A 2.6 m long pipe leans against a wall. If the bottom of the pipe is pulled away from the wall at a rate of 0.2 m/s, how fast is the area of the triangle formed by the pipe, wall, and ground changing

Let $L$ be the length of the pipe, $x$ be the distance of the lower end from the wall, and $y$ be the height of the top end on the wall.
Given:

• Pipe length $L = 2.6$ m (constant).
• Rate at which the lower end is pulled away $\frac{dx}{dt} = 0.2$ m/s.
• We need to find the rate of change of the area $\frac{dA}{dt}$ when $x = 1$ m.

The pipe, wall, and ground form a right-angled triangle, so by the Pythagorean theorem:

$$x^2 + y^2 = L^2$$

The area $A$ of this triangle is:

$$A = \frac{1}{2}xy$$

Substitute $x = 1$ m and $L = 2.6$ m into the Pythagorean theorem:

$$1^2 + y^2 = (2.6)^2$$ $$1 + y^2 = 6.76$$ $$y^2 = 6.76 - 1$$ $$y^2 = 5.76$$ $$y = \sqrt{5.76} = 2.4 \text{ m}$$

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$ $$x\frac{dx}{dt} + y\frac{dy}{dt} = 0$$

Now, substitute $x = 1$ m, $y = 2.4$ m, and $\frac{dx}{dt} = 0.2$ m/s:

$$(1)(0.2) + (2.4)\frac{dy}{dt} = 0$$ $$0.2 + 2.4\frac{dy}{dt} = 0$$ $$2.4\frac{dy}{dt} = -0.2$$ $$\frac{dy}{dt} = -\frac{0.2}{2.4} = -\frac{2}{24} = -\frac{1}{12} \text{ m/s}$$

Using the product rule:

$$\frac{dA}{dt} = \frac{1}{2} \left( \frac{dx}{dt}y + x\frac{dy}{dt} \right)$$

Substitute $x = 1$ m, $y = 2.4$ m, $\frac{dx}{dt} = 0.2$ m/s, and $\frac{dy}{dt} = -\frac{1}{12}$ m/s:

$$\frac{dA}{dt} = \frac{1}{2} \left( (0.2 \text{ m/s})(2.4 \text{ m}) + (1 \text{ m})\left(-\frac{1}{12} \text{ m/s}\right) \right)$$ $$\frac{dA}{dt} = \frac{1}{2} \left( 0.48 \text{ m}^2/\text{s} - \frac{1}{12} \text{ m}^2/\text{s} \right)$$

To simplify, convert 0.48 to a fraction: $0.48 = \frac{48}{100} = \frac{12}{25}$.

$$\frac{dA}{dt} = \frac{1}{2} \left( \frac{12}{25} - \frac{1}{12} \right)$$

Find a common denominator for 25 and 12, which is 300:

$$\frac{dA}{dt} = \frac{1}{2} \left( \frac{12 \times 12}{25 \times 12} - \frac{1 \times 25}{12 \times 25} \right)$$ $$\frac{dA}{dt} = \frac{1}{2} \left( \frac{144}{300} - \frac{25}{300} \right)$$ $$\frac{dA}{dt} = \frac{1}{2} \left( \frac{144 - 25}{300} \right)$$ $$\frac{dA}{dt} = \frac{1}{2} \left( \frac{119}{300} \right)$$ $$\frac{dA}{dt} = \frac{119}{600} \text{ m}^2/\text{s}$$

The rate at which the area is changing is $\frac{119}{600} \text{ m}^2/\text{s}$.