Answer:
[tex]-0.224[/tex] ft/sec.
Step-by-step explanation:
We have been given that a 18-foot ladder is leaning against a building. The bottom of the ladder is sliding along the pavement directly away from the building at 2 ft/sec. We are asked to find find how fast is the top of the ladder moving down when the foot of the ladder is 2 feet from the wall.
First of all, we will draw a graph to represent the given information as shown in the attachment.
Now, we will use Pythagoras theorem as:
[tex]x^2+y^2=18^2[/tex]
Now, we will find derivative with respect to time.
[tex]2x\cdot \frac{dx}{dt}+2y\cdot \frac{dy}{dt}=0[/tex]
Let us find value of x, when y is 2 using Pythagoras theorem.
[tex]x^2+2^2=18^2[/tex]
[tex]x^2+4-4=324-4[/tex]
[tex]x^2=320[/tex]
[tex]x=8\sqrt{5}[/tex]
Upon substituting our given values in derivative function, we will get:
[tex]2(8\sqrt{5})\cdot \frac{dx}{dt}+2(2)\cdot2=0[/tex]
[tex]16\sqrt{5}\cdot \frac{dx}{dt}+8=0[/tex]
[tex]\frac{dx}{dt}=-\frac{8}{16\sqrt{5}}[/tex]
[tex]\frac{dx}{dt}=-\frac{1}{2\sqrt{5}}[/tex]
[tex]\frac{dx}{dt}=-\frac{1\cdot \sqrt{5}}{2\sqrt{5}\cdot \sqrt{5}}[/tex]
[tex]\frac{dx}{dt}=-\frac{\sqrt{5}}{10}[/tex]
[tex]\frac{dx}{dt}=-0.22360[/tex]
[tex]\frac{dx}{dt}\approx -0.224[/tex]
Therefore, the top of the ladder is moving down at a rate of [tex]-0.224[/tex] feet per second.
