A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position, as shown in the figure. The winch pulls in rope at a rate of 
−0.4
 meter per second 
(ds/dt = −0.4 m/sec)
. Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y = 4. (Round your answers to three decimal places.)

Let's break down the problem step by step.

### Step 1: Define the scenario and variables.
We have a winch at the top of a 12-meter building pulling a 12-meter pipe. The pipe is being pulled in by the winch, so the winch reduces the length of the rope at a rate of \(\frac{ds}{dt} = -0.4 \, \text{m/sec}\).

- \(s\) is the length of the rope being pulled in.
- \(x\) is the horizontal distance from the bottom of the pipe to the base of the building.
- \(y\) is the vertical height from the bottom of the pipe to the ground.

Given:
- The total length of the pipe is 12 meters, so by the Pythagorean theorem:
  \[
  s^2 = x^2 + y^2
  \]
  Since the pipe remains 12 meters in length, the equation becomes:
  \[
  12^2 = x^2 + y^2
  \]
  Simplifying:
  \[
  144 = x^2 + y^2
  \]
  
- The rate at which the rope is being pulled is \(\frac{ds}{dt} = -0.4 \, \text{m/sec}\).
- We need to find the rate of vertical change (\(\frac{dy}{dt}\)) and horizontal change (\(\frac{dx}{dt}\)) when \(y = 4\).

### Step 2: Solve for \(x\) when \(y = 4\).
From the equation \(144 = x^2 + y^2\), when \(y = 4\), we get:
\[
x^2 + 4^2 = 144
\]
\[
x^2 + 16 = 144
\]
\[
x^2 = 128
\]
\[
x = \sqrt{128} = 8\sqrt{2} \approx 11.314
\]

### Step 3: Differentiate the equation \(x^2 + y^2 = 144\).
To find the rates of change, differentiate both sides of the equation with respect to time \(t\):
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]
This simplifies to:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \tag{1}
\]

### Step 4: Relating \(ds/dt\) to \(dx/dt\) and \(dy/dt\).
Since \(s^2 = x^2 + y^2\) and \(s = 12\), we can differentiate \(s = \sqrt{x^2 + y^2}\) with respect to \(t\):
\[
\frac{ds}{dt} = \frac{1}{2s}(2x \frac{dx}{dt} + 2y \frac{dy}{dt})
\]
Simplifying:
\[
\frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s}
\]
Given that \(s = 12\) and \(\frac{ds}{dt} = -0.4\), we have:
\[
-0.4 = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{12} \tag{2}
\]

### Step 5: Solve for \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\).
Now, we have two equations:
- Equation (1): \(x \frac{dx}{dt} + y \frac{dy}{dt} = 0\)
- Equation (2): \(-0.4 = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{12}\)

Substitute \(x \frac{dx}{dt} + y \frac{dy}{dt} = 12 \times (-0.4)\):
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = -4.8
\]

From Equation (1), express \(\frac{dx}{dt}\) in terms of \(\frac{dy}{dt}\):
\[
\frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt}
\]
Substitute this into Equation (2):
\[
x \left(-\frac{y}{x} \frac{dy}{dt}\right) + y \frac{dy}{dt} = -4.8
\]
Simplify:
\[
-y \frac{dy}{dt} + y \frac{dy}{dt} = -4.8
\]
Since the terms involving \(\frac{dy}{dt}\) cancel out, we are left with:

Thus, the rate of

A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position, as shown in the figure. The winch pulls in rope at a rate of -0.4 meter per second (ds/dt = -0.4 m/sec). Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y = 4. (Round your answers to three decimal places.)

Solve a related rates problem involving a winch pulling a 12-meter pipe. Learn how to calculate the rate of vertical and horizontal change using differentiation and the Pythagorean theorem. This problem applies calculus concepts and is suitable for high school students in Grades 11-12.

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