A 15 ft ladder is leaning against a wall.If the top of the ladder slips down the wall at a rate of 5ft/s, how fast will the foot be moving away from the wall when the top is 9ft above the ground.

3.75

To find how fast the foot of the ladder is moving away from the wall, we need to use related rates. Let's assign variables to the given values:

Let x represent the distance from the foot of the ladder to the base of the wall.
Let y represent the distance from the top of the ladder to the base of the wall.

We are given that dy/dt = -5 ft/s since the top of the ladder is slipping down the wall. We need to find dx/dt.

From the given information, we have the Pythagorean theorem: x^2 + y^2 = 15^2

Taking the derivative of both sides with respect to time:
2x(dx/dt) + 2y(dy/dt) = 0

We are given that y = 9 ft when the top of the ladder is 9 ft above the ground, so we can substitute that in:
2x(dx/dt) + 2(9)(-5) = 0

Simplifying, we have:
2x(dx/dt) = 90

Dividing both sides by 2x:
dx/dt = 90 / 2x

Since we want to find how fast the foot of the ladder is moving away from the wall when the top is 9 ft above the ground, we can substitute x = 12 ft (15 ft - 9 ft) into the equation:
dx/dt = 90 / (2(12))
dx/dt = 90 / 24
dx/dt = 3.75 ft/s

Therefore, the foot of the ladder is moving away from the wall at a rate of 3.75 ft/s when the top is 9 ft above the ground.

To determine how fast the foot of the ladder is moving away from the wall, we can use related rates.

Let's assign variables to the given information:
- The heigh of the ladder on the wall is h (in this case, h = 9ft).
- The distance between the foot of the ladder and the wall is x.
- The rate at which the top of the ladder is moving down the wall is dx/dt = -5 ft/s (negative because it is moving downward).

We want to find dx/dt when h = 9 ft.

We can relate the variables x and h using the Pythagorean theorem.

x^2 + h^2 = 15^2

Differentiating both sides with respect to time (t), we get:

2x(dx/dt) + 2h(dh/dt) = 0

Now we plug in the known values:
x = ?
h = 9 ft
dx/dt = -5 ft/s
dh/dt = ?
15^2 = x^2 + (9)^2

(2x)(-5) + (2)(9)(dh/dt) = 0

Simplifying the equation:

-10x + 18(dh/dt) = 0

Now we need to solve for dh/dt when x = ?.

Using the equation: (15^2 = x^2 + (9)^2), we can find x:

225 = x^2 + 81

x^2 = 225 - 81

x^2 = 144

x = √144

x = 12 ft

Substituting x = 12 ft into the equation:

-10(12) + 18(dh/dt) = 0

-120 + 18(dh/dt) = 0

18(dh/dt) = 120

dh/dt = 120 / 18

dh/dt ≈ 6.67 ft/s

Therefore, when the top of the ladder is 9 ft above the ground, the foot of the ladder is moving away from the wall at a rate of approximately 6.67 ft/s.

x^2 + y^2 = 225

when y = 9
x^2 + 81 = 225
x^2 = 144
x = 12

2 x dx/dt + 2 y dy/dt = 0
24 dx/dt = 18 * 5