1. **State the problem:** We have a cable from a pulley to a person's hand and from the pulley vertically down to a weight. The cable length from the person's hand to the weight is fixed at 7.6 m, and the person's hand is always 3.6 m below the pulley. 2. **Find y as a function of x:** Given $x$ is the horizontal distance from the pulley to the person's hand, and $h=3.6$ m is the vertical distance from the pulley to the person's hand. The length of cable from pulley to weight is $y$, and the cable from hand to weight is fixed at 7.6 m. Using the Pythagorean theorem for the triangle formed by $x$, $h$, and the cable from hand to pulley: $$\text{Cable length from hand to pulley} = \sqrt{x^2 + h^2} = \sqrt{x^2 + 3.6^2}$$ The total cable length is 7.6 m, so: $$y + \sqrt{x^2 + 3.6^2} = 7.6$$ Rearranged: $$y = 7.6 - \sqrt{x^2 + 3.6^2}$$ The problem states $y = 4 - \sqrt{x^2 + 3.6^2}$, which seems inconsistent with the total length 7.6 m. We will use the correct formula: $$y = 7.6 - \sqrt{x^2 + 3.6^2}$$ 3. **Find $x$ when $y=1.6$ m:** Set $y=1.6$: $$1.6 = 7.6 - \sqrt{x^2 + 3.6^2}$$ Rearranged: $$\sqrt{x^2 + 3.6^2} = 7.6 - 1.6 = 6.0$$ Square both sides: $$x^2 + 3.6^2 = 6.0^2$$ $$x^2 + 12.96 = 36$$ $$x^2 = 36 - 12.96 = 23.04$$ $$x = \sqrt{23.04} = 4.80\text{ m}$$ 4. **Find how fast the weight is moving when $y=1.6$ m and the person moves horizontally at 0.8 m/s:** Differentiate $y = 7.6 - \sqrt{x^2 + 3.6^2}$ with respect to time $t$: $$\frac{dy}{dt} = - \frac{1}{2\sqrt{x^2 + 3.6^2}} \cdot 2x \frac{dx}{dt} = - \frac{x}{\sqrt{x^2 + 3.6^2}} \frac{dx}{dt}$$ At $x=4.80$ m and $\frac{dx}{dt} = -0.8$ m/s (moving towards pulley means $x$ decreases): $$\frac{dy}{dt} = - \frac{4.80}{6.0} \times (-0.8) = -0.8 \times (-0.8) = 0.64\text{ m/s}$$ The positive sign means the weight is moving upward at 0.64 m/s. 5. **Find how fast the person is moving when $y=1.6$ m and the weight is moving downward at 0.3 m/s:** Given $\frac{dy}{dt} = -0.3$ m/s (downward is negative direction), solve for $\frac{dx}{dt}$: $$\frac{dy}{dt} = - \frac{x}{\sqrt{x^2 + 3.6^2}} \frac{dx}{dt}$$ Rearranged: $$\frac{dx}{dt} = - \frac{\sqrt{x^2 + 3.6^2}}{x} \frac{dy}{dt}$$ Substitute values: $$\frac{dx}{dt} = - \frac{6.0}{4.80} \times (-0.3) = \frac{6.0}{4.80} \times 0.3 = 0.375\text{ m/s}$$ The positive value means the person is moving horizontally towards the pulley at 0.375 m/s. **Final answers:** - (b) $x = 4.80$ m - (c) The weight is moving upward at 0.64 m/s - (d) The person is moving horizontally towards the pulley at 0.375 m/s