Rates of Change — Question 8

Question 8 - Solution

Let:

• $x(t)$: distance from the wall to the base of the ladder

• $y(t)$: height of the top of the ladder on the wall

• Given: $\frac{dx}{dt} = 2 \, \text{ft/s}, \quad x = 6$

Using the Pythagorean theorem:

$$x^2 + y^2 = 10^2 = 100$$

Differentiate both sides with respect to time $t$:

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

We know: $$x = 6, \quad \frac{dx}{dt} = 2$$

Solve for $y$:

$$36 + y^2 = 100 \Rightarrow y^2 = 64 \Rightarrow y = 8$$

Substitute into the differentiated equation:

$$6(2) + 8 \frac{dy}{dt} = 0 \Rightarrow 12 + 8 \frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{12}{8} = -\frac{3}{2}$$

Answer: $$\boxed{\frac{dy}{dt} = -\frac{3}{2} \text{ ft/s}}$$

The negative sign indicates that the top of the ladder is sliding downward.