Let's solve each part step by step. ### (i) **Expression for \( P(x) \):** Let \( x \) be the height in meters from the base of the reservoir. The weight of a thin water layer between heights \( x \) and \( x + \Delta x \) is approximately \( P(x) \Delta x \). - **Density of water:** \( \rho = 1000 \ \text{kg/m}^3 \) - **Acceleration due to gravity:** \( g = 9.8 \ \text{m/s}^2 \) #### **Step 1: Area of cross-section at height \( x \)** The reservoir is a prism with isosceles triangle ends. - **Base width of triangle:** 4 m - **Height of triangle:** 7 m At height \( x \), the width of the triangle is proportional to \( x \): \[ \text{Width at } x = \frac{4}{7}x \] So, the area at height \( x \): \[ A(x) = \text{width at } x \times \text{length of reservoir} \] But actually, the triangular cross-section area at height \( x \) is: - The triangle has base \( 4 \) m and height \( 7 \) m. The area of a horizontal slice at height \( x \) (parallel to the base): \[ \text{At height } x, \text{ width } = \frac{4}{7}x \] So, area at this height (since it extends the length of the reservoir, 9 m): \[ \text{Area of slice} = (\text{width at } x) \times (\text{length}) = \frac{4}{7}x \cdot 9 = \frac{36}{7}x \] But actually, the area of a thin horizontal slice of thickness \( \Delta x \) at height \( x \) in a triangle is: \[ \text{Area of thin slice} = \text{width at } x \times (\Delta x) \] But since the triangle is isosceles, the width at height \( x \) is proportional to \( x \). But the volume of the thin layer: \[ \text{Volume} = \text{Area of slice at height } x \times dx \] But the area of the triangle at height \( x \) is: - The full triangle has base 4 and height 7, so slope of each side: \( m = \frac{4/2}{7} = \frac{2}{7} \) - At height \( x \), the width is \( 2 \times \frac{2}{7}x = \frac{4}{7}x \) So, the area of the slice at height \( x \) is the width (at \( x \)) times the length (9 m): \[ A(x) = \frac{4}{7}x \times 9 = \frac{36}{7}x \] #### **Step 2: Weight of the thin layer** \[ P(x) = \rho g \cdot \left[\text{Volume of thin layer at } x\right] \] \[ \text{Volume of thin layer} = \text{Area at } x \cdot dx = \frac{36}{7} x \cdot dx \] So, \[ P(x) = \rho g \cdot \frac{36}{7} x \] **Final Formula:** \[ \boxed{P(x) = \rho g \cdot \frac{36}{7} x} \] --- ### (ii) **Work to pump to 2 m above the reservoir:** The work required to lift the thin layer at height \( x \) to \( 2 \) meters above the reservoir: - The distance it needs to be lifted: \( 2 + (7 - x) = 9 - x \) meters (since the top is at 7 m, and the hose is 2 m above that). - Work = weight \(\times\) height lifted. \[ w(x) = P(x) \cdot (9 - x) \] \[ w(x) = \rho g \cdot \frac{36}{7} x \cdot (9 - x) \] **Final Formula:** \[ \boxed{w(x) = \rho g \cdot \frac{36}{7} x (9 - x)} \] --- ### (iii) **Total Work:** Integrate \( w(x) \) from \( x = 0 \) to \( x = 7 \): \[ \text{Total Work} = \int_0^7 w(x) dx = \int_0^7 \rho g \cdot \frac{36}{7} x (9 - x) dx \] Substitute \( \rho = 1000 \), \( g = 9.8 \): \[ = 1000 \cdot 9.8 \cdot \frac{36}{7} \int_0^7 x (9 - x) dx \] \[ = 35280 \int_0^7 (9x - x^2) dx \] Compute the integral: \[ \int_0^7 (9x - x^2) dx = \left[ \frac{9}{2}x^2 - \frac{1}{3}x^3 \right]_0^7 \] Plug in \( x = 7 \): \[ \frac{9}{2}(7^2) - \frac{1}{3}(7^3) = \frac{9}{2} \cdot 49 - \frac{1}{3} \cdot 343 = 220.5 - 114.333 = 106.167 \] So, \[ \text{Total Work} = 35280 \times 106.167 = 3,743,798.76 \text{ Joules} \] Roundoff to 3 significant figures: \[ \boxed{3,740,000 \text{ Joules}} \] --- ## **Summary of Answers:** 1. \( P(x) = \rho g \cdot \frac{36}{7} x \) 2. \( w(x) = \rho g \cdot \frac{36}{7} x (9 - x) \) 3. Work \( \approx 3,740,000 \) Joules Let me know if you need these answers formatted for direct input!