Question

The directional derivative of \( \nabla \cdot (\nabla f) \) at the point (1, -2, 1) in the direction of the normal to the surface \( xy^2z = 3x + z^2 \) where \( f = 2x^3y^2z^4 \) and \( \nabla = \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \) is

• A. \( \frac{1724}{\sqrt{21}} \)
• B. \( \frac{1724}{\sqrt{23}} \)
• C. \( \frac{1724}{\sqrt{19}} \)
• D. \( \frac{1724}{\sqrt{29}} \)

Hint:

Break down complex vector calculus problems into smaller, manageable steps: 1. Identify the function you're differentiating. 2. Identify the direction vector. 3. Compute the gradient of the function. 4. Normalize the direction vector. 5. Take the dot product. Be methodical with partial derivatives to avoid errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem asks for the directional derivative of a scalar function at a given point in a specified direction. The scalar function is \( \phi = \nabla \cdot (\nabla f) = \nabla^2 f \), which is the Laplacian of \(f\). The direction is that of the normal to a given surface.
Step 2: Key Formula or Approach:
1. Calculate the scalar function \( \phi = \nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} \). 2. Find the gradient of this new scalar function, \( \nabla \phi \). 3. Find the normal vector to the surface \( S(x,y,z) = xy^2z - 3x - z^2 = 0 \) by calculating its gradient, \( \nabla S \). 4. Find the unit vector \( \hat{u} \) in the direction of the normal. 5. The directional derivative is \( D_{\hat{u}}\phi = \nabla \phi \cdot \hat{u} \).
Step 3: Detailed Explanation:
1. Calculate \( \phi = \nabla^2 f \):
\( f = 2x^3y^2z^4 \) \( \frac{\partial f}{\partial x} = 6x^2y^2z^4 \implies \frac{\partial^2 f}{\partial x^2} = 12xy^2z^4 \) \( \frac{\partial f}{\partial y} = 4x^3yz^4 \implies \frac{\partial^2 f}{\partial y^2} = 4x^3z^4 \) \( \frac{\partial f}{\partial z} = 8x^3y^2z^3 \implies \frac{\partial^2 f}{\partial z^2} = 24x^3y^2z^2 \) \( \phi = \nabla^2 f = 12xy^2z^4 + 4x^3z^4 + 24x^3y^2z^2 \)
2. Find \( \nabla \phi \) at (1, -2, 1):
First, find \( \phi(1, -2, 1) = 12(1)(-2)^2(1)^4 + 4(1)^3(1)^4 + 24(1)^3(-2)^2(1)^2 = 48 + 4 + 96 = 148 \). Now find the gradient 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): \( 48(1)(4)(1) + 16(1)(1) + 48(1)(4)(1) = 192 + 16 + 192 = 400 \). So, \( \nabla \phi(1, -2, 1) = 348\hat{i} - 144\hat{j} + 400\hat{k} \).
3. Find the normal vector to the surface:
Let \( S(x,y,z) = xy^2z - 3x - z^2 \). \( \nabla S = (y^2z - 3)\hat{i} + (2xyz)\hat{j} + (xy^2 - 2z)\hat{k} \). At (1, -2, 1): \( \nabla S = ((-2)^2(1) - 3)\hat{i} + (2(1)(-2)(1))\hat{j} + ((1)(-2)^2 - 2(1))\hat{k} \) \( \nabla S = (4 - 3)\hat{i} - 4\hat{j} + (4 - 2)\hat{k} = 1\hat{i} - 4\hat{j} + 2\hat{k} \). This is the direction vector \(\vec{v}\).
4. Find the unit vector \( \hat{u} \):
\( |\vec{v}| = \sqrt{1^2 + (-4)^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21} \). \( \hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{\sqrt{21}}(1\hat{i} - 4\hat{j} + 2\hat{k}) \).
5. Calculate the directional derivative:
\( D_{\hat{u}}\phi = \nabla \phi \cdot \hat{u} = (348\hat{i} - 144\hat{j} + 400\hat{k}) \cdot \frac{1}{\sqrt{21}}(1\hat{i} - 4\hat{j} + 2\hat{k}) \) \( = \frac{1}{\sqrt{21}} [348(1) + (-144)(-4) + 400(2)] \) \( = \frac{1}{\sqrt{21}} [348 + 576 + 800] = \frac{1724}{\sqrt{21}} \). Step 4: Final Answer:
The directional derivative is \( \frac{1724}{\sqrt{21}} \).