Let's break down the problem step by step. --- ### (i) **Find \( P(x) \): The weight of a thin water layer at height \( x \) of thickness \( \Delta x \)** The reservoir is a right prism with isosceles triangular faces, height \( 5 \) m, width (base of triangle) \( 4 \) m, and length \( 9 \) m. #### **Step 1: Find the area of the cross-section at height \( x \)** The cross-sectional area is constant at every height \( x \) because it is a prism. - The triangle has base \( b = 4 \) m, height \( h = 5 \) m. \[ A = \frac{1}{2} b h = \frac{1}{2} \times 4 \times 5 = 10 \text{ m}^2 \] But the reservoir is \( 9 \) m long, so the volume of a thin slab of water at height \( x \) and thickness \( \Delta x \): \[ \text{Volume} = (\text{area of cross-section}) \times (\text{thickness}) \times (\text{length}) \] But the length is already included as the prism's length; the cross-section is the triangle, and the prism extends along the 9m direction. But for a prism, the area perpendicular to the height \( x \) is the area of the triangle, and the length \( L \) is along the prism, so for a thin "slab" at height \( x \), the volume is: \[ \text{Volume of thin layer} = (\text{area of triangular face}) \times (\text{prism length}) \times \Delta x \] But this is not correct—actually, for a prism, the volume between heights \( x \) and \( x+\Delta x \) is: - The area of the triangle face (\( 10 \) m\(^2\)) - Times the **length** (\( 9 \) m) - Times the **thickness** (\( \Delta x \)), but the thickness is already "built in" if you integrate over the height. But let's clarify: - The "height" \( x \) goes from \( 0 \) (base) to \( 5 \) (top). - The triangular face is oriented so that the height of the triangle is vertical. But since the prism's cross section is the triangle, and the length is 9m, each horizontal "slice" at height \( x \) is a rectangle of length 9m and width equal to the width of the triangle at height \( x \). But the width of the triangle at height \( x \) **varies** linearly from 0 at the base to 4 m at the top (since it's isosceles). Let’s define: Let’s put the base of the triangle at \( x = 0 \), and the top at \( x = 5 \). At height \( x \), the width is: \[ w(x) = \frac{4}{5}x \] (because at \( x = 5 \), \( w(5) = 4 \); at \( x = 0 \), \( w(0) = 0 \); so it increases linearly). So the area of a thin strip at height \( x \), thickness \( \Delta x \): \[ \text{Area at height } x = w(x) \cdot 1 \] But for a thin "slab" of thickness \( \Delta x \), we want the **volume**: \[ \text{Volume} = (\text{width at } x) \times (\text{length}) \times \Delta x \\ = w(x) \cdot 9 \cdot \Delta x \] So the **weight** of this thin layer is: \[ P(x) \Delta x = (\text{density}) \cdot (\text{volume}) \cdot (\text{gravity}) \] \[ P(x) \Delta x = \rho \cdot g \cdot (\text{volume}) \\ = 1000 \cdot 9.8 \cdot w(x) \cdot 9 \cdot \Delta x \] So, \[ P(x) = 1000 \cdot 9.8 \cdot 9 \cdot w(x) \] But as above, \[ w(x) = \frac{4}{5}x \] So, \[ P(x) = 1000 \cdot 9.8 \cdot 9 \cdot \frac{4}{5}x \] Or, \[ \boxed{ P(x) = 1000 \cdot 9.8 \cdot 9 \cdot \frac{4}{5}x } \] Or simplify: - \( 1000 \times 9.8 = 9800 \) - \( 9 \times \frac{4}{5} = \frac{36}{5} \) So, \[ P(x) = 9800 \cdot \frac{36}{5} x = (9800 \times 7.2)x = 70,560x \] Final answer (either form is acceptable): \[ \boxed{ P(x) = 70,560x } \] --- ### (ii) **Find \( w(x) \): The work to pump the thin layer at height \( x \) to 3 m above the reservoir** The thin layer at height \( x \) must be lifted a distance of \( (3 + 5 - x) \) meters (since the reservoir is 5m high, and the hose is 3m above the top). So, \[ \text{Distance to lift} = 3 + (5 - x) = 8 - x \] So the work to lift the thin layer is: \[ w(x) \Delta x = P(x) \cdot (\text{distance}) \Delta x \] \[ w(x) = P(x) \cdot (8 - x) \] From above, \[ w(x) = 70,560 x \cdot (8 - x) \] or in the earlier, expanded form, \[ w(x) = 1000 \cdot 9.8 \cdot 9 \cdot \frac{4}{5} x \cdot (8-x) \] \[ \boxed{ w(x) = 70,560 x (8 - x) } \] --- ### (iii) **Total work required** Integrate \( w(x) \) from \( x = 0 \) to \( x = 5 \): \[ W = \int_0^5 w(x) dx = \int_0^5 70,560 x (8-x) dx \] Calculate: \[ \int_0^5 x (8-x) dx = \int_0^5 (8x - x^2) dx = 8 \int_0^5 x dx - \int_0^5 x^2 dx \] - \( \int_0^5 x dx = \left. \frac{1}{2}x^2 \right|_0^5 = \frac{1}{2}(25) = 12.5 \) - \( \int_0^5 x^2 dx = \left. \frac{1}{3}x^3 \right|_0^5 = \frac{1}{3}(125) = 41.\overline{6} \) So \[ \int_0^5 x(8-x) dx = 8 \cdot 12.5 - 41.666... = 100 - 41.666... = 58.333... \] So \[ W = 70,560 \cdot 58.333... = 70,560 \cdot \frac{175}{3} \] Calculate: - \( 70,560 \times 58.3333 = 4,115,200 \) Or, more precisely: \[ 70,560 \times \frac{175}{3} = \frac{70,560 \times 175}{3} = \frac{12,348,000}{3} = 4,116,000 \] So, \[ \boxed{ W = 4,116,000 \text{ Joules} } \] --- ## **Summary of Answers** --- ### (i) \( P(x) \): \[ \boxed{ P(x) = 70,560 x } \] where \( x \) is the height above the base in meters. --- ### (ii) \( w(x) \): \[ \boxed{ w(x) = 70,560 x (8 - x) } \] --- ### (iii) Total work required: \[ \boxed{ W = 4,116,000 \text{ Joules} } \] (Rounded to the nearest integer, well within 1% error.) --- **If you need the answers in terms of variables (\( \rho, g, L, b, h \)) instead of numbers, let me know!**