Answer:
[tex]\frac{16}{5}[/tex]ft/s
Explanation:
We are given that
Length of ladder,z=17 ft
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{dy}{dt}=-6ft/s[/tex]
We have to find the rate at which the base of the ladder sliding away from the wall when x=15 ft
[tex]x^2+y^2=z^2[/tex]
[tex]x^2+y^2=(17)^2=289[/tex]
[tex](15)^2+y^2=289[/tex]
[tex]y^2=289-(15)^2=289-225=64[/tex]
[tex]y=\sqrt{64}=8[/tex]ft
Differentiate w.r. t time
[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt}=0[/tex]
[tex]15\frac{dx}{dt}-8\times 6=0[/tex]
[tex]15\frac{dx}{dt}=48[/tex]
[tex]\frac{dx}{dt}=\frac{48}{15}=\frac{16}{5}[/tex]ft/s
