Answer :
The angle between the ladder and the ground changes when the bottom of the ladder is 8 ft from the wall is -1/12 radians per second.
Denoting the distance in feet between the wall and the base of the ladder by x and the angle in radians between the ladder and the ground by y, it is noted;
cos(y) = x/10
which implies;
y = arccos(x/10)
Denoting time in seconds by t, it is further noted that;
[tex]\frac{dx}{dy} = \frac{dy}{dx} \frac{dx}{dt}[/tex] (chain rule)
Noting (using a standard table of derivatives for convenience)
[tex]\frac{dy}{dx}= -\frac{1}{\sqrt{1-(0.1x)^2} }(0.1)[/tex] (also by chain rule)
The above equation changes as;
[tex]\frac{dy}{dx}= -\frac{0.1}{\sqrt{1-0.1x^2} }[/tex]
It is noted from the question that in this particular system;
[tex]\frac{dx}{dt} = 0.5[/tex] feet per second
So (denoting the derivative as a function of x)
[tex]\frac{dy}{dt}(x)=\frac{dy}{dx} \frac{dx}{dt} = - \frac{0.05}{\sqrt{1-0.01x^2} }[/tex]
At last;
[tex]\frac{dy}{dt}(8)=\frac{dy}{dx} \frac{dx}{dt} = - \frac{0.05}{\sqrt{1-0.01(64)} }[/tex]
= [tex]\frac{0.05}{\sqrt{1-0.64} }[/tex]
= [tex]\frac{-0.05}{\sqrt{0.36} }[/tex]
= -5/60
= -1/12 radians per second
The angle between the ladder and the ground changes when the bottom of the ladder is 8 ft from the wall is -1/12 radians per second.
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