answer tutorial 3.4 page 212 (MAT235 Chapter 3)

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

Tutorial 3.4 : Partial Differentiation for Implicit Functions

1. Jun 2019/Q3a/ 6 marks

Find for 3 + 2 2 = 1 at the point (1, 1) by using implicit differentiation. Write



your answer in four decimal places. Ans : = −4
1+4 2


2. Dec 2018/Q2d/ 5 marks

For an implicit function 2 2 + 4 = 2, evaluate by using partial differentiation.



Ans : = 2−4 −4
2 2+4 −2


3. Jun 2018/Q3a/ 6 marks
Given 2 2 + 8 √ = 2 find at the point (2, 1) by using partial differentiation.



Ans : = − 9

16+8

4. Jan 2018/Q3a/ 6 marks

Use partial derivatives to find of (2 + ) = 3√ − 2 . Hence, find the gradient of the
2

tangent line to the curve at the point (−1, 1). 5
3 3( 2+4 +2 2)
Ans : =
2

(2+ )( 3+6 2 3)

= − 3

7

5. Mar 2017/Q3a/ 6 marks

Use partial differentiation to find the equation of the tangent line to the curve

( − 2)2 + 2 √ − 13 − 4 = 0 at the point (4, 3). Ans : = 17 + 7
102 3

6. Oct 2016/Q3c/ 5 marks

Use partial differentiation to find the equation of the tangent line to the curve

2 3 + 2 + 2 2 + − 6 = 0 at the point (1, −2). Ans : = − 4 − 2
3 3

7. Sep 2013 /Q3b/ 4 marks

Find the slope of the tangent line to the curve (3 + ) 4 + − = 0 at the point

(1, 0) by using partial differentiation. Ans : = 1
13

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

8. Oct 2010 /Q4c/ 5 marks

Given + √ − 4 − 2 = 0 , evaluate at the point (0, 4) using partial differentiation.


Ans : = 15


9. Oct 2009 /Q4a/ 6 marks

Find the equation of the tangent line to the curve + ( 2) = −3 2 at the point (1, 0).

Ans : = −6 + 6

10. Oct 2006 /Q3b/ 4 marks

Using partial differentiation, evaluate at the point (2, 3) for the function
( , ) = 3 2√ .


Ans : = −24


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

ANSWER Tutorial 3.4 : Partial Differentiation for Implicit Functions

1. Jun 2019/Q3a/ 6 marks
Find for 3 + 2 2 = 1 at the point (1, 1) by using implicit differentiation. Write



your answer in four decimal places.

Solution :

Step 1 : write the given function as f

= 3 + 2 2 − 1


Step 2 : differentiate using partial differentiation (find and )

= 3 + 2 2 − 1 = 3 + 2 2 − 1


= 3 + 2 2 − −1 = 3 + 2 2 − 1

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

At (1, 1) At (1, 1)
= (1)3 + 4 2(1)
= 3(1)2(1) + 1 = 1 + 4 2
(1)2

= 3 + 1

= 4

Step 3 : use formula = −



= − = −4
1 + 4 2

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

2. Dec 2018/Q2d/ 5 marks
For an implicit function 2 2 + 4 = 2, evaluate by using partial differentiation.



Solution :

Step 1 : write the given function as f
= 2 2 + 4 − 2

Step 2 : differentiate using partial differentiation (find and )

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

Step 3 : use formula = −



= − = −(4 + 4 − 2) = 2 − 4 − 4
2 2 + 4 − 2 2 2 + 4 − 2

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

3. Jun 2018/Q3a/ 6 marks
Given 2 2 + 8 √ = 2 find at the point (2, 1) by using partial differentiation.



Solution :

Step 1 : write the given function as f

= 2 2 + 8 √ − 2


Step 2 : differentiate using partial differentiation (find and )

= 2 2 + 8 √ − 2 = 2 2 + 8 √ − 2


= 2 2 + 8 √ − 2 −1 = 1 − 2

2 2 + 8 2

= 2(2 ) + 0 − 2(− −2) 1 1 −21
2
= 2 2 + 2 = 2(2 ) + 8 2 ∙
2
4 √
= 2 2 + √

At (2, 1)

= 2(2)(1)2 + 2 At (2, 1)
(2)2

1 = 2(2)2(1) + 4 √1
= 4 + 2 √1

9 = 8 + 4
= 2

Step 3 : use formula = −



= − = − (29) 91 9
8 + 4 = (− 2) ∙ (8 + 4 ) = − 16 + 8 #

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

4. Jan 2018/Q3a/ 6 marks

Use partial derivatives to find of (2 + ) = 3√ − 2 . Hence, find the gradient of the
2


tangent line to the curve at the point (−1, 1).

Solution :

Step 1 : write the given function as f

= (2 + ) − 3√ + 2
2

Step 2 : differentiate using partial differentiation (find and )

= (2 + ) − 3√ + 2 = (2 + ) − 3√ + 2
2 2

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

11 = 0 − 1 −32 + 2(−2 −3)
= 2 + ∙ (0 + 1) − 0 + 2 ∙ (2 ) 3

1 2 = − 1 −32 − 2 2
= 2 + + 2 3 3

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

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

2 3

3 3

− 3 − 6 2 2

= 3

At (−1, 1) 11
1 2(−1) 3 3

= 2 + (−1) + (1)2

= 1 − 2

= −1 At (−1, 1)

Step 3 : use formula = − = − 1 (1)−32 − 2(−1)2
3 (1)3


−(−1) 1 3 1
= = − = = − = −= 7−#3 − 2
(− 73) (73) 7

= − 3

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

= −



= − ÷

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

11

3 3

−( 2 + 4 + 2 2) 11
= ∙
3 3

(2 + ) 2 − ( 3 + 6 2 2

3)

3 5 ( 2 + 4 + 2 2)
3
= #
(2 + ) ( 3 + 6 2 2

3)

(−1, 1)

3(1)53((1)2 + 4(−1) + 2(1))
= (2 − 1) ((1)3 + 6(1)(1)32)
3(1 − 4 + 2)
= (1)(1 + 6)
3
= − 7

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

5. Mar 2017/Q3a/ 6 marks
Use partial differentiation to find the equation of the tangent line to the curve
( − 2)2 + 2 √ − 13 − 4 = 0 at the point (4, 3).

Solution :
Step 1 : write the given function as f
= ( − 2)2 + 2 √ − 13 − 4

Step 2 : differentiate using partial differentiation (find and )

= ( − 2)2 + 2 √ − 13 − 4 = ( − 2)2 + 2 √ − 13 − 4
= ( − 2)2 + 2√ ∙ − 13 − 4
= ( − 2)2 + 2 ∙ 1 − 13 − 4 = 2( − 2)(0 − 2 ) + 2√ ∙ (1) − 13(1) − 0
2 = −4 ( − 2) + 2√ − 13

= 2( − 2)(1 − 0) + 2 ∙ 1 −12 − 0 − 0
2

= 2( − 2) +
√

At (4, 3) At (4, 3)
= −4(3)(4 − (3)2) + 2√4 − 13
= 2(4 − 32) + 3 = −12(−5) + 2(2) − 13
√4 = 60 + 4 − 13
= 51
3
= 2(4 − 9) + 2

3
= −10 + 2

17
= − 2

Step 3 : use formula = −



= = − = − (− 127) = 17 ∙ 1 = 17
51 2 51 102

, = +

(4, 3) = 17 +

102

→ 3 = 17 (4) +

102

= 3 − 68 = 7

102 3

= 17 + 7 #

102 3

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

6. Oct 2016/Q3c/ 5 marks
Use partial differentiation to find the equation of the tangent line to the curve
2 3 + 2 + 2 2 + − 6 = 0 at the point (1, −2).

Solution :
Step 1 : write the given function as f

= 2 3 + 2 + 2 2 + − 6

Step 2 : differentiate using partial differentiation (find and )

= 2 3 + 2 + 2 2 + − 6 = 2 3 + 2 + 2 2 + − 6
= 2 3 + 2 + 2 2 + − 6 = 2 3 + 2 + 2 2 + − 6
= 2(3 2) + (2 ) + 2 2(1) + 0 − 0 = 0 + 2(1) + 2 (2 ) + 1 − 0
= 6 2 + 2 + 2 2 = 2 + 4 + 1
(1, −2)
(1, −2) = (1)2 + 4(1)(−2) + 1
= 6(1)2 + 2(1)(−2) + 2(−2)2 = 1 − 8 + 1
= 6 − 4 − 8 = −6
= −8

Step 3 : use formula = −



= = − = −(−8) 4
−6 = − 3

, = +

= − 4 +

3

(1, −2) → −2 = − 4 (1) +

3

= −2 + 4 = − 2

33

= − 4 − 2 #

33

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

7. Sep 2013 /Q3b/ 4 marks

Find the slope of the tangent line to the curve (3 + ) 4 + − = 0 at the point

(1, 0) by using partial differentiation. Ans : = 1
13

Solution :
Step 1 : write the given function as f

= (3 + ) 4 + −

Step 2 : differentiate using partial differentiation (find and )

= 3 + = 4
′ = 0 + 1 ′ = 4 ∙ 4(1)
′ = 1 ′ = 4 4

= (3 + ) 4 + − = (3 + ) 4 + −
= (3 + ) 4 + −
= ( 4 )(3 + ) + − = ′ + ′ + 1 − 0
1 = 4 + (3 + ) ∙ 4 4 + 1
= 4 + 4(3 + ) 4 + 1
= ( 4 )(3(1) + 0) + 0 −
1

= 3 4 −

(1, 0) (1, 0)
= 4(0) + 4(3(1) + 0) 4(0) + 1
1 = 0 + 12(1) + 1
= 3 4(0) − 1 = 13
= 0 − 1
= −1

Step 3 : use formula = −



= = − = −(−1) 1
13 = 13

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

8. Oct 2010 /Q4c/ 5 marks

Given + √ − 4 − 2 = 0 , evaluate at the point (0, 4) using partial differentiation.


Ans : = 15


Solution :

Step 1 : write the given function as f


= + √ − 4 − 2

Step 2 : differentiate using partial differentiation (find and )


= + √ − 4 − 2 = + √ − 4 − 2

1 = −1 + 1 − 4 − 2
= ∙ + √ − 4 − 2 2

1 = (− −2) + 1 −12 − 0 − 0
= ∙ (1) + 0 − 4(1) − 0 2

1 1
= − 4 = − 2 + 2√

(0, 4) (0, 4)
1 01

= 4 − 4 = − 42 + 2√4
15 11

= − 4 = 0 + 2(2) = 4

Step 3 : use formula = −



= − = − (− 145) = 15 1 15 4 #
(14) 4 ÷4= 4 ∙ 1 = 15

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

9. Oct 2009 /Q4a/ 6 marks
Find the equation of the tangent line to the curve + ( 2) = −3 2 at the point (1, 0).

Ans : = −6 + 6

Solution :

Step 1 : write the given function as f

= + ( 2) + 3 2

Step 2 : differentiate using partial differentiation (find and )

= + ( 2) + 3 2 = + ( 2) + 3 2
= + ( 2) + 3 2 = + ( 2) + 3 2
= ∙ (1) + 0 + 3(2 ) = ∙ (1) + ( 2) ∙ 2 + 0
= + 6 = + 2 ( 2)

(1, 0) (1, 0)
= (0) (1)(0) + 6(1) = (1) (1)(0) + 2(0) (0)
= 0 + 6 = 0 + 0
= 6 = 1

Step 3 : use formula = −



= = − = 6
− 1 = −6

, = +
= −6 +

(1, 0) → 0 = −6(1) +
= 6
= −6 + 6 #

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

10. Oct 2006 /Q3b/ 4 marks

Using partial differentiation, evaluate at the point (2, 3) for the function
( , ) = 3 2√ .

Ans : = −24


Solution :
Step 1 : write the given function as f

= 3 2√

Step 2 : differentiate using partial differentiation (find and )

= 3 2√ = 3 2√
= 3√ ∙ 2
= 3√ ∙ 2(2 ) () = 3 2 ∙ 1
= 6 2√ 2

= 3 2 ∙ 1 −21
2

3 2
= 2√

(2, 3) (2, 3)

= 6(2) (2)2√3 3 (2)2
= 12√3 4 = 2√3

3 4
= 2√3

Step 3 : use formula = −



= −


−(12√3 4)
= (3 4 )

2√3
= −12√3 4 ÷ 3 4

2√3

= −12√3 4 ∙ 2√3
3 4

= −4√3 ∙ 2√3

= −8(3)

= −24 #

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