Step 1: Problem Definition:
The objective is to compute the directional derivative of a scalar function at a specified point in a particular direction. The scalar function is the Laplacian of \(f\), defined as \( \phi = abla \cdot (abla f) = abla^2 f \). The direction is determined by the normal vector to a given surface.
Step 2: Methodology:
1. Compute the Laplacian of \(f\), denoted as \( \phi = abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} \).
2. Determine the gradient of the scalar function \( \phi \), represented as \( abla \phi \).
3. Calculate the gradient of the surface equation \( S(x,y,z) = xy^2z - 3x - z^2 = 0 \) to find the normal vector \( abla S \).
4. Obtain the unit vector \( \hat{u} \) corresponding to the normal direction.
5. The directional derivative is calculated using the formula \( D_{\hat{u}}\phi = abla \phi \cdot \hat{u} \).
Step 3: Execution:
1. Calculation of \( \phi = abla^2 f \):
Given \( f = 2x^3y^2z^4 \).
Partial derivatives:
\( \frac{\partial f}{\partial x} = 6x^2y^2z^4 \implies \frac{\partial^2 f}{\partial x^2} = 12xy^2z^4 \)
\( \frac{\partial f}{\partial y} = 4x^3yz^4 \implies \frac{\partial^2 f}{\partial y^2} = 4x^3z^4 \)
\( \frac{\partial f}{\partial z} = 8x^3y^2z^3 \implies \frac{\partial^2 f}{\partial z^2} = 24x^3y^2z^2 \)
Therefore, \( \phi = abla^2 f = 12xy^2z^4 + 4x^3z^4 + 24x^3y^2z^2 \).
2. Gradient of \( \phi \) at (1, -2, 1):
Evaluate \( \phi \) at the point (1, -2, 1):
\( \phi(1, -2, 1) = 12(1)(-2)^2(1)^4 + 4(1)^3(1)^4 + 24(1)^3(-2)^2(1)^2 = 48 + 4 + 96 = 148 \).
Compute the partial derivatives of \( \phi \):
\( \frac{\partial \phi}{\partial x} = 12y^2z^4 + 12x^2z^4 + 72x^2y^2z^2 \). At (1,-2,1): \( 12(4)(1) + 12(1)(1) + 72(1)(4)(1) = 48 + 12 + 288 = 348 \).
\( \frac{\partial \phi}{\partial y} = 24xyz^4 + 48x^3yz^2 \). At (1,-2,1): \( 24(1)(-2)(1) + 48(1)(-2)(1) = -48 - 96 = -144 \).
\( \frac{\partial \phi}{\partial z} = 48xy^2z^3 + 16x^3z^3 + 48x^3y^2z \). At (1,-2,1): \( 48(1)(4)(1) + 16(1)(1) + 48(1)(4)(1) = 192 + 16 + 192 = 400 \).
The gradient is \( abla \phi(1, -2, 1) = 348\hat{i} - 144\hat{j} + 400\hat{k} \).
3. Normal Vector to the Surface:
For the surface \( S(x,y,z) = xy^2z - 3x - z^2 \), the gradient is:
\( abla S = (y^2z - 3)\hat{i} + (2xyz)\hat{j} + (xy^2 - 2z)\hat{k} \).
Evaluate \( abla S \) at (1, -2, 1):
\( abla S = ((-2)^2(1) - 3)\hat{i} + (2(1)(-2)(1))\hat{j} + ((1)(-2)^2 - 2(1))\hat{k} \)
\( abla S = (4 - 3)\hat{i} - 4\hat{j} + (4 - 2)\hat{k} = 1\hat{i} - 4\hat{j} + 2\hat{k} \).
This vector, \( \vec{v} = \hat{i} - 4\hat{j} + 2\hat{k} \), indicates the direction.
4. Unit Normal Vector \( \hat{u} \):
The magnitude of \( \vec{v} \) is \( |\vec{v}| = \sqrt{1^2 + (-4)^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21} \).
The unit vector is \( \hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{\sqrt{21}}(\hat{i} - 4\hat{j} + 2\hat{k}) \).
5. Directional Derivative Calculation:
The directional derivative is \( D_{\hat{u}}\phi = abla \phi \cdot \hat{u} \).
\( D_{\hat{u}}\phi = (348\hat{i} - 144\hat{j} + 400\hat{k}) \cdot \frac{1}{\sqrt{21}}(\hat{i} - 4\hat{j} + 2\hat{k}) \)
\( = \frac{1}{\sqrt{21}} [348(1) + (-144)(-4) + 400(2)] \)
\( = \frac{1}{\sqrt{21}} [348 + 576 + 800] = \frac{1724}{\sqrt{21}} \).
Step 4: Conclusion:
The computed directional derivative is \( \frac{1724}{\sqrt{21}} \).