SOLUTION: 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

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Question 1193538: 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.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.
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.
~~~~~~~~~~~~~~~ 

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%2813%5E2+-+5%5E2%29 = sqrt%28169-25%29 = sqrt%28144%29 = 12.


You will get  y'(t) = - %285%2A1%29%2F12 = - 5/12 ft/minute.    


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

Solved.



Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The ladder, the ground, and the wall form a right triangle with hypotenuse 13. The side lengths of the triangle and the hypotenuse are related by the equation

x%5E2%2By%5E2=169

Differentiate with respect to time:

2x%28dx%2Fdt%29%2B2y%28dy%2Fdt%29=0

At the specified time, when x (the distance from the foot of the ladder to the wall) is 5 ft, the vertical distance from the top of the ladder to the ground is 12 ft. Use those values of x and y and the given value of dx/dt to find dy/dt.

2%285%29%281%29%2B2%2812%29%28dy%2Fdt%29=0
10%2B24%28dy%2Fdt%29=0
24%28dy%2Fdt%29=-10
dy%2Fdt=-10%2F24=-5%2F12

Since the foot of the ladder is sliding away from the foot of the wall (x is increasing), the top of the ladder is obviously moving down the wall, so dy/dt is negative. But since the answer choices are all positive, we can assume that we can ignore the sign of dy/dt.

ANSWER: 5/12 ft/min