A 10 metre plank of wood is propped against a wall (see Figure 1).
The bottom end of the plank is sliding away from the wall as the top end of the plank slips
down the wall. When the bottom of the plank is 3 metres from the wall, the top of the plank
is slipping down the wall at a rate of 1m/sec.
What is the rate at which the bottom of the plank is sliding along the ground
when the bottom of the plank is 3 metres from the wall? (Remember to provide all
working and to present your mathematics clearly and correctly

To solve this problem, we'll apply related rates, a technique in calculus that deals with the rates at which two or more related variables change.

### Step 1: Set up the problem

Let:
- \( x \) be the distance of the bottom of the plank from the wall.
- \( y \) be the height of the top of the plank on the wall.
- \( L = 10 \) meters be the length of the plank, which remains constant.

Given:
- \( x = 3 \) meters (at the specific moment we are analyzing).
- \( \frac{dy}{dt} = -1 \) m/s (the rate at which the top of the plank is slipping down the wall; negative because \( y \) is decreasing).

We need to find:
- \( \frac{dx}{dt} \) when \( x = 3 \) meters.

### Step 2: Relate the variables using the Pythagorean theorem

Since the plank forms a right triangle with the wall and the ground:
\[
x^2 + y^2 = L^2
\]
\[
x^2 + y^2 = 10^2
\]
\[
x^2 + y^2 = 100
\]

### Step 3: Differentiate with respect to time \( t \)

Differentiate both sides of the equation with respect to \( t \):
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]
We can simplify this by dividing the entire equation by 2:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0
\]

### Step 4: Substitute the known values

We know \( x = 3 \), \( \frac{dy}{dt} = -1 \), and we need to find \( \frac{dx}{dt} \). First, we need to determine \( y \) when \( x = 3 \).

Using the Pythagorean theorem again:
\[
3^2 + y^2 = 100
\]
\[
9 + y^2 = 100
\]
\[
y^2 = 91
\]
\[
y = \sqrt{91} \text{ meters}
\]

Now, substitute into the differentiated equation:
\[
3 \frac{dx}{dt} + \sqrt{91} \cdot (-1) = 0
\]
\[
3 \frac{dx}{dt} - \sqrt{91} = 0
\]
\[
3 \frac{dx}{dt} = \sqrt{91}
\]
\[
\frac{dx}{dt} = \frac{\sqrt{91}}{3} \text{ m/s}
\]

### Step 5: Simplify the expression

Calculating \( \sqrt{91} \):
\[
\sqrt{91} \approx 9.54 \text{ (approximately)}
\]

Thus:
\[
\frac{dx}{dt} \approx \frac{9.54}{3} \approx 3.18 \text{ m/s}
\]

### Final Answer

The rate at which the bottom of the plank is sliding along the ground when it is 3 meters from the wall is approximately \( 3.18 \) m/s.

---

Would you like further details on any part of this solution, or do you have any questions?

Here are some related questions you might find interesting:
1. What would happen to the rate of sliding if the plank was longer?
2. How would the problem change if the rate of sliding at the top was increasing?
3. Can you derive a general formula for the rate at which the bottom of the plank slides given any length?
4. What if the wall was not vertical—how would that affect the rates?
5. How does the rate of change of the angle between the plank and the ground relate to this problem?

**Tip:** When working with related rates, always express the relationship between the variables using an equation before differentiating with respect to time.

Learn how to solve a related rates problem involving a sliding plank. Understand the application of calculus and the Pythagorean theorem to find the rate at which the bottom of the plank slides along the ground. Suitable for undergraduate students and anyone interested in calculus problems.

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