assignment 11- mathematics 1 - aastmt

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1 | Page AASTMT ARAB ACADEMY FOR SCIENCE TECHNOLOGY AND MARITIME TRANSPORT Problem. 1: (a) Find the first three terms of the Maclaurin series for ln (1 + e x ). (b) Hence, determine the value of . 2 0 4 ln ) e 1 ln( 2 lim x x x x Assignment 11- Mathematics 1 MacLaurin series& partial derivatives Name: Specialization: Registration No: Date: / /2019

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Page 1: Assignment 11- Mathematics 1 - AASTMT

1 | P a g e A A S T M T

ARAB ACADEMY FOR SCIENCE TECHNOLOGY AND

MARITIME TRANSPORT

Problem. 1: (a) Find the first three terms of the Maclaurin series for ln (1 + ex).

(b) Hence, determine the value of .

20

4ln)e1ln(2lim

x

xx

x

Assignment 11- Mathematics 1

MacLaurin series& partial derivatives

Name:

Specialization:

Registration No:

Date: / /2019

Page 2: Assignment 11- Mathematics 1 - AASTMT

2 | P a g e A A S T M T

Problem. 2. (a) Find the value of

(b) By using the series expansions for and cos x evaluate

.2sin

lnlim

1

x

x

x

2e x

.cos1

e1lim

2

0

x

x

x

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Problem. 3. (a) (i) Find the first four derivatives with respect to x of y = ln(1

+ sin x)

(ii) Hence, show that the Maclaurin series, up to the term in x4, for y is

y = x –

...12

1

6

1

2

1 432 xxx

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(b) Deduce the Maclaurin series, up to and including the term in x4, for

(i) y = ln(1 – sin x);

(ii) y = ln cosx;

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(iii) y = tan x.

(c) Hence calculate .

x

x

x cosln

)tan(lim

2

0

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Problem. 4. (a) Given that y = ln cos x, show that the first two non-zero terms of

the Maclaurin series for y are

(b) Use this series to find an approximation in terms of for ln 2.

.122

42 xx

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Partial Derivatives

Given a function of two variables z = f (x, y). Then

h

yxfyhxfyxf

x hx

),(),(lim),(

respect towith

Derivative Partial

0

h

yxfhyxfyxf

y hy

),(),(lim),(

respect towith

Derivative Partial

0

Notations For Partial Derivatives

Given z = f (x, y).

. respect to with derivative partial),(),( xx

z

x

fyxfyxf

xx

. respect to with derivative partial),(),( yy

z

y

fyxfyxf

yy

),(point at the evaluated sderivative partial

),(

),(

),(

),(

ba

bafdy

z

bafdx

z

y

ba

xba

*Note: In partial differentiation, we treat every variable as a constant except for the one we are

differentiating with respect to.

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Problem. 5 Find the first partial derivatives of the function 221),( yxyxf .

Solution:

Problem. 6: Find the first partial derivatives of the function yexxyxyxf 2323 54),( .

Solution:

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Problem. 7: Find the first partial derivatives of the function 22

)ln(),( 22 yxeyxyxf .

Solution:

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Problem. 8: Find )1,1(xf and )1,1(yf for 22

),(yx

xyyxf

Solution:

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Note: We can also differentiate functions of more than 2 variables.

Problem. 9: Find the first partial derivatives of the function

)sin(),,( 22222 yxzyxzyxf .

Solution:

Geometric Interpretation of the Partial Derivative

Given a surface z = f (x, y). We want to consider the point ( a, b, f (a, b) )

direction in the )),(,,(point

at the surface the toline tangent theof Slope),(

),( xbafbax

zbaf

bax

direction in the )),(,,(point

at the surface the toline tangent theof Slope),(

),(ybafbay

zbaf

ba

y

Page 12: Assignment 11- Mathematics 1 - AASTMT

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Problem. 10: Find the slope of the surface 22),( yxyxf at the point (-2, 1, 5).

Solution:

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Second Order Partial Derivatives

Just as we can find second order derivatives for functions of one variable, we can do the same for

functions of two variables.

Notation for Second Order Derivatives

again respect to then with, respect to with partial Take :),(2

2

2

2

xxz

z

x

f

x

f

dxyxf xx

. respect to then with, respect to with partial Take :),(22

yxxy

z

xy

f

x

f

dyyxf yx

. respect to then with, respect to with partial Take :),(22

xyyx

z

yx

f

y

f

dxyxf xy

again respect to then with, respect to with partial Take :),(2

2

2

2

yyy

z

y

f

y

f

dyyxf yy

Note: In general,

),(),( yxfyxf yxxy

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Problem. 11: Find all of the second derivatives for 22 23),( xxyxyyxf .

Solution:

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Problem .12:

a) If 𝒛 = 𝐥𝐧(𝒙𝟐 + 𝒚𝟐) , 𝒔𝒉𝒐𝒘 𝒕𝒉𝒂𝒕 𝝏𝟐𝒛

𝝏𝒙𝟐 +𝝏𝟐𝒛

𝝏𝒚𝟐 = 𝟎

Solution:

b) If 𝒖 = 𝐭𝐚𝐧−𝟏 𝒚

𝒙 , 𝒔𝒉𝒐𝒘 𝒕𝒉𝒂𝒕

1. 𝒙𝝏𝒖

𝝏𝒚 − 𝒚

𝝏𝒖

𝝏𝒙 = 𝟏

2. 𝝏𝟐𝒖

𝝏𝒙𝟐 + 𝝏𝟐𝒖

𝝏𝒚𝟐 =0

Solution:

c) 𝒛 =𝒙𝒚

𝒙−𝒚 , 𝒔𝒉𝒐𝒘 𝒕𝒉𝒂𝒕 𝒙𝟐

𝝏𝟐𝒛

𝝏𝒙𝟐 + 𝟐𝒙𝒚 𝝏𝟐𝒛

𝝏𝒙𝝏𝒚+ 𝒚𝟐

𝝏𝟐𝒛

𝝏𝒚𝟐 = 𝟎

Solution:

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d) If 𝒖 = 𝒇(𝒙𝟐 + 𝒚𝟐), 𝒔𝒉𝒐𝒘 𝒕𝒉𝒂𝒕 𝒙𝝏𝒖

𝝏𝒚− 𝒚

𝝏𝒖

𝝏𝒙= 𝟎

Solution:

e) If 𝒖 = 𝒙𝟐𝒚 + 𝒚𝟐𝒛 + 𝒛𝟐𝒙 , 𝒔𝒉𝒐𝒘 𝒕𝒉𝒂𝒕 𝝏𝒖

𝝏𝒙+

𝝏𝒖

𝝏𝒚+

𝝏𝒖

𝝏𝒛= (𝒙 + 𝒚 + 𝒛)𝟐

Solution:

Good luck