Let's break down and solve the problem step by step: --- ### **Given Data** - **Control Volume (V₀):** 1.0 ft³ - **Flow Area (A):** 1.0 ft² - **Length (ℓ₀):** 1.0 ft - **Velocity at center (V₀):** 1.0 ft/s - **Density at center (ρ₀):** 1.800 slug/ft³ - **Density gradient:** decreases at 0.25 slug/(ft⁴) in the axial direction (∂ρ/∂x = –0.25 slug/ft⁴) - **No local change in density:** ∂ρ/∂t = 0 --- ## **A. System (Lagrangian) Approach** For a system, you follow the same fluid particles as they move. The rate of change of density for a particle is given by the **material derivative:** \[ \frac{D\rho}{Dt} = \frac{\partial \rho}{\partial t} + V \frac{\partial \rho}{\partial x} \] Given: - \(\frac{\partial \rho}{\partial t} = 0\) - \(V = V_0 = 1.0\) ft/s - \(\frac{\partial \rho}{\partial x} = -0.25\) slug/ft⁴ Plug in the values: \[ \frac{D\rho}{Dt} = 0 + (1.0)\times (-0.25) = -0.25 \text{ slug/ft}^3\text{/s} \] --- ## **B. Control Volume (Eulerian) Approach** The average rate of change of density in the control volume is: \[ \frac{d}{dt}\int_{CV} \rho \, dV = \int_{CV} \frac{\partial \rho}{\partial t} dV + \oint_{CS} \rho \vec{V} \cdot d\vec{A} \] But since the control volume is fixed and \(\frac{\partial \rho}{\partial t} = 0\), the time rate of change inside the CV is zero. Instead, we focus on the **flux** across the CV boundaries. #### **Net outflow of mass (density flux):** \[ \text{Net flux} = (\rho V A)_{out} - (\rho V A)_{in} \] Since area is the same on both sides, but density varies linearly in \(x\): - At the **inlet** (left): \(x = x_0 - \frac{\ell_0}{2}\) - At the **outlet** (right): \(x = x_0 + \frac{\ell_0}{2}\) So, \[ \rho_{in} = \rho_0 + \frac{\partial \rho}{\partial x} \left(-\frac{\ell_0}{2}\right) = 1.800 + (-0.25) \times \left(-\frac{1}{2}\right) = 1.800 + 0.125 = 1.925 \text{ slug/ft}^3 \] \[ \rho_{out} = \rho_0 + \frac{\partial \rho}{\partial x} \left(\frac{\ell_0}{2}\right) = 1.800 + (-0.25) \times \left(\frac{1}{2}\right) = 1.800 - 0.125 = 1.675 \text{ slug/ft}^3 \] Assume the velocity is constant (V₀): \[ \text{Net mass flux} = (\rho_{out} V_0 A) - (\rho_{in} V_0 A) \] \[ = [1.675 \times 1.0 \times 1.0] - [1.925 \times 1.0 \times 1.0] = 1.675 - 1.925 = -0.25 \text{ slug/s} \] Divide by the control volume \(V_0\) to get the rate of change of average density: \[ \frac{d\overline{\rho}}{dt} = \frac{\text{Net mass flux}}{V_0} = \frac{-0.25}{1.0} = -0.25 \text{ slug/ft}^3\text{/s} \] --- ## **C. Comparison** Both the **system (material derivative)** and the **control volume (flux)** approaches yield: \[ \boxed{ \frac{d\rho}{dt} = -0.25 \text{ slug/ft}^3\text{/s} } \] This shows consistency between the Lagrangian (system) and Eulerian (control volume) analyses. --- ### **Summary Table** | Approach | Result for \( \frac{d\rho}{dt} \) | |---------------|:---------------------------------:| | System/Lagrangian | \(-0.25 \text{ slug/ft}^3\text{/s}\) | | Control Volume | \(-0.25 \text{ slug/ft}^3\text{/s}\) | If you need a step-by-step derivation or further explanation of any step, let me know!

Let's carefully analyze the function and find all the required features step-by-step. Given: \[ f(x) = 4x^3 + 2x^2 - \frac{9}{4}x + 2 \] --- ### **(a) Find the intervals where \(f\) is increasing and decreasing** **Step 1: Find the first derivative \(f'(x)\):** \[ f'(x) = \frac{d}{dx} [4x^3 + 2x^2 - \frac{9}{4}x + 2] \] \[ f'(x) = 12x^2 + 4x - \frac{9}{4} \] --- **Step 2: Find critical points where \(f'(x) = 0\):** Set \(f'(x) = 0\): \[ 12x^2 + 4x - \frac{9}{4} = 0 \] Multiply through by 4 to clear denominators: \[ 4 \times 12x^2 + 4 \times 4x - 4 \times \frac{9}{4} = 0 \] \[ 48x^2 + 16x - 9 = 0 \] **Step 3: Solve quadratic:** \[ 48x^2 + 16x - 9 = 0 \] Using quadratic formula: \[ x = \frac{-16 \pm \sqrt{(16)^2 - 4 \times 48 \times (-9)}}{2 \times 48} \] Calculate discriminant: \[ D = 16^2 - 4 \times 48 \times (-9) = 256 + 4 \times 48 \times 9 \] \[ 4 \times 48 = 192,\quad 192 \times 9 = 1728 \] \[ D = 256 + 1728 = 1984 \] \[ x = \frac{-16 \pm \sqrt{1984}}{96} \] Simplify \(\sqrt{1984}\): \[ \sqrt{1984} = \sqrt{16 \times 124} = 4 \sqrt{124} \] And \(\sqrt{124} = \sqrt{4 \times 31} = 2 \sqrt{31}\): \[ \sqrt{1984} = 4 \times 2 \sqrt{31} = 8 \sqrt{31} \] Thus, \[ x = \frac{-16 \pm 8 \sqrt{31}}{96} \] Divide numerator and denominator by 8: \[ x = \frac{-2 \pm \sqrt{31}}{12} \] --- **Critical points:** \[ x_1 = \frac{-2 - \sqrt{31}}{12} \] \[ x_2 = \frac{-2 + \sqrt{31}}{12} \] --- **Step 4: Determine the sign of \(f'(x)\) in intervals:** Intervals are: \[ (-\infty, x_1), \quad (x_1, x_2), \quad (x_2, \infty) \] Choose test points: - For \(x < x_1\), pick \(x = -1\) - For \(x_1 < x < x_2\), pick \(x = 0\) - For \(x > x_2\), pick \(x = 1\) Calculate \(f'(x)\): 1. \(x = -1\): \[ f'(-1) = 12(-1)^2 + 4(-1) - \frac{9}{4} = 12 - 4 - 2.25 = 5.75 > 0 \] 2. \(x = 0\): \[ f'(0) = 0 + 0 - 2.25 = -2.25 < 0 \] 3. \(x = 1\): \[ f'(1) = 12(1) + 4(1) - 2.25 = 12 + 4 - 2.25 = 13.75 > 0 \] **Answer:** - \(f\) **increases** on \((-\infty, x_1)\) and \((x_2, \infty)\) - \(f\) **decreases** on \((x_1, x_2)\) --- ### **(b) Find the local minimum and maximum values** - **Local maximum** at \(x = x_1 = \frac{-2 - \sqrt{31}}{12}\) (since \(f'\) changes from positive to negative) - **Local minimum** at \(x = x_2 = \frac{-2 + \sqrt{31}}{12}\) (since \(f'\) changes from negative to positive) **Calculate \(f(x)\) at these points:** 1. \(x_1 = \frac{-2 - \sqrt{31}}{12}\) 2. \(x_2 = \frac{-2 + \sqrt{31}}{12}\) **(Approximate \(\sqrt{31} \approx 5.57\):** - \(x_1 \approx \frac{-2 - 5.57}{12} = \frac{-7.57}{12} \approx -0.631\) - \(x_2 \approx \frac{-2 + 5.57}{12} = \frac{3.57}{12} \approx 0.297\) **Evaluate \(f(x)\):** \[ f(x) = 4x^3 + 2x^2 - \frac{9}{4}x + 2 \] - At \(x \approx -0.631\): \[ f(-0.631) \approx 4(-0.631)^3 + 2(-0.631)^2 - 2.25(-0.631) + 2 \] Calculate step-by-step: \(-0.631)^3 \approx -0.251\) \[ 4 \times -0.251 \approx -1.004 \] \((-0.631)^2 \approx 0.398\) \[ 2 \times 0.398 \approx 0.796 \] \[ -2.25 \times -0.631 \approx 1.419 \] Sum: \[ -1.004 + 0.796 + 1.419 + 2 \approx 3.211 \] - At \(x \approx 0.297\): \[ (0.297)^3 \approx 0.0262 \] \[ 4 \times 0.0262 \approx 0.1048 \] \[ (0.297)^2 \approx 0.0882 \] \[ 2 \times 0.0882 \approx 0.1764 \] \[ -2.25 \times 0.297 \approx -0.668 \] Sum: \[ 0.1048 + 0.1764 - 0.668 + 2 \approx 1.613 \] **Thus:** - **Local maximum value:** \(\boxed{\approx 3.211}\) at \(x \approx -0.631\) - **Local minimum value:** \(\boxed{\approx 1.613}\) at \(x \approx 0.297\) --- ### **(c) Find the inflection point** **Step 1: Find the second derivative \(f''(x)\):** \[ f''(x) = \frac{d}{dx} [12x^2 + 4x - \frac{9}{4}] = 24x + 4 \] **Step 2: Set \(f''(x) = 0\):** \[ 24x + 4 = 0 \Rightarrow x = -\frac{4}{24} = -\frac{1}{6} \approx -0.167 \] **Step 3: Find \(f(-1/6)\):** \[ f(-\frac{1}{6}) = 4 \left(-\frac{1}{6}\right)^3 + 2 \left(-\frac{1}{6}\right)^2 - \frac{9}{4} \left(-\frac{1}{6}\right) + 2 \] Calculate step-by-step: \[ \left(-\frac{1}{6}\right)^3 = -\frac{1}{216} \] \[ 4 \times -\frac{1}{216} = -\frac{4}{216} = -\frac{1}{54} \] \[ \left(-\frac{1}{6}\right)^2 = \frac{1}{36} \] \[ 2 \times \frac{1}{36} = \frac{2}{36} = \frac{1}{18} \] \[ -\frac{9}{4} \times -\frac{1}{6} = \frac{9}{4} \times \frac{1}{6} = \frac{9}{24} = \frac{3}{8} \] Sum: \[ -\frac{1}{54} + \frac{1}{18} + \frac{3}{8} + 2 \] Express all with common denominator 216: - \(-\frac{1}{54} = -\frac{4}{216}\) - \(\frac{1}{18} = \frac{12}{216}\) - \(\frac{3}{8} = \frac{81}{216}\) - \(2 = \frac{432}{216}\) Sum: \[ -\frac{4}{216} + \frac{12}{216} + \frac{81}{216} + \frac{432}{216} = \frac{-4 + 12 + 81 + 432}{216} = \frac{521}{216} \approx 2.412 \] --- ### **(d) Intervals of concavity** Since: \[ f''(x) = 24x + 4 \] - **Concave up** where \(f''(x) > 0\): \[ 24x + 4 > 0 \Rightarrow x > -\frac{1}{6} \] - **Concave down** where \(f''(x) < 0\): \[ x < -\frac{1}{6} \] --- ### **Final Answers:** **(a) Intervals of increase/decrease:** - **Increasing:** \(\boxed{(-\infty, \frac{-2 - \sqrt{31}}{12}) \cup (\frac{-2 + \sqrt{31}}{12}, \infty)}\) - **Decreasing:** \(\boxed{\left(\frac{-2 - \sqrt{31}}{12}, \frac{-2 + \sqrt{31}}{12}\right)}\) **(b) Local extrema:** - **Maximum:** \(\boxed{\approx 3.211}\) at \(x \approx \frac{-2 - \sqrt{31}}{12}\) - **Minimum:** \(\boxed{\approx 1.613}\) at \(x \approx \frac{-2 + \sqrt{31}}{12}\) **(c) Inflection point:** \[ \boxed{\left( -\frac{1}{6}, \frac{521}{216} \approx 2.412 \right)} \] **(d) Concavity:** - **Concave up:** \(\boxed{\left(-\frac{1}{6}, \infty\right)}\) - **Concave down:** \(\boxed{\left(-\infty, -\frac{1}{6}\right)}\) --- Let me know if you'd like further explanation!