A 165 inch rod is leaning up against a wall, but is sliding down the wall so that the vertical distance between the top of the rod and the floor is decreasing at a rate of 2 inches per second. How fast the is the horizontal distance between the bottom of the rod and the base of the wall changing when the vertical distance is 153 inches

the horizontal distance is increasing at a rate of 4.953 in/s

Step-by-step explanation:

since the rod is leaning up against the wall and its length L is constant , then the relationship between y ( vertical distance ) , x (horizontal distance ) and L is

x² + y² = L²

then using derivatives with respect to the time t in both sides of the equation

2*x*dx/dt + 2*y*dy/dt = 0

dx/dt = y/x*dy/dt

since we know that x² + y² = L² → x= √(L² -y²) , then

dx/dt = -y/√(L² -y²)*dy/dt

replacing values

dx/dt = -153 in /√[(165 in)² -(153 in)²]* (-2 in/s) = 4.953 in/s

therefore the horizontal distance is increasing at a rate of 4.953 in/s

£56

step-by-step explanation:

i believe the correct answer is b.