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

Given:

f(x, y, z) = x² + 2y² − 2√11 z
Point P = (1, 1, √2)

Direction is along the tangent to the curve of intersection of:
2x² + 3y² − z² = 3
x² + y² = z²

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Step 1: Gradient of the given function

∇f = (2x, 4y, −2√11)

At P(1, 1, √2):

∇f = (2, 4, −2√11)

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Step 2: Direction of the tangent to the curve

Let
φ₁ = 2x² + 3y² − z²
φ₂ = x² + y² − z²

∇φ₁ = (4x, 6y, −2z)
∇φ₂ = (2x, 2y, −2z)

At P(1, 1, √2):

∇φ₁ = (4, 6, −2√2)
∇φ₂ = (2, 2, −2√2)

A vector along the tangent is:

t = ∇φ₁ × ∇φ₂ = (−8√2, 4√2, −4)

|t| = √176

Unit direction vector:

t̂ = (−8√2/√176, 4√2/√176, −4/√176)

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Step 3: Directional derivative

D = ∇f · t̂

D = [2(−8√2) + 4(4√2) + (−2√11)(−4)] / √176

D = 8√11 / √176 = 2

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Final Answer:

Absolute value of the directional derivative,
|D| = 2