Answer:
Step-by-step explanation:
The question has typographical errors. The correct question is:
"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 3 feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the wall?
Solution:
The ladder forms a right angle triangle with the ground. The length of the ladder represents the hypotenuse.
Let x represent the distance from the top of the ladder to the ground(opposite side)
Let y represent the distance from the foot of the ladder to the base of the wall(adjacent side)
The bottom of the ladder is sliding along the pavement directly away from the building at 3ft/sec. This means that y is increasing at the rate of 3ft/sec. Therefore,
dy/dt = 3 ft/s
The rate at which x is reducing would be
dx/dt
Applying Pythagoras theorem which is expressed as
Hypotenuse² = opposite side² + adjacent side², it becomes
x² + y² = 12²- - - - - - - -1
Differentiating with respect to time, it becomes
2xdx/dt + 2ydy/dt = 0
2xdx/dt = - 2ydy/dt
Dividing through by 2x, it becomes
dx/dt = - y/x ×dy/dt- - - - - - - - - - 2
Substituting y = 5 into equation 1, it becomes
x² + 5 = 144
x² = 144 - 25 = 119
x = √119 = 10.91
Substituting x = 10.91, dy/dt = 3 and y = 5 into equation 2, it becomes
dx/dt = - 5/10.91 × 3
dx/dt = - 1.37 ft/s
x r (radius)