Question 1194771
.
The top of a ladder 10 meters long rests on a vertical wall of a residential building 
while the bottom rests on a horizontal ground. If the top slides down at the rate 
of 20 meters per minute, how fast is the lower end moves along the ground 
when the lower end is 8 meters from the wall?
~~~~~~~~~~~~~~~~~~~


<pre>
Let x = x(t) be horizontal distance from the wall of one lsadder endpontt 
and y = y(t) be vertical coordinate of the other endpoint.


Then from Pythagoras, the length of the ladder is

    x^2 + y^2 = 10^2.    (1)


Here x = x(t) and y = y(t) are functions of time, t.


Differentiate equation  (1)  over t.  You will get

    2x*x'(t) + 2y*y'(t) = 0,

hence

    y'(t) = - (x*x'(t))/y.


Evaluate it at the given values  x = 8 m,  x'(t) = 20 m/minute,  y = {{{sqrt(100 - 8^2)}}} = {{{sqrt(100-64)}}} = {{{sqrt(36)}}} = 6.


You will get  y'(t) = - {{{(8*20)/6}}} = {{{160/6}}} = {{{80/3}}} = 26 {{{2/3}}} meters per minute.     <U>ANSWER</U>
</pre>

Solved.