Answer:
[tex]-\frac{9}{4}ft/s[/tex]
Explanation:
We are given that
Length of ladder,z=10ft
Let x be the distance of bottom of ladder from the wall and y be the distance of the top of ladder from the bottom of wall.
[tex]\frac{dx}{dt}=3ft/s[/tex]
We have to find the rate at which the top of the ladder sliding down the wall when the bottom of the ladder is 6ft from the wall.
i.e.x=6 ft
By Pythagoras theorem
[tex]x^2+y^2=z^2[/tex]
[tex]x^2+y^2=(10)^2=100[/tex]
[tex]6^2+y^2=(10)^2=100[/tex]
[tex]y^2=100-6^2=64[/tex]
[tex]y=\sqrt{64}=8 ft[/tex]
Differentiate w.r.t time
[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt}=0[/tex]
[tex]6\times 3+8\frac{dy}{dt}=0[/tex]
[tex]8\frac{dy}{dt}=-6\times 3=-18[/tex]
[tex]\frac{dy}{dt}=-\frac{18}{8}=-\frac{9}{4}ft/s[/tex]
