Directional derivative of - -2xz-y2 at the piont 1-3-2 becomes maximum in the direction of

Grad ϕ=▽ϕ=∂ϕ∂ xî+∂ϕ∂ yĵ+∂ϕ∂ zk̂ =2zî 2yĵ+2xk̂ So, (grad ϕ)_(1,3,2)=(4î 6ĵ+2k̂) and we know that directional derivative is max in the direction of grad ϕ so Norm ...

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Standard IX

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The October Revolution

Directional d...

Question

Directional derivative of $ϕ = 2 x z - y^{2}$ at the piont $(1, 3, 2)$ becomes maximum in the direction of

4^i + 2^j - 3^k

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4^i - 6^j + 2^k

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2^i - 6^j + 2^k

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4^i - 6^j - 2^k

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Solution

The correct option is B $4^i - 6^j + 2^k$
$g r a d ϕ = ▽ ϕ = \frac{\partial ϕ}{\partial x}^i + \frac{\partial ϕ}{\partial y}^j + \frac{\partial ϕ}{\partial z}^k$
$= 2 z^i - 2 y^j + 2 x^k$
So, $(g r a d ϕ)_{(1, 3, 2)} = (4^i - 6^j + 2^k)$ and we know that directional derivative is max in the direction of $g r a d ϕ$ so Normal vector is the required vector.

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Directional derivative of ϕ=2xz−y2 at the piont (1,3,2) becomes maximum in the direction of

The correct option is B 4^i−6^j+2^kgradϕ=▽ϕ=∂ϕ∂x^i+∂ϕ∂y^j+∂ϕ∂z^k =2z^i−2y^j+2x^k So, (gradϕ)(1,3,2)=(4^i−6^j+2^k) and we know that directional derivative is max in the direction of gradϕ so Normal vector is the required vector.

Directional derivative of $ϕ = 2 x z - y^{2}$ at the piont $(1, 3, 2)$ becomes maximum in the direction of