eBook Calculus Differentiation for Science And Engineering Students And

Calculus: Differentiation Example 1

How to solve total derivative ?? Find total derivative for = 2 3 − 2 + 3
5
STEPS
1. Extract ALL given info from question. .
Including WHAT to find.
2. Differentiate function using partial Step 2 = 2 3 − 2
differentiation method.
3. Derive related Chain Rule of total derivative.
4. Substitute the partial differentiation into the = 3 2 2 − 3
Chain Rule. 5 2
5. Then substitute all related values and solve
the differentiation Step 3
Step 4 = ∙ + ∙

= 2 3 − 2 ∙ + 3 2 2 − 3 ∙
5 2

96

Calculus: Differentiation
Example 2

The radius and height of a cylinder are both 2 cm. The radius is decreased at 1 cmΤsec and the height
is increasing at 2 cmΤsec . Calculate the change in volume with respect to time at this instant?

= 2

ℎ = 2 Step 4 2 ℎ ∙ + ( 2) ∙ ℎ
=

Step 1 ℎ = ℎ = 2 Step 5 Substitute r = 2, ℎ =2, = −1 and
Step 2 = −1 = 2
ℎ into step 4
Step 3 = 2

The volume of cylinder is: = 2 2 2 ∙ −1 + 2 2 ∙ 2

= 2ℎ = −8 + 8
ℎ
= 2 ℎ = 2 =0

The total derivative of this with respect to time is ∴ there is no changes in volume of the cylinder

ℎ 97
= ∙ + ℎ ∙

Calculus: Differentiation ℎ &

change 1, 2 change

of 0.8, 2.3 of

= 0.8 − 1 = 2.3 − 2

= −0.2 = 0.3

Example 3

If = 1 − 2 2 + 3 + 3, Find the total derivative of
when ( , ) changes from (1, 2) to (0.8, 2.3).

Step 1 = −0.2 = 0.3 Step 4 = −4 + 3 ∙ + (−3 + 3 2) ∙
Step 2
Step 5 Substitute = 1, =2, = −0.2 and
Step 3 = −4 + 3 = 0.3 into step 4

= −3 + 3 2 = −4 1 + 3 2 ∙ −0.2 + +(−3 1 + 3 2 2) ∙ 0.3


= 2.9

The total derivative is ∴ there is 2.9 changes in


= ∙ + ∙

98

Calculus: Differentiation

Example 4 Find the total derivative of volume of a cone when the height decrease by 0.03 Τ and the radius
increase by 0.05 Τ . The height and radius of the cone is 12 and 10 ?

= 10

ℎ = 12

Step 4 2 1 2 ℎ
= 3 ℎ ∙ + 3 ∙
ℎ ℎ = 12
Step 1 = 0.05 = −0.03 = 10 Substitute and
Step 2
Step 5 r = 10, ℎ = 12, = 0.05
Step 3
The volume of cone is: ℎ = −0.03 into step 4


= 1 2ℎ 2 12 ∙ 0.05 1 10 2∙ −0.03
3 = 3 10 + 3

2 = 1 2 = 4 −
= 3 ℎ ℎ 3

The total derivative of this with respect to time is = 3

∴ there is 3 3Τ changes in volume of the cone

ℎ 99
= ∙ + ℎ ∙

Calculus: Differentiation

Example 5 Consider a cuboid where all the edges are increasing. Given that the height is 15 , length is

Step 1 12 and the width is 7 . Evaluate when = 0.3, ℎ = 0.5 and = 0.8?


ℎ = 7 = 7
= 0.3 = 0.5 = 0.8 = 12 = 12
ℎ = 15
ℎ = 15

Step 2 The volume of cuboid is:
Step 3
= ℎ ∙ ∙ Step 4 ℎ
= ℎ ∙ + ∙ + ∙

= ℎ = ℎ = Substitute = 12, = 7, ℎ = 15, = 0.3,

Step 5

The total derivative of this with respect to time is ℎ = 0.5 and = 0.8 into step 4

ℎ
= ∙ + ∙ + ℎ ∙ = 15 7 ∙ 0.3 + 12 7 ∙ 0.8 + 7 12 ∙ 0.5

= 140.7

∴ there is 140 3Τ changes in volume of the cuboid

100

EXERCISE

101

REFERENCES

Abd Wahid Md Raji, Hamisan Rahmat, Ismail Kamis, Mohd Nor Mohamad & Ong
Chee Tiong (2013). The First Course of Calculus for Science &
Engineering Students. Universiti Teknologi Malaysia

Calculus 1 (article). (n.d.). Khan Academy. Retrieved 2021, from
https://www.khanacademy.org/math/calculus-1

Suhana Ramli, Khairul Faizal Abdul Hamid & Intan Noor Dzalika Azis (2017).
Engineering Mathematics 2 For Polytechnics (Volume 1). Politeknik
Mukah

Zuraini Ibrahim, Myzatul Mansor, Suria Masnin, Baharudin Azit & Fatin Hamimah

Mohamed Salleh(2018). Polytechnic Series Engineering Mathematics 2.

Oxford Fajar. 102

www.polipd.edu.my
2022