The value of the directional derivative of the - function x-y-z-xy2-yz2-zx2 at the point 2-1-1 in the direction of the vector p-i-2j-2k is

Q=xy^2+yz^2+zx^2 then Grad ϕ=î∂ϕ∂ x+ĵ∂ϕ∂ y+k̂∂ϕ∂ z i.e. Grad ϕ=(y^2+2xz)î+(2xy+z^2)ĵ+(2yz+x^2)k̂ so (grad ϕ)_p = [ ( 1)^2+2(2)(1)k̂ ]+ [ 2(2)( 1)+(1)^2ĵ ]+ [ 2 ...

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Engineering Mathematics

Directional Derivative

The value of ...

Question

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 $p = i + 2 j + 2 k$ is

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Solution

The correct option is A $1$
$Q = x y^{2} + y z^{2} + z x^{2}$ then
$G r a d ϕ =^i \frac{\partial ϕ}{\partial x} +^j \frac{\partial ϕ}{\partial y} +^k \frac{\partial ϕ}{\partial z}$
$i . e .$
$G r a d ϕ = (y^{2} + 2 x z)^i + (2 x y + z^{2})^j + (2 y z + x^{2})^k$
so $(g r a d ϕ)_{p}$
$= [(- 1)^{2} + 2 (2) (1)^k] + [2 (2) (- 1) + (1)^{2}^j] + [2]^k$
$= 5^i - 3^j + 2^k$
So required $D D$
$(g r a d ϕ)_{p} .^p = (5^i - 3^j + 2^k) . (\frac{^i + 2^j + 2^k}{\sqrt{9}})$
$= \frac{5 - 6 + 4}{3} = 1$

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The value of the directional derivative of the ϕ function (x,y,z)=xy2+yz2+zx2 at the point (2,−1,1) in the direction of the vector p=i+2j+2k is

The correct option is A 1Q=xy2+yz2+zx2 then Gradϕ=^i∂ϕ∂x+^j∂ϕ∂y+^k∂ϕ∂z i.e. Gradϕ=(y2+2xz)^i+(2xy+z2)^j+(2yz+x2)^k so (gradϕ)p =[(−1)2+2(2)(1)^k]+[2(2)(−1)+(1)2^j]+[2]^k =5^i−3^j+2^k So required DD (gradϕ)p.^p=(5^i−3^j+2^k).(^i+2^j+2^k√9) =5−6+43=1

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 $p = i + 2 j + 2 k$ is