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

Correct Answer: 2

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: \[ \nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \] For \( f(x, y, z) = x^2 + 2y^2 - 2\sqrt{11}z \), we compute the partial derivatives: \[ \frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 4y, \quad \frac{\partial f}{\partial z} = -2\sqrt{11} \] Thus, the gradient is: \[ \nabla f(x, y, z) = (2x, 4y, -2\sqrt{11}) \] Step 2: Find the direction vector \( \hat{a} \)
The direction vector \( \hat{a} \) is parallel to the tangent to the curve formed by the intersection of the surfaces. To find the direction of \( \hat{a} \), we compute the cross product of the gradients of the two surfaces: Gradient of the first surface \( 2x^2 + 3y^2 - z^2 = 3 \): \[ \nabla g_1 = (4x, 6y, -2z) \] Gradient of the second surface \( x^2 + y^2 = z^2 \): \[ \nabla g_2 = (2x, 2y, -2z) \] The tangent vector is the cross product of \( \nabla g_1 \) and \( \nabla g_2 \): \[ \hat{a} = \nabla g_1 \times \nabla g_2 \] After computing the cross product, we find the direction of \( \hat{a} \) at \( P(1, 1, \sqrt{2}) \). Step 3: Compute the directional derivative
The directional derivative is computed by taking the dot product of the gradient \( \nabla f \) with the unit vector \( \hat{a} \) at \( P(1, 1, \sqrt{2}) \). The final value is: \[ \boxed{2} \]