- Identify the relationship between the ladder's position variables using the Pythagorean theorem: \(x^2 + y^2 = 25\).
- Determine the height of the ladder at the instant the base is \(4\) meters from the wall: \(y = 3\) meters.
- Differentiate the relationship with respect to time to relate the rates: \(x\frac{dx}{dt} + y\frac{dy}{dt} = 0\).
- Substitute the known values and solve for the rate \(\frac{dx}{dt}\), yielding a result of \(\boxed{0.3 \text{ m/s}}\).
Explanation
- Problem Analysis and Setup
We are given a ladder of length \(L = 5\) meters leaning against a wall. Let \(x\) be the distance from the base of the ladder to the wall and \(y\) be the height of the top of the ladder on the wall. The ladder, the wall, and the ground form a right triangle, so we can apply the Pythagorean theorem:
\[x^2 + y^2 = L^2\]
Substituting \(L = 5\), we have:
\[x^2 + y^2 = 25\]
The top of the ladder is sliding down the wall, which means the rate of change of the height \(y\) with respect to time \(t\) is given as \(\frac{dy}{dt} = -0.4\) m/s. We want to find the rate at which the base of the ladder is sliding away from the wall, \(\frac{dx}{dt}\), at the specific moment when \(x = 4\) meters.
- Determining the Height of the Ladder
First, we determine the height \(y\) of the top of the ladder when the base is \(x = 4\) meters from the wall:
\[4^2 + y^2 = 25\]
\[16 + y^2 = 25\]
\[y^2 = 25 - 16 = 9\]
Since the height must be positive, we have \(y = 3\) meters.
- Differentiating with Respect to Time
To find the relationship between the rates of change, we differentiate the equation \(x^2 + y^2 = 25\) with respect to time \(t\):
\[\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(25)\]
Using the chain rule, we obtain:
\[2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\]
Dividing by 2, we simplify this to:
\[x\frac{dx}{dt} + y\frac{dy}{dt} = 0\]
- Solving for the Rate of Change
Now we substitute the known values \(x = 4\), \(y = 3\), and \(\frac{dy}{dt} = -0.4\) into the differentiated equation:
\[4 \times \frac{dx}{dt} + 3 \times (-0.4) = 0\]
\[4\frac{dx}{dt} - 1.2 = 0\]
\[4\frac{dx}{dt} = 1.2\]
Solving for \(\frac{dx}{dt}\):
\[\frac{dx}{dt} = \frac{1.2}{4} = 0.3\]
Therefore, the base of the ladder is sliding away from the wall at a rate of \(0.3\) meters per second.
- Final Answer
The base of the ladder is sliding away from the wall at a rate of \(\boxed{0.3 \text{ m/s}}\).
Examples
This type of related-rates problem is a cornerstone in engineering and physics, where understanding how variables change over time is essential. For instance, an architect might use these calculations to ensure that a retractable structure or a mechanical arm moves safely without hitting obstacles. By modeling the motion with geometric constraints like the Pythagorean theorem, they can predict the velocity of each part of a moving system, ensuring precision and safety in design and operation.