MAT2100 Assignment 2, University Calculus: Comprehensive Solutions

This document presents a comprehensive solution set for Assignment 2 in a Calculus course, likely MAT2100, at a university. The solutions cover a range of calculus topics, beginning with basic problem-solving and progressing to more complex concepts. Specific problems addressed include Fourier ...Read More

Assignment 2, MAT2100
Name of the Student
Name of the University
Author Note

Table of Contents
Answer to question 1:.................................................................................................................................3
Answer to question 2:.................................................................................................................................6
Answer to question 3:.................................................................................................................................7
Answer to question 4:...............................................................................................................................10
Answer to question 5:...............................................................................................................................16
Answer to question 6:...............................................................................................................................16
Answer to question 7:...............................................................................................................................18
Answer to question 8:...............................................................................................................................18
Answer to question 9:...............................................................................................................................24
Answer to question 10:.............................................................................................................................27

Answer to question 1:

Answer to question 2:
Simplifying

[b]
Answer to question 3:
[a]

[b]

[c]

Answer to question 4:
[a]
[b]
The function is defined as
[i] not defined in the period of (-10, -2];
[ii] –t^2 in the period of (-2,0]
[iii] t^2 in the period of [0,2)
[iv] not defined in the period of [2, -10)
[c]
The function is an even function.

[d]
Average value of the function on its period
= 1
0+2 ∫
−2
0
−t2 dt + 1
2−0 ∫
0
2
t2 dt
= −1
2 [ t3
3 ]−2
0
+ 1
2 [ t3
3 ]0
2
=-4/3 + 4/3
=0
[e]
The Fourier Series representation is
xT (t )=a 0+∑
n=1
∞
(an cos ( nπ t
2 )+bn sin( nπ t
2 ))
Since, the function is even,
xT (t )=a 0+∑
n=1
∞
¿ ¿
= ∑
n=0
∞
¿ ¿
The average value is 0
And the value of an is 1
2 ¿
= 1
2 ¿
Now,

Also,

Thus,
We have an = ½(-8πcos(πn)/π^3n^3)
=-4 πcos(πn)/(π^3n^3)
So,
The Fourier series will be = ∑
n=0
∞
(−4 π cos ( π n )
π3 n3 cos ( nπ t
2 ))
Now,
First term = −4 cos ( π )
π2 cos ( π t
2 )
Second term = −cos ( 2 π )
2 π2 cos ( πt )
Third term = −4 cos ( π )
27 π2 cos ( 3 π t
2 )

Answer to question 5:
The total differential at the point (0, 0) is dw = wx(0, 0) dx + wy(0, 0) dy
In the first case, wx = (2 cos x – sin x) exp (2x + 3y)
and
Wy = 2 exp (2x + 3y) cos x
Hence, at (0, 0), wx =(2*1 -0)exp(0) = 2
And Wy = 2exp(0)*1 = 2
Hence, dw = 2(dx +dy)
In the second case, wx = (2 cos x – sin x) exp (2x + 3y)
and
Wy = 3 exp (2x + 3y) cos x
Hence, at (0, 0), wx =(2*1 -0)exp(0) = 2
And Wy = 3exp(0)*1 = 3
Hence, dw = 2dx +3dy
Therefore, as case one fulfils U(0, 0) = 2; it can be said that first one is the total differential equation.

Answer to question 6:

=5.79
Answer to question 7:
Given
z (x, y) = 6 tan^(-1)(y/x)
Bounded by x^2 + y^2 = 1
Bounded by x^2 + y^2 = 9
Bounded by y=x/sqrt(3)
And Bounded by y=sqrt(3)x
Plotting all the boundaries, we have
X varies from 0 to 1 and y varies from 1/sqrt(3) to sqrt(3).

Therefore,

Answer to question 8:
Work done = ∫ F∗dr
= ∫
0
1
F∗dr
dt dt
= ∫
0
1
[ ( 5 x−8 y √ x2 + y2 ) i+ ( 4 x +10 y √ x2 + y2 ) j+ zk ] . d
dt ¿ ¿
=∫
0
1
[ (5 x−8 y √x2 + y2 ) i+ ( 4 x +10 y √x2 + y2 ) j+ zk ] . ¿ ¿
=∫
0
1
¿ ¿

=∫
0
1
¿ ¿
=∫
0
1
¿ ¿

=2.2632 [approx.]
Answer to question 9:
Total mass, M = ∫
3 /2
2
ρ( x ( t ) , y ( t ) , z ( t )) √ ( x' ( t ) )2
+ ( y' ( t ) )2
+ ( z' ( t ) )2
dt
=∫
3 /2
2
24 / √16 cos2 t +16 sin2 t−16 ( t−1 )2 √16 sin2 t+16 cos2 t +9dt
=∫
3
2
2
( 24
4 ( t−1 ) +5 ) dt
=∫
3
2
2
( 6
t−1 +5 ) dt

=6.659 approx.
Answer to question 10:

Since, curl F(1,2,3) = 0
It is a solenoidal vector field.