Question 1193538
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a 13 foot ladder leans against a wall. The foot of a ladder begins to slide away from the wall 
at the rate of 1 foot per minute. When the foot is 5ft from the wall, at what rate is the top 
of the ladder is falling?
a.5/12 ft/min.
b.4/3 ft/min.
c.3/4 ft/min.
d.12/5 ft/min.
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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 = 13^2.    (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 = 5 ft,  x'(t) = 1 ft/minute,  y = {{{sqrt(13^2 - 5^2)}}} = {{{sqrt(169-25)}}} = {{{sqrt(144)}}} = 12.


You will get  y'(t) = - {{{(5*1)/12}}} = - 5/12 ft/minute.    


<U>ANSWER</U>.  The top of the ladder moves vertically down at the rate of 5/12 ft per minute.
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Solved.