Find the directional derivative of V^2 = V ⋅ V, where V = xy^2 i + zy^2 j + xz^2 k at the point (2,0,3) in the direction of the outward normal to the sphere x^2 + y^2 + z^2 = 14 at... Find the directional derivative of V^2 = V ⋅ V, where V = xy^2 i + zy^2 j + xz^2 k at the point (2,0,3) in the direction of the outward normal to the sphere x^2 + y^2 + z^2 = 14 at the point (3,2,1). Read more

Steps to Solve

1. Calculate the Gradient of the Vector Field

First, we need to compute the gradient of the scalar function represented by the vector field ( V ). Here, we have:

$$ V = xy^2 , \hat{i} + zy^2 , \hat{j} + xz^2 , \hat{k} $$

To find the directional derivative, we first find ( \nabla V^2 ).

Using the product rule:

$$ V^2 = V \cdot V = (xy^2)^2 + (zy^2)^2 + (xz^2)^2 $$

We can calculate the gradient:

$$ \nabla V^2 = \left( \frac{\partial V^2}{\partial x}, \frac{\partial V^2}{\partial y}, \frac{\partial V^2}{\partial z} \right) $$

Calculating each component at the point (2, 0, 3):

• For ( x ):

$$ \frac{\partial V^2}{\partial x} = 2xy^2 + 2xz^2 = 2(2)(0^2) + 2(2)(3^2) = 0 + 2(2)(9) = 36 $$

• For ( y ):

$$ \frac{\partial V^2}{\partial y} = 2y(x^2 + z^2) = 2(0)((2)^2 + (3)^2) = 0 $$

• For ( z ):

$$ \frac{\partial V^2}{\partial z} = 2z(y^2 + x^2) = 2(3)(0^2 + 2^2) = 2(3)(4) = 24 $$

Thus:

$$ \nabla V^2(2, 0, 3) = (36, 0, 24) $$

2. Find the Outward Normal to the Sphere

The equation of the sphere is:

$$ x^2 + y^2 + z^2 = 14 $$

The outward normal vector to the surface at a point ( (3, 2, 1) ) is given by:

$$ \mathbf{N} = \nabla(x^2 + y^2 + z^2) = (2x, 2y, 2z) $$

At the point (3, 2, 1):

$$ \mathbf{N} = (2(3), 2(2), 2(1)) = (6, 4, 2) $$

3. Normalize the Normal Vector

To find the unit normal vector, we normalize ( \mathbf{N} ):

$$ |\mathbf{N}| = \sqrt{6^2 + 4^2 + 2^2} = \sqrt{36 + 16 + 4} = \sqrt{56} = 2\sqrt{14} $$

Thus:

$$ \mathbf{u} = \frac{1}{|\mathbf{N}|} \mathbf{N} = \left( \frac{6}{2\sqrt{14}}, \frac{4}{2\sqrt{14}}, \frac{2}{2\sqrt{14}} \right) = \left( \frac{3}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{1}{\sqrt{14}} \right) $$

4. Calculate the Directional Derivative

The directional derivative in the direction of ( \mathbf{u} ) is given by:

$$ D_{\mathbf{u}} V^2 = \nabla V^2 \cdot \mathbf{u} $$

Calculating the dot product:

$$ D_{\mathbf{u}} V^2 = (36, 0, 24) \cdot \left( \frac{3}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{1}{\sqrt{14}} \right) $$

This yields:

$$ D_{\mathbf{u}} V^2 = 36 \cdot \frac{3}{\sqrt{14}} + 0 \cdot \frac{2}{\sqrt{14}} + 24 \cdot \frac{1}{\sqrt{14}} $$

Simplifying:

$$ D_{\mathbf{u}} V^2 = \frac{108 + 24}{\sqrt{14}} = \frac{132}{\sqrt{14}} $$

5. Final Simplification

To express it neatly:

$$ D_{\mathbf{u}} V^2 = \frac{66}{7} \sqrt{14} $$