Question

Let \( \hat{a} \) be a unit vector parallel to the tangent at the point \( P(1, 1, \sqrt{2}) \) to the curve of intersection of the surfaces \( 2x^2 + 3y^2 - z^2 = 3 \) and \( x^2 + y^2 = z^2 \). Then, the absolute value of the directional derivative of \[ f(x, y, z) = x^2 + 2y^2 - 2\sqrt{11} z \] at P in the direction of \( \hat{a} \) is _________ (in integer).

Hint:

To compute the directional derivative, find the gradient of the function and the unit vector in the direction of the tangent, then compute their dot product.

Answer 1

We are tasked with finding the absolute value of the directional derivative of the function \( f(x, y, z) = x^2 + 2y^2 - 2\sqrt{11}z \) at the point \( P(1, 1, \sqrt{2}) \) in the direction of the unit vector \( \hat{a} \), which is parallel to the tangent to the curve of intersection of the surfaces \( 2x^2 + 3y^2 - z^2 = 3 \) and \( x^2 + y^2 = z^2 \).

Step 1: Compute the gradient of \( f(x, y, z) \)

The gradient of \( f(x, y, z) \) is:
\[ \nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (2x, 4y, -2\sqrt{11}) \]
At the point \( P(1, 1, \sqrt{2}) \), the gradient becomes:
\[ \nabla f(1, 1, \sqrt{2}) = (2, 4, -2\sqrt{11}) \]

Step 2: Find the direction vector \( \hat{a} \)

Let:

• \( g_1(x, y, z) = 2x^2 + 3y^2 - z^2 \)
• \( g_2(x, y, z) = x^2 + y^2 - z^2 \)

Then: \[ \nabla g_1 = (4x, 6y, -2z), \quad \nabla g_2 = (2x, 2y, -2z) \] At point \( (1, 1, \sqrt{2}) \), we compute: \[ \nabla g_1 = (4, 6, -2\sqrt{2}), \quad \nabla g_2 = (2, 2, -2\sqrt{2}) \] Now take the cross product: \[ \hat{a} = \nabla g_1 \times \nabla g_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 6 & -2\sqrt{2} \\ 2 & 2 & -2\sqrt{2} \end{vmatrix} = \hat{i}(6 \cdot -2\sqrt{2} - (-2\sqrt{2} \cdot 2)) - \hat{j}(4 \cdot -2\sqrt{2} - (-2\sqrt{2} \cdot 2)) + \hat{k}(4 \cdot 2 - 6 \cdot 2) \] \[ = \hat{i}(-12\sqrt{2} + 4\sqrt{2}) - \hat{j}(-8\sqrt{2} + 4\sqrt{2}) + \hat{k}(8 - 12) = \hat{i}(-8\sqrt{2}) + \hat{j}(4\sqrt{2}) - 4\hat{k} \] Thus, the direction vector is: \[ \vec{a} = (-8\sqrt{2}, 4\sqrt{2}, -4) \] Normalize it to get unit vector \( \hat{a} \): \[ |\vec{a}| = \sqrt{(-8\sqrt{2})^2 + (4\sqrt{2})^2 + (-4)^2} = \sqrt{128 + 32 + 16} = \sqrt{176} \] \[ \hat{a} = \left( \frac{-8\sqrt{2}}{\sqrt{176}}, \frac{4\sqrt{2}}{\sqrt{176}}, \frac{-4}{\sqrt{176}} \right) \]

Step 3: Compute the directional derivative

Use the dot product \( D_{\hat{a}}f = \nabla f \cdot \hat{a} \):
\[ \nabla f = (2, 4, -2\sqrt{11}), \quad \hat{a} = \left( \frac{-8\sqrt{2}}{\sqrt{176}}, \frac{4\sqrt{2}}{\sqrt{176}}, \frac{-4}{\sqrt{176}} \right) \] \[ D_{\hat{a}}f = 2 \cdot \frac{-8\sqrt{2}}{\sqrt{176}} + 4 \cdot \frac{4\sqrt{2}}{\sqrt{176}} + (-2\sqrt{11}) \cdot \frac{-4}{\sqrt{176}} \] \[ = \frac{-16\sqrt{2} + 16\sqrt{2} + 8\sqrt{11}}{\sqrt{176}} = \frac{8\sqrt{11}}{\sqrt{176}} = \frac{8\sqrt{11}}{4\sqrt{11}} = 2 \] So the absolute value is: \[ \boxed{2} \]