Answer :
Answer:
The rate of change of the area when the bottom of the ladder (denoted by [tex]b[/tex]) is at 36 ft. from the wall is the following:
[tex]\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s[/tex]
Explanation:
The Area of the triangle is given by [tex]A=h\times b[/tex] where [tex]h=\sqrt{l^2-b^2}[/tex] (by using the Pythagoras' Theorem) and [tex]b[/tex] is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.
The area is then
[tex]A=\sqrt{l^2-b^2}b[/tex]
The rate of change of the area is given by its time derivative
[tex]\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)[/tex]
[tex]\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}[/tex]
[tex]\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}[/tex] Product rule
[tex]\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}[/tex] Chain rule
[tex]\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}[/tex]
[tex]\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)[/tex]
In here we can identify [tex]b=36\, ft[/tex], [tex]l=39[/tex] and [tex]\frac{db}{dt}=8\,ft/s[/tex].
The result is then
[tex]\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s[/tex]