Complete question :
a ladder 5 meters long is leaning against a wall. the base of the ladder is sliding away from the wall at a rate of 1 meter per second. how fast is the top of the ladder sliding down the wall at the instant when the base is 4 meters from the wall?
Answer:
-4/3 m/s
Step-by-step explanation:
Using Pythagoras :
a² = x² + c²
5² = x² + y² - - - (1)
The horizontal distance x with respect to time t = 1m/sec
dx/dt = 1m/sec
To obtain the vertical height 'y' with respect to time t when x = 4
5² = x² + y²
Wen x = 4
5² = 4² + y²
25 = 16 + y²
y =√(25 - 16)
y = 3
Differentiate (1) with respect to t
x² + y² = 25 - ---- - (1)
2x * dx /dt + 2y dy/dt = 0
dx/dt = 1
x = 4.
y = 3
2(4)(1) + 2(3) * dy/dt = 0
8 + 6dy/dt = 0
6 dy/dt = - 8
dy/dt = - 8/6
dy/dt = - 4/3
