Groundwater Contaminant Plume in Radial Aquifer (Environmental Engineering) Solution

An environmental engineer is modeling the chemical contaminant concentration $C(x, y)$ (in milligrams per liter, $\text{mg/L}$) leaking from an underground storage tank into a uniform, porous sand aquifer:

where $x$ and $y$ represent Cartesian coordinates in meters measured relative to the source leak.

To monitor groundwater remediation and design circular extraction well barriers, the engineering team transforms the analytical model from Cartesian coordinates to polar coordinates ($r, \theta$) via:

1. State the Case 2 Multivariable Chain Rule formulas for $\frac{\partial C}{\partial r}$ and $\frac{\partial C}{\partial \theta}$.
2. Compute the Cartesian partial derivatives $\frac{\partial C}{\partial x}$ and $\frac{\partial C}{\partial y}$.
3. Evaluate the radial concentration gradient $\frac{\partial C}{\partial r}$ at a monitoring radius of $r = 10\text{ m}$.
4. Show analytically that $\frac{\partial C}{\partial \theta} = 0$, and interpret the environmental significance of this result for circular barrier placement.

Step-by-Step Solution:

1. State the Governing Multivariable Chain Rule (Case 2): Because $C$ depends on intermediate variables $x$ and $y$, which each depend on independent polar coordinates $r$ and $\theta$:

   $$\frac{\partial C}{\partial r} = \frac{\partial C}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial C}{\partial y}\frac{\partial y}{\partial r}$$
   $$\frac{\partial C}{\partial \theta} = \frac{\partial C}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial C}{\partial y}\frac{\partial y}{\partial \theta}$$
   (Defined in Multivariable Chain Rule.)

2. Compute Cartesian Partial Derivatives: Using the single-variable exponential chain rule:

   $$\frac{\partial C}{\partial x} = 100 e^{-0.02(x^2+y^2)} \cdot \frac{\partial}{\partial x}[-0.02(x^2+y^2)] = 100 e^{-0.02(x^2+y^2)} (-0.04x) = -4x e^{-0.02(x^2+y^2)}$$
   $$\frac{\partial C}{\partial y} = 100 e^{-0.02(x^2+y^2)} \cdot \frac{\partial}{\partial y}[-0.02(x^2+y^2)] = 100 e^{-0.02(x^2+y^2)} (-0.04y) = -4y e^{-0.02(x^2+y^2)}$$

3. Compute Polar Intermediate Derivatives:

   • With respect to $r$:
     $$\frac{\partial x}{\partial r} = \cos\theta, \qquad \frac{\partial y}{\partial r} = \sin\theta$$
   • With respect to $\theta$:
     $$\frac{\partial x}{\partial \theta} = -r\sin\theta, \qquad \frac{\partial y}{\partial \theta} = r\cos\theta$$

4. Assemble and Simplify $\frac{\partial C}{\partial r}$:

   $$\frac{\partial C}{\partial r} = \left(-4x e^{-0.02(x^2+y^2)}\right)(\cos\theta) + \left(-4y e^{-0.02(x^2+y^2)}\right)(\sin\theta)$$
   Factor out $-4 e^{-0.02(x^2+y^2)}$:
   $$\frac{\partial C}{\partial r} = -4 e^{-0.02(x^2+y^2)} (x\cos\theta + y\sin\theta)$$
   Substitute $x = r\cos\theta$, $y = r\sin\theta$, and $x^2 + y^2 = r^2$:
   $$x\cos\theta + y\sin\theta = r\cos^2\theta + r\sin^2\theta = r(\cos^2\theta + \sin^2\theta) = r$$
   $$\frac{\partial C}{\partial r} = -4r e^{-0.02r^2}$$
   • Evaluate at $r = 10\text{ m}$:
     $$\frac{\partial C}{\partial r}\Big|_{r=10} = -4(10) e^{-0.02(10)^2} = -40 e^{-2.0} \approx -40(0.1353) = -5.41\text{ (mg/L)/m}$$
     The concentration decreases at $5.41\text{ mg/L}$ per meter as one moves outward from the source.

5. Assemble and Simplify $\frac{\partial C}{\partial \theta}$:

   $$\frac{\partial C}{\partial \theta} = \left(-4x e^{-0.02(x^2+y^2)}\right)(-r\sin\theta) + \left(-4y e^{-0.02(x^2+y^2)}\right)(r\cos\theta)$$
   Factor out $-4 e^{-0.02 r^2}$:
   $$\frac{\partial C}{\partial \theta} = -4 e^{-0.02 r^2} \left( -x r\sin\theta + y r\cos\theta \right)$$
   Substitute $x = r\cos\theta$ and $y = r\sin\theta$:
   $$-x r\sin\theta + y r\cos\theta = -(r\cos\theta)(r\sin\theta) + (r\sin\theta)(r\cos\theta) = -r^2\sin\theta\cos\theta + r^2\sin\theta\cos\theta = 0$$
   $$\frac{\partial C}{\partial \theta} = 0$$

6. Environmental Engineering Interpretation: Because $\frac{\partial C}{\partial \theta} = 0$, the plume exhibits perfect radial symmetry (isotropic dispersion). All monitoring wells drilled at radius $r = 10\text{ m}$ will measure identical pollutant concentrations ($C(10) = 100e^{-2} \approx 13.53\text{ mg/L}$) regardless of angular orientation $\theta$, allowing engineers to build concentric extraction rings that treat the entire perimeter equally.