]> Exercise 4

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007)

Chapter 3: Many Variables; Section 1: Differentiation; page 2

Partial Derivatives, Exercise 4

Question

The function z of two variables x and y is presented in the implicit form:

F( x,y,z )=x​ z 5 +y​ z 2 ​−x​ y 2 ​z−1=0
  1. Calculate ∂F ∂z ,  ∂F ∂y  and  ∂F ∂x

  2. What are ∂z ∂x  and  ∂z ∂y

  3. What is the value of z for the point (x, y)=(1, 0) ?
  4. Obtain the tangent plane at

    ( x,y ) 0 =( 1,0 )

Reminder

We define the partial derivative of a function

y=f( x 1 ,.., x k ,.., x n ) (3.1.2.1)

with respect to the variable xk, as the limit

∂y ∂ x k = ∂f ∂ x k = lim⁡ Δ x k →0 f( x 1 ,.., x k−1 , x k +Δ x k , x k+1 ,.., x n )−f( x 1 ,.., x k ,.., x n ) Δ x k (3.1.2.2)

In the case of a function of two variables,

z=z( x,y ) (3.1.2.34)

the first approximation is the tangent plane:

z 1 = z 0 + ( ∂z ∂x ) 0 ( x− x 0 )+ ( ∂z ∂y ) 0 ( y− y 0 ) (3.1.2.35)
F[ x,y,z( x,y ) ]=0 { ∂z ∂x =− ∂F ∂x ∂F ∂z ∂z ∂y =− ∂F ∂y ∂F ∂z } (3.1.2.37)

With the help of (3.1.2.37), one can transform the expression of the tangent plane (3.1.2.35) in the following symmetric form:

( ∂F ∂z ) 0 ( z− z 0 )+ ( ∂F ∂y ) 0 ( y− y 0 )+ ( ∂F ∂x ) 0 ( x− x 0 )=0 (3.1.2.38)

Parts 1-3

Solution of question 1

  1.  
    ∂F ∂z =5x z 4 +2yz−x y 2
  2.  
    ∂F ∂y = z 2 −2xyz=z( z−2xy )
  3.  
    ∂F ∂x = z 5 − y 2 z=z( z 4 − y 2 )

Parts 4-5

Solution of question 2

  1.  
    ∂z ∂x =− ∂F ∂x ∂F ∂z =− z( z 4 − y 2 ) 5x z 4 +2yz−x y 2 = z( y 2 − z 4 ) 5x z 4 +2yz−x y 2
  2.  
    ∂z ∂y =− ∂F ∂y ∂F ∂z =− z( z−2xy ) 5x z 4 +2yz−x y 2 = z( 2xy−z ) 5x z 4 +2yz−x y 2

Part 6

Solution of question 3

  1. The substitution of x=1 and y=0 in the implicit form of the function yields:

    z 5 +0−0−1=0
    therefore z=1

Parts 7-9

Solution of question 4

  1. The required tangent plane should be obtained by the substitution of

    ( x,y,z ) 0 =( 1,0,1 ) in (3.1.2.38)
  2. The partial derivatives become

    ( ∂F ∂z ) 0 =5 ( ∂F ∂y ) 0 =1 ( ∂F ∂x ) 0 =1 }

  3. and the tangent plane:

    5( z−1 )+y+( x−1 )=0 or 5z+y+x=6

Score

By parts.

Parts 1, 2, 3, 4, 5, 7, 8, 9 are worth 1 point each.
Part 6 is worth 2 points.

By questions.

Questions 1 and 4 are worth 3 points each.
Questions 2 and 3 are worth 2 points each.