Answer:
6.87 ft/s is the rate at which the top of ladder slides down.
Explanation:
Given:
Length of the ladder is, [tex]L=20\ ft[/tex]
Let the top of ladder be at height of 'h' and the bottom of the ladder be at a distance of 'b' from the wall.
Now, from triangle ABC,
AB² + BC² = AC²
[tex]h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1[/tex]
Differentiating the above equation with respect to time, 't'. This gives,
[tex]\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2[/tex]
In the above equation the term [tex]\frac{dh}{dt}[/tex] is the rate at which top of ladder slides down and [tex]\frac{db}{dt}[/tex] is the rate at which bottom of ladder slides away.
Now, as per question, [tex]h=8\ ft, \frac{db}{dt}=3\ ft/s[/tex]
Plug in [tex]h=8[/tex] in equation (1) and solve for [tex]b[/tex]. This gives,
[tex]8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft[/tex]
Now, plug in all the given values in equation (2) and solve for [tex]\frac{dh}{dt}[/tex]
[tex]8\times \frac{dh}{dt}+18.33\times 3=0\\8\times \frac{dh}{dt}+54.99=0\\8\times \frac{dh}{dt}=-54.99\\ \frac{dh}{dt}=-\frac{54.99}{8}=-6.87\ ft/s[/tex]
Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.
