answer tutorial 3.3 page 207 Module 1 (MAT235 Chapter 3)

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

Tutorial 3.3 : Chain Rule and Rate of Change Problems

1. Jun 2019/Q2c/ 5 marks

Consider a cylindrical tank with radius = 0.3 and ℎ = 1 . Find the rate at which the

total volume of the tank is changing if is increasing at the rate of 0.01 / and ℎ is

decreasing at 0.05 / . Ans : = 0.0015 3/ s
Hint: Volume of the cylinder is = 2ℎ.


2. Dec 2018/Q3b/ 5 marks

Find if = 3 − , = 1 and = 4 by using the chain rule.



Ans : = 3 (1)
2


3. Jun 2018/Q2c/ 5 marks

Consider a conical tank with the radius at the top is 4 and the height ℎ is 10 .

Given the volume of the tank is = 1 2ℎ. Find the rate at which the total volume of the

3

tank is changing if is increasing at the rate of 0.1 and ℎ is decreasing at

0.2 . Ans : = 1.6 3/



4. Jan 2018/Q2c/ 5 marks

The total surface area of a container is given by = 2 2 + ( + 1) ℎ, where is the radius

and ℎ is the height of the container. When = 5 a and ℎ = 10 , find the rate at which

the total surface area is changing if is increasing at the rate of 0.05 −1 while ℎ is

decreasing at the rate of 0.1 −1. Ans : = 2 −1



5. Mar 2017 2019/Q2c/ 5 marks

The volume of a tank is given by = ℎ2 − 1 ℎ3, where is the radius and ℎ is the

3

height of the tank. Find the rate at which the volume is changing when the radius is 10 m

and increasing at the rate of 0.5 −1 while the height is 12 m and decreasing at the rate of

rate 0.2 −1. Ans : = 52.8 3 −1



© Amirah 2022 207

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

6. Mar 2016 /Q4a/ 6 marks (refer page 206)

The pressure, volume and temperature of a mole of an ideal gas are related by the equation

P = 8.31 T
V

where P is measured in kilopascals (kPa), V in liters (L), T in kelvins (K). Find the rate of

change in pressure when the temperature is 350K and increasing at a rate of 0.3 K/s and the

volume is 200L and increasing at a rate of 0.5 L/s. Ans : = −0.02389 /



7. Sep 2014/ Q1c/ 6 marks

Given = 2 , = 2 = 2 , use chain rule and partial differentiation to find at
1+

= . Ans : = 2
( +2)2

8. Mar 2014/ Q3a/ 5 marks

The surface area of a solid is given by = 2 + ( + 2) ℎ where r is the radius and h is

the height. Find the rate at which the area is changing when the radius is 6 cm and

increasing at the rate of 0.1 cm/sec while the height is 10 cm and decreasing at the rate of

0.2 cm/sec. Ans : = 2.74 2/



9. Mar 2013/ Q3b/ 4 marks

If = ( + ) , = + = ( ) , use chain rule and partial differentiation to

4

find the value of when = . Ans : = 4

3

10. Oct 2012/ Mar 2013/ Q1b/ 5 marks

If = (2 − 3)2 , = 1 = + , evaluate when = 0.

+1

Ans : = 5



© Amirah 2022

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

Answer Tutorial 3.3 : Chain Rule and Rate of Change Problems

1. Jun 2019/Q2c/ 5 marks
Consider a cylindrical tank with radius = 0.3 and ℎ = 1 . Find the rate at which the
total volume of the tank is changing if is increasing at the rate of 0.01 / and ℎ is
decreasing at 0.05 / .
Hint: Volume of the cylinder is = 2ℎ.

Solution :


= 0.01 /
ℎ
= −0.05 /

= 0.3
ℎ = 1

=?

= 2ℎ = 2ℎ

= ℎ 2 = 2ℎ

= ℎ(2 ) = 2(1)
ℎ

= 2
= 2 ℎ ℎ

2 (0.3)(1) = (0.3)2
= ℎ


= 0.6 ℎ = 0.09

Using chain rule,

ℎ
= ∙ + ℎ ∙

= (0.6 ) ∙ (0.01) + (0.09 ) ∙ (−0.05)


= 0.0015 3/ s #


© Amirah 2022

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

2. Dec 2018/Q3b/ 5 marks
Find if = 3 − , = 1 and = 4 by using the chain rule.



Solution :

= 3 − ( ) → = −3 − ( )
= 3 − ( ) →


= − ( )



= 1 = −1 → = − −2 = − 1
2

→ = 4
= 4


Using chain rule,


= • + •

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

3 1
= 2 + 2 ( ) − 4 ( ) =
= 4
3 1 4 1 11
= 2 ( ) + 2 ( ∙ 4 ) − 4 ( ) ( ∙ 4 )

3 1 4 4
= 2 ( ) + (4) − ( ) (4)

3 1
= 2 ( ) #

© Amirah 2022

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

3. Jun 2018/Q2c/ 5 marks
Consider a conical tank with the radius at the top is 4 and the height ℎ is 10 .
Given the volume of the tank is = 1 2ℎ. Find the rate at which the total volume of the

3

tank is changing if is increasing at the rate of 0.1 and ℎ is decreasing
at 0.2 .

Solution :


= 0.1 /
ℎ
= −0.2 /

= 4
ℎ = 10

=?

= 1 2 ℎ = 1 2 ℎ
3 3

= 1 ℎ 2 = 1 2 ℎ
3 3

1 = 1 2 (1)
= 3 ℎ(2 ) ℎ 3

2 = 1 2
= 3 ℎ ℎ 3

2 = 1 (4)2
= 3 (4)(10) ℎ 3

80 16
= 3 ℎ = 3

Using chain rule,

ℎ
= ∙ + ℎ ∙

80 16
= ( 3 ) ∙ (0.1) + ( 3 ) ∙ (−0.2)

= 1.6 3/ s #


© Amirah 2022

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

4. Jan 2018/Q2c/ 5 marks
The total surface area of a container is given by = 2 2 + ( + 1) ℎ, where is the
radius and ℎ is the height of the container. When = 5 a and ℎ = 10 , find the rate at
which the total surface area is changing if is increasing at the rate of 0.05 −1 while ℎ
is decreasing at the rate of 0.1 −1.

Solution :

= 0.05 −1


ℎ = −0.1 −1


= 5

ℎ = 10


=?

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

ℎ = 0 + ( + 1) (1)
= 2 (2 ) + ( + 1)ℎ(1)
ℎ = +
= 4 + ( + 1)ℎ
ℎ = (5) + (5)

ℎ = 5 + 5
= 4 (5) + ( + 1)(10)


= 20 + 10 + 10


= 30 + 10

Using chain rule,

ℎ
= ∙ + ℎ ∙

(30 + 10) ∙ (0.05) + (5 + 5) ∙ (−0.1)
=

= 1.5 + 0.5 − 0.5 − 0.5


2 −1 #
=

© Amirah 2022

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

5. Mar 2017 2019/Q2c/ 5 marks

The volume of a tank is given by = ℎ2 − 1 ℎ3, where is the radius and ℎ is the
3

height of the tank. Find the rate at which the volume is changing when the radius is 10 m

and increasing at the rate of 0.5 −1 while the height is 12 m and decreasing at the rate

of rate 0.2 −1.

Solution :

= 0.5 −1


ℎ = −0.2 −1


= 10

ℎ = 12


=?

= ℎ2 − 1 ℎ3 = ℎ2 − 1 ℎ3
3 3

= ℎ2 − 1 ℎ3 = ℎ2 − 1 ℎ3
3 3

= ℎ2(1) − 0 = (2ℎ) − 1 (3ℎ2)
ℎ 3

= ℎ2 = 2 ℎ − ℎ2
ℎ

= (12)2 = 2 (10)(12) − (12)2
ℎ


= 144 ℎ = 240 − 144


ℎ = 96

Using chain rule,

ℎ
= ∙ + ℎ ∙

(144 ) ∙ (0.5) + (96 ) ∙ (−0.2)
=

= 52.8 3 −1 #


© Amirah 2022

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

6. Mar 2016 /Q4a/ 6 marks (refer page 39)

The pressure, volume and temperature of a mole of an ideal gas are related by the
equation

P = 8.31 T
V

where P is measured in kilopascals (kPa), V in liters (L), T in kelvins (K). Find the rate of
change in pressure when the temperature is 350K and increasing at a rate of 0.3 K/s and
the volume is 200L and increasing at a rate of 0.5 L/s.
Solution :


= 0.3 /

= 0.5 /
= 350
= 200

=?


= 8.31 = 8.31
= 8.31 −1
1
= 8.31 = 8.31 ∙ (− −2)
1
= 8.31 (1)
8.31 8.31
= = − 2
8.31
= 200 8.31(350)
= − (200)2
= 0.04155

= −0.0727125

Using chain rule,


= ∙ + ∙

(0.04155)(0.3) + (−0.0727125)(0.5)
=


= −0.02389 / s #

© Amirah 2022

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

7. Sep 2014/ Q1c/ 6 marks

Given = 2 , = 2 = 2 , use chain rule and partial differentiaon to find at
1+

= .

Solution :

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

1+

= 2 → = − 2 ∙ 2 = −2 2
= 2 = 2 −1


→ = 2 ∙ − −2 = − 2
2

Using chain rule,


= • + •

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

−4 2 2 2
= 1 + + 2(1 + )2
= 2

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


−4 2 2 2 2 2
= + 2 (1 + 2 )2
1 + 2


−4(1)(0) 2(1)2
= + 2 (1 + 2 )2
1 + 2


2 2 22
= 0 + 2 (1 + 2 )2 = = = ( + 2)2 #
( + 2)2 ( + 2)2
2 2 ∙ 2

© Amirah 2022

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

8. Mar 2014/ Q3a/ 5 marks
The surface area of a solid is given by = 2 + ( + 2) ℎ where r is the radius and h is
the height. Find the rate at which the area is changing when the radius is 6 cm and
increasing at the rate of 0.1 cm/sec while the height is 10 cm and decreasing at the rate of
0.2 cm/sec.

Solution :

= 0.1 /
ℎ
= −0.2 /
= 6

ℎ = 10

=?

= 2 + ( + 2) ℎ = 2 + ( + 2) ℎ

= 2 + ( + 2)ℎ = 2 + ( + 2) ℎ

= (2 ) + ( + 2)ℎ(1) = 0 + ( + 2) (1)
ℎ


= 2 + ( + 2)ℎ ℎ = + 2

= 2 (6) + ( + 2)(10) = (6) + 2(6)
ℎ


= 12 + 10 + 20 ℎ = 6 + 12


= 22 + 20

Using chain rule,

ℎ
= ∙ + ℎ ∙

(22 + 20)(0.1) + (6 + 12)(−0.2)
=

= 2.2 + 2 − 1.2 − 2.4


= − 0.4 = 2.74 2/ #


© Amirah 2022

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

9. Mar 2013/ Q3b/ 4 marks

If = ( + ) , = + = ( ) , use chain rule and partial differentiation to

4

find the value of when = . Ans : = 4
3

Solution : → = 1 ∙ (1 + 0) = 1
= ( + ) + +
= ( + )
→ = 1 ∙ (0 + 1) = 1
+ +

= + → = 1 + (− ) = 1 −
= ( )

4
→ = 2 ( ) ∙ (1) = 1 2 ( )
44 4 4

Using chain rule,

= +
= • + • = +

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

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

1 1 1 4 = (4)

= + ( ) (4 ∙ 3) = 1

1 1 1 2 ( ) = 245° = 1 = 1 = 1 = 4
= + ( ) (3) 245° (√23)2 (43) 3
3 1 4 4
= 3 + 3 = 3 #

© Amirah 2022

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

10. Oct 2012/ Mar 2013/ Q1b/ 5 marks

If = (2 − 3)2 , = 1 = + , evaluate when = 0.

+1

Solution : Ans : = 5


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


→ = (2 − 3)2 ∙ (1) = (2 − 3)2



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


Using chain rule,


= • + •

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

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

−4(2 − 3) (2 − 3)2 (1 − 0)
= +
1

= −4(−1) + (−1)2 = 1 = 1 = 1

+1 0+1

= 4 + = + = 0 + (0) = 0 + 1 = 1


= 5 #

© Amirah 2022