A 18 ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 3 ft/s, how fast (in ft/s) is the bottom moving along the ground when the bottom of the ladder is 9 ft from the wall?

Also, this one. I have done the pathag theorem and then dy/dt = -x/y * dy/dt
Then, plugged stuff in to solve for that, but the answer I got which was 26sqrt3 which is the wrong answer, so I am just confused

no, dy/dx = -x/y

x^2+y^2=18^2
so, when x=9, y=9√3

2x dx/dt + 2y dy/dt = 0
Now plug in your values and you have
2*9 dx/dt + 2*9√3*(-3) = 0
dx/dt = 54√3/18 = 3√3 ft/s

Well, it seems like this ladder really wants to go places! Let's figure out its speed.

We can use the Pythagorean theorem to relate the distances involved. Let's call the distance from the bottom of the ladder to the wall "x," and the distance from the top of the ladder to the ground "y."

According to the problem, we know that dx/dt (the rate at which x is changing) is 3 ft/s. We want to find dy/dt (the rate at which y is changing) when x = 9 ft.

Now, we can use the Pythagorean theorem to relate x and y. It states that x^2 + y^2 = 18^2 (since the ladder is 18 ft long).

Differentiating this equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0

Now, plugging in the known values:
2(9)(3) + 2y(dy/dt) = 0
54 + 2y(dy/dt) = 0
2y(dy/dt) = -54
dy/dt = -54 / (2y)

Now, we can substitute the value of x when x = 9 ft:
dy/dt = -54 / (2y)
dy/dt = -27 / y

To find the value of y, we can again use the Pythagorean theorem:
x^2 + y^2 = 9^2
81 + y^2 = 324
y^2 = 243
y = sqrt(243) = 9 sqrt(3)

Substituting this back into our equation for dy/dt:
dy/dt = -27 / (9 sqrt(3))
dy/dt = -3 / sqrt(3)
dy/dt = -3 sqrt(3) / 3
dy/dt = -sqrt(3) ft/s

So, the bottom of the ladder is moving along the ground at a speed of -sqrt(3) ft/s. By the way, it's important to note that the negative sign indicates that the bottom of the ladder is moving away from the wall.

To solve this problem, we can use the concept of related rates. Let's denote the distance of the bottom of the ladder from the wall as x, the distance of the top of the ladder from the ground as y, and the length of the ladder as L = 18 ft.

We are given that the top of the ladder slides down the wall at a rate of 3 ft/s. This means that dy/dt (the rate of change of y) is equal to -3 ft/s, since the ladder is sliding down.

We are asked to find dx/dt (the rate of change of x) when x = 9 ft.

To solve this, we can use the Pythagorean theorem, which states that x^2 + y^2 = L^2.

Taking the derivative of both sides with respect to time (t), we get:

2x(dx/dt) + 2y(dy/dt) = 0

Since dx/dt is what we're looking for, let's isolate it:

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

Plugging in the given values (dy/dt = -3 ft/s, x = 9 ft), we can find dx/dt:

dx/dt = -(y/9) * (-3)

Since L = 18 ft, we can use the Pythagorean theorem to find the value of y. When x = 9 ft, we have:

9^2 + y^2 = 18^2
81 + y^2 = 324
y^2 = 243
y = √(243) = 9√3

Plugging in y = 9√3 into the equation for dx/dt:

dx/dt = -(9√3/9) * (-3)
dx/dt = 3√3 * 3
dx/dt = 9√3 ft/s

Therefore, the bottom of the ladder is moving along the ground at a rate of 9√3 ft/s when the bottom of the ladder is 9 ft from the wall.

To solve this problem, we can use related rates. Let's define the following variables:

x = distance from the wall to the bottom of the ladder
y = distance from the ground to the top of the ladder

We need to find dy/dt, the rate at which the bottom of the ladder is moving along the ground.

Given information:
dx/dt = unknown (what we're trying to solve)
dy/dt = -3 ft/s (given in the problem)

We can apply the Pythagorean theorem to relate x, y, and the length of the ladder:

x^2 + y^2 = 18^2
x^2 + y^2 = 324

Differentiating both sides of this equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0

Now let's solve for dx/dt:

2x(dx/dt) = -2y(dy/dt)
dx/dt = (-2y(dy/dt)) / (2x)
dx/dt = -y(dy/dt) / x

We are given that dy/dt = -3 ft/s, and we need to find dx/dt when x = 9 ft.

Substituting the given values:
dx/dt = -y(dy/dt) / x
dx/dt = -9(-3) / 9
dx/dt = 27 / 9
dx/dt = 3 ft/s

Therefore, the bottom of the ladder is moving along the ground at a rate of 3 ft/s when the bottom of the ladder is 9 ft from the wall.