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Use Theorem 12.32 to find the indicated derivatives in Exercises

21–26. Express your answers as functions of a single variable

z=x2+y2,P=(-3,4),v=(4,-3)

Short Answer

Expert verified

The required directional derivative of the function is ∇z·u=-0.96

Step by step solution

01

Given information

Think about the following function.

z=x2+y2

02

The objective is to find the directional derivative of function at the point P=(-3,4) 

∇z=i∂z∂x+j∂z∂yz=x2+y2∂z∂x=2x2x2+y2∂z∂x=xx2+y2

Then,

∇z=i∂z∂x+j∂z∂y⇒∇z=ixx2+y2+jyx2+y2

Consider the vector below.

v=4i-3j

Along the vectorv=4i-3j, there is a unit vector

u=4i-3j(4)2+(-3)2⇒u=4i-3j16+9⇒u=4i-3j5

03

The directional derivatives of the function z in the direction

∇z·u=ixx2+y2+jyx2+y2·4i-3j5∇z·u=4x5x2+y2-3y5x2+y2

At point P=(-3,4)

∇z·u=4(-3)5(-3)2+(4)2-3(4)(-3)2+(4)25∇z·u=-0.96

Hence, the directional derivative of the function is

∇z·u=-0.96

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