Question

A 12 foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 3 feet from the wall

Answer 1

Using Pythagoras theorem, the top of the ladder moving down when the foot of the ladder is 3 feet from the wall is of -0.518 feet/sec.                      

Let distance from the wall to the foot of the ladder is 'x' feet and the height of the top of the ladder is 'y' feet.

Pythagoras theorem, $x^{2} + y^{2} = (12)^{2}$       --->(1)

Given,$\frac{dx}{dt}= 2feet/second$   at x=3

Put x=3 in Pythagoras theorem equation (1)

$(3)^{2} + y^{2} = 144$

         $y^{2} = 144 - 9$

        $y^{2}$  =  135

        y = 11.61

Derive equation (1) w.r.t to 't'

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$                ---->(2)

substitute the value of 'x', 'dx/dt' and 'y' in equation (2), we get the fast of the top of the ladder moving down when the foot of the ladder is 3 feet from the wall

$2(3)(2) + 2 (11.61)\frac{dy}{dt} = 0$

12 + 23.22 $\frac{dy}{dt}$  = 0

                  $\frac{dy}{dt}= \frac{-12}{23.22}$

                  $\frac{dy}{dt} = -0.518$

Hence,  using Pythagoras theorem the top of the ladder moving down when the foot of the ladder is 3 feet from the wall is of -0.518 feet/sec.  

Learn more about Pythagoras theorem here

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