Question 1193573
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A ladder 10 meters long is leaning against a wall. If the bottom of the ladder is being pushed
horizontally towards the wall at 2 m/s, how fast is the top of the ladder moving when the bottom 
is 6 meters from the wall?
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<pre>
Let x be horizontal distance from the wall and y be vertical coordinate.

Then from Pythagoras

    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 = -6 m,  x'(t) = 2 m/s,  y = {{{sqrt(100 - (-6)^2)}}} = {{{sqrt(100-36)}}} = {{{sqrt(64)}}} = 8.


You will get  y'(t) = - {{{((-6)*2)/8}}} = {{{12/8}}} = {{{3/2}}} m/s = 1{{{1/2}}} m/s = 1.5 m/s.    <U>ANSWER</U>
</pre>

Solved.