Question

a 13 ft ladder is leaning against a vertical wall when the foot of the ladder begins to move away from the wall at a constant rate of 0.5 ft/sec. How fast is the top of the ladder sliding down the wall when the foot of the ladder is 5ft from the wall?

Answer 1

x^2 + y^2 = 13^2=169

y^2 = 169-x^2

2 y dy/dt = 0 - 2 x dx/dt

dy/dt = - (x/y) dx/dt

right triangle is 13^2 = 12^2 + 5^2, so x/y = 5/12

dy/dt = - (5/12)(0.5 ft/sec)

But check my math. It's early here.

Answer 2

Glad to have helped.

Answer 3

thankyou thankyou thankyou :)
@douglaswinslowcooper