Question

the top of a ladder 6 m long is resting against a vertical wall on a level pavement, when the ladder begins to slide outwards. At the moment when the foot of the ladder is 4 m from the wall, it is sliding away from the wall @ 0.5m/sec. How far is the foot from the wall when it and the top are moving at the same rate>

Answer 1

since x^2+y^2 = 36
x dx/dt + y dy/dt = 0
since we want dy/dt = -dx/dt,
(x-y) dx/dt = 0
true when x=y, as might be expected.
Or, when x = 6/√2

All that stuff about 4m and .5 m/s is just noise, unless there are other parts to the problem.

Answer 2

I guess this is a calculus problem.

When I drew this out, I got a right triangle

x = length the foot is from the wall = 4

6 = length of ladder = hypothenuse

h = unknown height of wall

dx/dt = .5 m/sec

use pythag theorem to find h:

x^2 + h^2 = 6^2

4^2+h^2 = 6^2

h =4.47


Now find the rate of h with respect to time ( how fast the ladder is sliding down) so I guess differentiate with respect to time:

d/dt [ x^2 + h^2 ] = d/dt(36)

This is where the chain rule applies:

2x*dx/dt + 2h * dh/dt = 0 (derivative of constant is 0)

so now plug in everything:

(2*6*.5) + (2*4.47 * dh/dt) = 0

6 + 8.94*dh/dt = 0

8.94*dh/dt = -6

dh/dt = -.671 m/s (notice that it's negative, because its sliding down!)

I'm not sure what to do problem here, I might be completely wrong here…