In Exercises 21–28, find the directional derivative of the given

function at the specified point Pand in the direction of the

given unit vectoru.

f(x,y,z)=x2+y2z3atP=(2,2,2),u=35,0,45

Short Answer

Expert verified

The directional derivative of function is365

Step by step solution

01

Given data

The function is f(x,y,z)=x2+y2z3

The given points isP=x0,y0,z0=(2,2,2)andu=(α,β,γ)=35,0,45

02

Solution

Consider directional derivative

Dufx0,y0,z0=Limh0fx0+αh,y0+βh,z0+γhfx0,y0,z0h

Duf(2,2,2)=Limh0f2+35h,2+0h,2+45hf(2,2,2)h 1

Therefore,

f2+35h,2+0h,2+45h=2+35h2+(2+0h)22+45h3

=4+125h+925h2+48485h9625h264125h3

=365h8725h264125h3 2

And

fx0,y0,z0=f(2,2,2)=22+2223

f(2,2,2)=0 3

03

Substitute

Substituting

Dwf(2,2,2)=Limh0365h8725h264125h30h

=Limh03658725h64125h2

Dwf(2,2,2)=365

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