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## answer tutorial 3.2 page 195 Module 1 (MAT235 Chapter 3)

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# answer tutorial 3.2 page 195 Module 1 (MAT235 Chapter 3)

### answer tutorial 3.2 page 195 Module 1 (MAT235 Chapter 3)

Tutorial 3.2 Total Differentials and Approximation

1. Jun 2019/Q3b/ 7 marks (Answer : 0.0467 (4 . ))

Use total differentials to approximate the value of √15.98
(3.04)4

2. Dec 2018/Q3a/ 6 marks (answer : 0.578)
Use total differentials to approximate the value of (1.10, 1.98) if ( , ) = √1 + . Give
the answer correct to three (3) decimal places.

3. Jun 2018/Q3b/ 7 marks (answer : 0.3315)

Use total differentials to approximate the value of √8.995
1+(2.004)3

4. Jan 2018/Q3b/ 7 marks (refer page 189, 190)
Use total differentials to approximate the value of (3.99)5 √(9.02)3

5. Mar 2017/Q3b/ 7 marks (answer : 0.3994)

If f (x,y ) = x +1 use differentials to estimate f (2.998,1.003) .
4+ y2 ,

6. Mar 2016/Q3b/ 8 marks (refer page 191)

Given f (x, y ) = ln (2x − 3y ) and (x, y) changes from (5, 3) to (4.98, 3.07). Use differential to

approximate f(4.98, 3.07).

7. Sep 2015 /Q3c/ 8 marks (answer : 5.01 (2 . ))
Use total differentials to approximate √(3.04)2 + (3.98)2. Give the answer correct to there
(3) decimal places.

8. mar 2015 /Q3c/ 7 marks (answer :445.1576 (4 . ))

Let ( , ) = 1 + 2 + 2 . Use total differentials to approximate the change in ( , ) as
2

( , ) varies from point (1, 2) to the point (0.97, 2.04). Give the answer correct to there (3)

decimal places.

9. Sep 2014 /Q3a/ 7 marks (answer : 4.225 (3 . ) )
Approximate the value of √(3.98)2 + 2.01 using differentials.

10. mar 2014 /Q4a/ 6 marks (answer : 1.031 (3 . ) )

Let ( , , ) = (4 − 7)5 + . Approximate the change in ( , , ) from = 1, = 2, =

3 to = 1.12, = 1.98, = 3.2 using total differential. Give your answer correct to three

(3) decimal places.

195

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

(ANSWER) Tutorial 3.2 Total Differentials and Approximation

1. Jun 2019/Q3b/ 7 marks

Use total differentials to approximate the value of √15.98
(3.04)4

Solution

Step 1 : identify x, x0, y, y0 and dx, dy
x = 15.98, x0 = 16 → dx = −0.02

y = 3.04, y0 = 3 → dy = 0.04

Step 2 : find ( , ) and the value of ( 0, 0)

From √15.98 ,
(3.04)4

( , ) = √
4

( 0, 0) = (16, 3) = √16 4
34 = 81

Step 3 : find fx and fy and its value at ( , )

√ 1 1 = √ = √ −4
= 4 = 4 ∙ 2 4

= 1 ∙ 1 −12 = √ ∙ −4 −5
4 2
4√
1 = − 5
= 2 4√

At ( 0, 0 ) = (16, 3), At ( 0, 0 ) = (16, 3),
1 11 4√16 4(4) 16

= 2(3)4√16 = 2(81)(4) = 648 = − 35 = − (243) = − 243

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

Step 4 : use = +
= +
1 16
= (648) (−0.02) + (− 243) (0.04) = −0.00266

Step 5 : approximate

√15.98 ≈ ( 0, 0) +
(3.04)4

≈ 4 + (−0.00266)
81

4
≈ 81 − 0.00266

≈ 0.0467 (4 . ) #

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

2. Dec 2018/Q3a/ 6 marks (answer : 0.578)
Use total differentials to approximate the value of (1.10, 1.98) if ( , ) = √1 + . Give
the answer correct to three (3) decimal places.
Solution

Step 1 : identify x, x0, y, y0 and dx, dy
x = 1.10, x0 = 1 → dx = 0.1
y = 1.98, y0 = 2 → dy = −0.02
Step 2 : find ( , ) and the value of ( 0, 0)

( , ) = √1 +
( 0, 0) = (1, 2) = √1 + (1)(2) = √3

Step 3 : find fx and fy and its value at ( , )

= √1 + = (1 + )21 = 1 = √1 + = (1 + )12 = 1
2 (1 + ) 2 (1 + )

11 11
= 2 ∙ 1 + ∙ (0 + (1)) = 2 ∙ 1 + ∙ (0 + (1))

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

At ( 0, 0 ) = (1, 2), At ( 0, 0 ) = (1,2),
2 21 1 11

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

Step 4 : use = +

= +

= 1 (0.1) + 1 (−0.02) = 0.03
(3) (6)

Step 5 : approximate
(1.10, 1.98) ≈ ( 0, 0) +
≈ √3 + 0.03
≈ 0.579 (3 . ) #

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

3. Jun 2018/Q3b/ 7 marks (answer : 0.3315)

Use total differentials to approximate the value of √8.995
1+(2.004)3

Solution

Step 1 : identify x, x0, y, y0 and dx, dy
x = 8.995, x0 = 9 → dx = −0.005

y = 2.004, y0 = 2 → dy = 0.004

Step 2 : find ( , ) and the value of ( 0, 0)

From √8.995 ,
1+(2.004)3

( , ) = 1 √
+ 3

( 0, 0) = (9, 2) = 1 √9 31
+ 23 =9=3

Step 3 : find fx and fy and its value at ( , )

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

= (1 1 ∙ 1 −12 = √ ∙ −(1 + 3)−2 ∙ (0 + 3 2)
+ 3) 2
= − 3 2√
1 (1 + 3)2
= 2(1 + 3)√

At ( 0, 0 ) = (9, 2),

At ( 0, 0 ) = (9, 2), = − 3 2√
1 11 (1 + 3)2

= 2(1 + (2)3)√9 = 2(9)(3) = 54 3(2)2√9
= − (1 + (2)3)2

36
= − 81

4
= −9

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

Step 4 : use = +
= +
14
= (54) (−0.005) + (− 9) (0.004) = −0.00187

Step 5 : approximate

√8.995 ≈ ( 0, 0) +
1 + (2.004)3

1
≈ 3 + (−0.00187)

≈ 0.3315 (4 . ) #

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

4. Jan 2018/Q3b/ 7 marks (refer page 189, 190)
Use total differentials to approximate the value of (3.99)5 √(9.02)3
Solution

Step 1 : identify x, x0, y, y0 and dx, dy
x = 3.99, x0 = 4 → dx = −0.01
y = 9.02, y0 = 9 → dy = 0.02
Step 2 : find ( , ) and the value of ( 0, 0)
From (3.99)5√(9.02)3 ,

( , ) = 5√ 3
( 0, 0) = (4, 9) = (4)5√(9)3 = (1024)√729 = 27648

Step 3 : find fx and fy and its value at ( , ) = 5√ 3 = 5 3
= 5√ 3 = √ 3 5 2
= √ 3 ∙ 5 4
= 5 4√ 3 = 5 ∙ 3 1
2 2
At ( 0, 0 ) = (4, 9),
= 5(4)4√(9)3 = 5(256)√729 = 34560 = 3 5 √
2

At ( 0, 0 ) = (4, 9),

= 3 (4)5√9 = 3 (1024)(3) = 4608
2 2

Step 4 : use = +
= +
= (34560)(−0.01) + (4608)(0.02) = −253.44

Step 5 : approximate
(3.99)5√(9.02)3 ≈ ( 0, 0) +
≈ 27648 + (−253.44)
≈ 27394.56 (2 . ) #

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

5. Mar 2017/Q3b/ 7 marks (answer : 0.3994)

If f (x,y ) = x +1 use differentials to estimate f (2.998,1.003) .
4+ y2 ,

(7 marks)

Solution

Step 1 : identify x, x0, y, y0 and dx, dy
x = 2.998, x0 = 3 → dx = −0.002

y = 1.003, y0 = 1 → dy = 0.003

Step 2 : find ( , ) and the value of ( 0, 0)

( , ) = √ + 1
4 + 2

( 0, 0) = (3, 1) = √3 + 1 = 2
4 + 12 5

Step 3 : find fx and fy and its value at ( , )

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

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

At ( 0, 0 ) = (3, 1), At ( 0, 0 ) = (3, 1),
1 11 2(1)√3 + 1 4

= 2(4 + 12)√3 + 1 = 2(5)(2) = 20 = − (4 + (1)2)2 = − 25

Step 4 : use = +1

20

= +

14
= (20) (−0.002) + (− 25) (0.003) = −0.00058

Step 5 : approximate

(1.10, 1.98) ≈ ( 0, 0) +

≈ 2 + (−0.00058)
5

≈ 0.3994 (4 . ) #

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

6. Mar 2016/Q3b/ 8 marks (refer page 191)

Given f (x, y ) = ln (2x − 3y ) and (x, y) changes from (5, 3) to (4.98, 3.07). Use differential to

approximate f(4.98, 3.07).

Solution

Step 1 : identify x, x0, y, y0 and dx, dy
x = 4.98, x0 = 5 → dx = −0.02
y = 3.07, y0 = 3 → dy = 0.07
Step 2 : find ( , ) and the value of ( 0, 0)

( , ) = (2 − 3 )
( 0, 0) = (5, 3) = (2(5) − 3(3)) = (10 − 9) = (1) = 0

Step 3 : find fx and fy and its value at ( , ) = (2 − 3 )
1
= (2 − 3 )
1 = 2 − 3 ∙ (0 − 3(1))
−3
= 2 − 3 ∙ (2(1) − 0)
2 = 2 − 3

= 2 − 3

At ( 0, 0 ) = (5, 3), At ( 0, 0 ) = (5, 3),
2 22 −3 −3 −3

= 2(5) − 3(3) = 10 − 9 = 1 = 2 = 2(5) − 3(3) = 10 − 9 = 1 = −3

Step 4 : use = +1

20

= +

= (2)(−0.02) + (−3)(0.07) = −0.25

Step 5 : approximate
(4.98, 3.07) ≈ ( 0, 0) +
≈ 0 + (−0.25)
≈ −0.25 (2 . ) #

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

7. Sep 2015 /Q3c/ 8 marks

Use total differentials to approximate √(3.04)2 + (3.98)2. Give the answer correct to there
(3) decimal places.

Solution

Step 1 : identify x, x0, y, y0 and dx, dy
x = 3.04, x0 = 3 → dx = −0.01

y = 3.98, y0 = 4 → dy = 0.02

Step 2 : find ( , ) and the value of ( 0, 0)
From √(3.04)2 + (3.98)2 ,

( , ) = √ 2 + 2
( 0, 0) = (3, 4) = √(3)2 + (4)2 = √9 + 16 = √25 = 5

Step 3 : find fx and fy and its value at ( , )

= √ 2 + 2 = ( 2 + 2)12 = √ 2 + 2 = ( 2 + 2)21

= 1 ( 2 + 2 )−12 ∙ (2 + 0) = 1 ( 2 + 2 )−21 ∙ (0 + 2 )
2 2

= √ 2 + 2 = √ 2 + 2

At ( 0, 0 ) = (3, 4), At ( 0, 0 ) = (3, 4),
3 33 4 44

= √(3)2 + (4)2 = √25 = 5 = √(3)2 + (4)2 = √25 = 5

Step 4 : use = +
= +
34
= (5) (−0.01) + (5) (0.02) = 0.01

Step 5 : approximate
√(3.04)2 + (3.98)2 ≈ ( 0, 0) +
≈ 5 + 0.01
≈ 5.01 (2 . ) #

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

8. mar 2015 /Q3c/ 7 marks

Let ( , ) = 1 + 2 + 2 . Use total differentials to approximate the change in ( , ) as
2

( , ) varies from point (1, 2) to the point (0.97, 2.04). Give the answer correct to there (3)

decimal places.

Solution

Step 1 : identify x, x0, y, y0 and dx, dy
x = 0.97, x0 = 1 → dx = −0.03

y = 2.04, y0 = 2 → dy = 0.04

Step 2 : find ( , ) and the value of ( 0, 0)

( , ) = 1 + 2 + 2 = √ + 2 + 2

2

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

Step 3 : find fx and fy and its value at ( , )

= 1 + 2 + 2 = √ + 2 + 2 = 12 + 2 + 2
2

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

= + 2 2 + 2
2√

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

= 1 + 2(2) 2(1)+(2)2 = 1 + 4 6
2√2 2√2

Step 4 : use = +1

20

= +

= (√2 + 2 6)(−0.03) + ( 1 + 4 6) (0.04) = 40.31459
2√2

Step 5 : approximate
(0.97, 2.04) ≈ ( 0, 0) +
≈ √2 + 6 + 40.31459
≈ 445.1576 (4 . ) #

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

9. Sep 2014 /Q3a/ 7 marks (answer : 4.225 (3 . ) )
Approximate the value of √(3.98)2 + 2.01 using differentials.

Solution

Step 1 : identify x, x0, y, y0 and dx, dy
x = 3.98, x0 = 4 → dx = −0.02

y = 2.01, y0 = 2 → dy = 0.01

Step 2 : find ( , ) and the value of ( 0, 0)
From √(3.98)2 + 2.01 ,

( , ) = √ 2 +
( 0, 0) = (4, 2) = √(4)2 + 2 = √18

Step 3 : find fx and fy and its value at ( , )

= √ 2 + = ( 2 + )21 = √ 2 + = ( 2 + )12

= 1 ( 2 + )−21 ∙ (2 + 0) = 1 ( 2 + )−21 ∙ (0 + 1)
2 2

1
= √ 2 + = 2√ 2 +

At ( 0, 0 ) = (4, 2), At ( 0, 0 ) = (4, 2),
4 444 1 1 1 11

= √(4)2 + 2 = √18 = √(9)(2) = 3√2 = 2√(4)2 + 2 = 2√18 = 2√(9)(2) = 2(3)√2 = 6√2

Step 4 : use = +
= +
41
= ( ) (−0.02) + ( ) (0.01) = −0.01767
√18 2√18

Step 5 : approximate
√(3.98)2 + 2.01 ≈ ( 0, 0) +
≈ √18 + (−0.01767)
≈ 4.225 (3 . ) #

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

10. mar 2014 /Q4a/ 6 marks (answer : 1.031 (3 . ) )
Let ( , , ) = (4 − 7)5 + . Approximate the change in ( , , ) from = 1, = 2, =

3 to = 1.12, = 1.98, = 3.2 using total differential. Give your answer correct to three
(3) decimal places.

Solution :

Step 1 : identify x, x0, y, y0 and dx, dy
x = 1.12, x0 = 1 → dx = 0.12

y = 1.98, y0 = 2 → dy = −0.02

z = 3.2, z0 = 3 → dz = 0.2

Step 2 : find ( , ) and the value of ( 0, 0)

( , , ) = (4 − 7)5 +

( 0, 0, 0) = (1, 2, 3) = (4(2) − 7)5 + 1 = 1 + 1 = 4
3 3 3

Step 3 : find , and and its value at ( , , )

= (4 − 7)5 + = (4 − 7)5 + = (4 − 7)5 +

= (4 − 7)5 + 1 ∙ = 5(4 − 7)4 ∙ (4(1) − 0) + 0 = (4 − 7)5 + ∙ −1
= 20(4 − 7)4
= 0 + ∙ (− −2)
1
= 0 + ∙ (1)
= − 2
1 At ( 0, 0, 0) = (1, 2, 3),
= = 20(4(2) − 7)4 = 20

At ( 0, 0, 0) = (1, 2, 3), At ( 0, 0, 0) = (1, 2, 3),
1 11

= 1 = 1 = − (3)2 = − 9

Step 4 : use = + +

= + +

1
= (1)(0.12) + (20)(−0.02) + (− 9) (0.2) = −0.302229

Step 5 : approximate
(1.12, 1.98, 3.2) ≈ ( 0, 0, 0) +
4
≈ 3 + (−0.302229)
≈ 1.031 (3 . ) #