# Find the maximum value of the directional derivative at the given point? Let f(x,y,z)=x^2+y^2+z^2-3xy+2xz-yz. Find the maximum value of the directional derivative at the point (1,1,1). Give your...

Find the maximum value of the directional derivative at the given point?

Let f(x,y,z)=x^2+y^2+z^2-3xy+2xz-yz. Find the maximum value of the directional derivative at the point (1,1,1). Give your answer to 2 decimal places.

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Student Comments

pramodpandey | Student

The maximum occure in direction of gradF.

`F= x^2+y^2+z^2-3xy+2xz-yz`

`gradF=<2x-3y+2z,2y-3x-z,2z+2x-y>`

`Thus `

`gradF(1,1,1)=<1,-2,3>`

`u=(gradF(1,1,1))/|gradF(1,11)|`

`u=<1/sqrt(1^2+(-2)^2+(3)^2),-2/sqrt(14),3/sqrt(14))`

``So maximum value of directional derivative at (1,1,1)

`gradF(1,1,1)*u=<1,-2,3>*(1/sqrt(14))<1,-2,3>`

`=(1/sqrt(14))(1xx1+(-2)xx(-2)+3xx3)`

`=(1/sqrt(14))(14)`

`=sqrt(14)`

`=3.74`

ans.

This will occur in the direction of the gradient of f:

∇f = <2x - 3y + 2z, 2y - 3x - z, 2z + 2x - y>.

So, ∇f(1, 1, 1) = <1, -2, 3>.

Normalizing yields u = ∇f(1, 1, 1) / ||∇f(1, 1, 1)|| = (1/√14) <1, -2, 3>.

So, the maximal value of the directional derivative at (1, 1, 1) is

∇f(1, 1, 1) · u = <1, -2, 3> · (1/√14) <1, -2, 3> = 14/√14 = √14.