answer tutorial 3.1 page 184 Module 1 (MAT235 Chapter 3)

Tutorial 3.1 Partial Derivatives

1. Jun 2019/ Q2b/ 6 marks
Given ( , ) = (1 + 2), show that 2 + 3 = 0.

2. Dec 2018/Q2c
Let = (1 − 2 + 2)−21 show that − = 2 3 .



3. Jun 2018/ Q2b/ 6 marks
Given ( , ) = (2 )2, show that 2 − 2 = 0.

∶

4. Jan 2018/ Q2b/ 6 marks 5. = 2 2 −

( , ) = 2+ 2. = +3
1+2 2
Let Show that 2 = 2 2 + 2 3 2 − 3
( +3)2


5. Mar 2017/ Q2b/ 6 marks 6. = 4 2 2 − 4 (2 + 3 )
Let ( , ) = 2 − ( + 3). Find and 2 = 2 2 + 4 2 − 6 (2 + 3 )



7. + = 2+ 2
( + )2
6. Oct 2016/ Q3b/ 5 marks
If = 2 + (2 + 3 ) find and . 8. = 15 2 2 + 2 2+2
= 10 3 + 2 2+2
7. Oct 2016/ Q4a/ 6 marks
= 30 2 + 8 2+2

If ( , ) = , find + .
+
+ = 2 2 ( ) +
9a.

8. Mar 2016/ Q3a/ 7 marks 2 ( ) − 1 2 ( )
If ( , ) = 5 3 2 + 2+2 , find , and 2


10. + = 3 3 − + 2 3 −



9. Oct 2015/ Q3a,b

a) Let ( , ) = 2 ( ) . Find + . (4 marks)
(5 marks)


b) Show that − = 0 if ( , ) = ( 2− ) + 3

10. Mar 2015/Q3a/ 6 marks
Let ( , ) = 3 − (1 + ). Find + .



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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Answer Tutorial 3.1 Partial Derivatives

1. Jun 2019/ Q2b/ 6 marks
Given ( , ) = (1 + 2), show that 2 + 3 = 0.

Solution

( , ) = (1 + 2)

= (1 + 2 )

= 1 1 ∙ (0 + 2(1))
+ 2

2 = 2 = 1 + 2
= 1 + 2 ′ = 2 ′ = 0 + ∙ 2
′ = 2

= 1 2 = 1 = (1 + )−1 2 2
+ 2 + = 1 + 2 = 1 + 2

= ∙ −(1 + )−2 ∙ ′ − ′
= 2 ∙ −(1 + 2 )−2 ∙ 2 = 2
= − 4(1 + 2 )−2
2 = −2 4(1 + 2 )−2 = 2 (1 + 2) − 2 3

−2 4 (1 + 2)2
2 = (1 + 2)2
2 + 2 3 − 2 3
= (1 + 2)2

2
= (1 + 2)2

3 = (1 2 4
+ 2)2

2 + 3 = (1 −2 4 + (1 2 4
+ 2)2 + 2)2

= 0 # ( ℎ )

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

3. Jun 2018/ Q2b/ 6 marks
Given ( , ) = (2 )2, show that 2 − 2 = 0.

Solution ( , ) = (2 )2
= (2 )2 = 2 (2 )
( , ) = (2 )2 1
= (2 )2 = 2 (2 ) = 2 ∙ (2 ) ∙ 2 (1)
1 4
= 2 ∙ (2 ) ∙ 2 (1) = (2 )
4
= (2 )

= 4 = = ( ( ))−1 = 4 = = ( )−1
(2 ) ( ) (2 ) ( )

= ∙ −( ( ))−2 ∙ 1 ∙ = ∙ −( ( ))−2 ∙ 1 ∙


= 4 ∙ −( (2 ))−2 ∙ 1 ∙ 2 = 4 ∙ −( (2 ))−2 ∙ 1 ∙ 2
2 2

−1 1 −1 1
= 4 ∙ ( (2 ))2 ∙ = 4 ∙ ( (2 ))2 ∙

−4 −4
= ( (2 ))2 = ( (2 ))2

2 = −4 2 2 = −4 2
( (2 ))2 ( (2 ))2

2 = −4 2 = −4
( (2 ))2 ( (2 ))2

2 − 2 = −4 − −4
( (2 ))2 ( (2 ))2

−4 4
= ( (2 ))2 + ( (2 ))2

= 0 # ( ℎ )

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

5. Mar 2017/ Q2b/ 6 marks

Let ( , ) = 2 − ( + 3). Find and 2


Solution

( , ) = 2 − ( + 3)

= 2 − ( + 3)

= = 2 ∙ 2(1) − 1 3 ∙ ( (1) + 0)
+

= = 2 2 − 3
+

= = 2 2 − 3 = 2 = 2
+ ′ = 2 ∙ (2 )
′ = 2 2
= = 2 2 − ′ = 2
+ = + 3
3 ′ = (1) + 0

2 = ′ =
= = − ′ = 1

2 = = ( ′ + ′) − ′ − ′
( 2 )

2 = ( 2 ∙ 2 + 2 ∙ 2 2 ) − + 3 −
= ( ( + 3)2 )

2 = (2 2 + 2 3 2 ) − (( 3 3)2)
= +

2 = 2 2 + 2 3 2 − 3 3)2
= +
(

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

6. Oct 2016/ Q3b/ 5 marks
If = 2 + (2 + 3 ) find and .

Solution

= 2 + (2 + 3 ) = 2 = 2
= 2 + (2 + 3 )
= 2 ∙ 2 (1) − (2 + 3 ) ∙ (2(1) + 0) ′ = 2(1) ′ = 2 ∙ 2 (1)
= 2 2 − 2 (2 + 3 )
′ = 2 ′ = 2 2
= 2 2 − 2 (2 + 3 )
= 2 2 − 2 (2 + 3 )

= 2 2 − 2 (2 + 3 ) = 2 2 − 2 (2 + 3 )

= 2 2 2 − 2 ∙ (2 + 3 ) ∙ (2(1) + 0) = ′ + ′ − 2 ∙ (2 + 3 ) ∙ (0 + 3(1))

= 4 2 2 − 4 (2 + 3 ) = 2 2 + 4 2 − 6 (2 + 3 )

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

7. Oct 2016/ Q4a/ 6 marks

If ( , ) = , find + .
+

Solution

= = + = = +
′ = 1 + 0 ′ = (1) ′ = 0 + 1
′ = (1) ′ = 1 ′ = ′ = 1
′ =

( , ) =
+ ( , ) = +


= + = +
′ − ′
′ − ′ = = 2
= = 2 ( + ) −
= = ( + )2
( + ) − 2 + −
= = ( + )2 = = ( + )2
2
+ 2 − = = ( + )2
= = ( + )2

2
= = ( + )2

2 2
+ = ( + )2 + ( + )2

2 + 2
= ( + )2 #

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

8. Mar 2016/ Q3a/ 7 marks
If ( , ) = 5 3 2 + 2+2 , find , and

Solution

( , ) = 5 3 2 + 2+2 ( , ) = 5 3 2 + 2+2
= 5 2 3 + 2+2 = 5 3 2 + 2+2
= 5 2(3 2) + 2+2 ∙ (2 + 0) = 5 3(2 ) + 2+2 ∙ (0 + 2(1))
= 15 2 2 + 2 2+2 = 10 3 + 2 2+2

= 10 3 + 2 2+2
= 10 3(1) + 2 2+2 ∙ (0 + 2(1))
= 10 3 + 4 2+2

= 10 3 + 4 2+2
= 10(3 2) + 4 2+2 ∙ (2 + 0)
= 30 2 + 8 2+2

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

9. Oct 2015/ Q3a,b (4 marks)
(5 marks)
a) Let ( , ) = 2 ( ) . Find + .
(4 marks)


b) Show that − = 0 if ( , ) = ( 2− ) + 3

Solution
a) Let ( , ) = 2 ( ) . Find + .



= 2 = = ( −1)
′ = 2 ∙ 2(1) ( )
′ = 2 2
′ = − ( ) ∙ (− −2)

′ = ∙ −
− ( ) 2

′ = 2 ( )

( , ) = 2 ( , ) = 2
( ) ( )

= 2 = 2 1 ∙ )
( ) (

= = ′ + ′ = = 2 ∙ − ∙ 1 (1))
( ) (

= = 2 2 2 = = 2 ∙ − ∙ 1
( ) + 2 ( ) ( )

= = − 1 2
( )

= 2 2 2 + [− 1 2
+ ( ) + 2 ( ) ( )]

+ = 2 2 + 2 − 1 2 #
( ) 2 ( ) ( )

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

9.b) Show that − = 0 if ( , ) = ( 2− ) + 3 (5 marks)

Solution ( , ) = ( 2− ) + 3
= ( 2− ) + 3
( , ) = ( 2− ) + 3 = ( 2− ) ∙ (0 − 1) + (3 2)
= ( 2− ) + 3 = − ( 2− ) + 3 2
= ( 2− ) ∙ (2 − 0) + 3(1)
= 2 ( 2− ) + 3

= 2 ( 2− ) + 3 = − ( 2− ) + 3 2
= 2 ∙ ( 2− ) ∙ (0 − 1) + 3 2 = − ( 2− ) ∙ (2 − 0) + 3 2(1)
= −2 ( 2− ) + 3 2 = −2 ( 2− ) + 3 2

− = [−2 ( 2− ) + 3 2] − [−2 ( 2− ) + 3 2 ]
− = −2 ( 2− ) + 3 2 + 2 ( 2− ) − 3 2
− = 0 # ( ℎ )

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

10. Mar 2015/Q3a/ 6 marks
Let ( , ) = 3 − (1 + ). Find + .



Solution

= 3 − = 1 +
′ = 3 − ∙ (0 − 1) ′ = 0 + 1
′ = − 3 −
′ = 1

( , ) = 3 − (1 + ) ( , ) = 3 − (1 + )

= 3 − (1 + ) = (1 + ) 3 −

= = ′ + ′ = = (1 + ) 3 − ∙ (3(1) − 0)


= = − 3 − (1 + ) + 3 − = = 3(1 + ) 3 −


= = − 3 − − 3 − + 3 − = = (3 + 3 ) 3 −


= = − 3 − = = 3 3 − + 3 3 −


= − 3 − + 3 3 − + 3 3 −
+

= 3 3 − + 2 3 −

= 3 − (3 + 2 ) #

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