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 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

History

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4^i - 6^j + 2^k

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2^i - 6^j + 2^k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4^i - 6^j - 2^k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

Q. The directional derivative of

ϕ = 3 x^{2} y - 4 y z^{2} + 6 z^{2} x

at point (1,1,1) in the direction of line

\frac{x - 1}{2} = \frac{y - 4}{2} = \frac{z}{3}

Q. The directional derivative of the following function at

(1, 2)

in the direction

(4^i + 3^j)

is:

f (x, y) = x^{2} + y^{2}

Q. The value of the directional derivative of the

ϕ

function

(x, y, z) = x y^{2} + y z^{2} + z x^{2}

at the point

(2, - 1, 1)

in the direction of the vector

p = i + 2 j + 2 k

Q. The magnitude of the directional derivative of the function

f (x, y) = x^{2} + 3 y^{2}

in a direction normal to the circle

x^{2} + y^{2} = 2,

at the point

(1, 1),

Q. The directional derivative

f (x, y, z) = 2 x^{2} + 3 y^{2} + z^{2}

at point

P (2, 1, 3)

in the direction of the vector

\to a = \to i - \to 2 k

Join BYJU'S Learning Program

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