
#Rightarrow 2y{dy}/{dt}=-2x{dx}/{dt}#
by dividing by #2y#,
#Rightarrow {dy}/{dt}=-x/y{dx}/{dt}#
by plugging in the given values,
#=-{4sqrt{21}}/{8}cdot3=-{3sqrt{21}}/2#
Hence, the top of the ladder is sliding down at #{3sqrt{21}}/{2}# ft/s.
