Answer:
The ladder is moving at the rate of 0.65 ft/s
Explanation:
A 16-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second. We need to find the rate at which the top of the ladder moving down when the foot of the ladder is 5 feet from the wall.
The attached figure shows whole description such that,
[tex]x^2+y^2=256[/tex].........(1)
[tex]\dfrac{dx}{dt}=2\ ft/s[/tex]
We need to find, [tex]\dfrac{dy}{dt}[/tex] at x = 5 ft
Differentiating equation (1) wrt t as :
[tex]2x.\dfrac{dx}{dt}+2y.\dfrac{dy}{dt}=0[/tex]
[tex]2x+y\dfrac{dy}{dt}=0[/tex]
[tex]\dfrac{dy}{dt}=-\dfrac{2x}{y}[/tex]
Since, [tex]y=\sqrt{256-x^2}[/tex]
[tex]\dfrac{dy}{dt}=-\dfrac{2x}{\sqrt{256-x^2}}[/tex]
At x = 5 ft,
[tex]\dfrac{dy}{dt}=-\dfrac{2\times 5}{\sqrt{256-5^2}}[/tex]
[tex]\dfrac{dy}{dt}=0.65[/tex]
So, the ladder is moving down at the rate of 0.65 ft/s. Hence, this is the required solution.
