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>

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…

To find the distance between the foot of the ladder and the wall when it and the top of the ladder are moving at the same rate, we can use the concept of similar triangles.

Let's assume that the foot of the ladder is initially at point A, the top of the ladder is at point B, and the ladder makes an angle θ with the ground. The distance between the foot of the ladder and the wall is represented by x meters.

Since the ladder is sliding away from the wall at a rate of 0.5 m/sec, the rate at which the distance x is changing can be represented as dx/dt = 0.5 m/sec.

Using the given information, we have a right-angled triangle formed by the ladder, the wall, and the ground. The perpendicular drawn from point B (the top of the ladder) to the ground forms another right-angled triangle with the ladder, the perpendicular, and the ground.

Now, let's consider the similar triangles formed by these two right-angled triangles:

Triangle ABC (formed by the ladder, the wall, and the ground):
- Side AB represents the length of the ladder, which is 6 m.
- Side BC represents the distance between the foot of the ladder and the wall, which is x m.
- Side AC represents the distance between the top of the ladder and the ground, which is 4 m.

Triangle DBE (formed by the ladder, the perpendicular, and the ground):
- Side DE represents the length of the ladder, which is 6 m.
- Side BE represents the distance between the foot of the ladder and the wall, which is x m.
- Side BD represents the distance between the top of the ladder and the ground, which is changing at a rate of 0.5 m/sec.

Since the two triangles are similar, we can set up a proportion:

AC/BC = BD/BE

Substituting the known values, we get:

4/x = 0.5/6

Cross-multiplying, we have:

0.5x = 24

Dividing both sides by 0.5, we get:

x = 48

Therefore, the foot of the ladder is 48 meters from the wall when it and the top of the ladder are moving at the same rate.

To find the distance between the foot of the ladder and the wall when it and the top are moving at the same rate, we can use the concept of similar triangles.

Let's denote the distance between the foot of the ladder and the wall when they are moving at the same rate as "x" meters.

First, let's consider the two similar triangles formed: one triangle includes the ladder, the wall, and the floor, while the other triangle includes a smaller segment of the ladder, the wall, and the floor.

The ratio of corresponding sides of similar triangles is equal. Therefore, we can set up the following proportion:

x / (6 - 4) = (6 - x) / 0.5

Here, (6 - 4) represents the distance between the foot of the ladder and the top of the ladder, and (6 - x) represents the remaining portion of the ladder extending beyond the distance x.

Now, let's solve the proportion:

2x = 1.5(6 - x)
2x = 9 - 1.5x
2x + 1.5x = 9
3.5x = 9
x = 9 / 3.5
x ≈ 2.57

Therefore, the foot of the ladder is approximately 2.57 meters away from the wall when it and the top are moving at the same rate.

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.