Question 1 MST125 TMA 01 Question 1 Microsoft word The distance

Question 1
MST125 TMA 01 Question 1
Microsoft word
1. The distance between A(x1 , y1 ) and B(x2 , y2 ) is given by,
AB= √ ( x2−x1 )
2+ ( y2− y1 ) 2
2. Hence the distance between A(−1, 3) and B ( 3,1 ) is
AB= √ ( 3−(−1) ) 2+ ( 1−3 ) 2
¿ √ 42+ 22
¿ √ 20
3. The gradient M of the line through ( x1 , y1 ) and ( x2 , y2 ) is given by,
m= y2− y1
x2−x1
Hence the distance between (−1,3) and ( 3,1 ) is
m= 1−3
3− (−1 ) =−2
4 =−1
2
The gradient of the line is −1
2 . Hence tan α=−1
2 . Let φ be the acute angle that the line
makes with the negative direction of the x axis. Then,
tan φ= 1
2
So,
φ=tan−1 1
2 =0.463
Hence,
α=π−0.463=2.677
Therefore the angle α is 2.68 radians (to 2 d.p.)
Question 2
a)
254 254 254 254 254
Sum of all the digits is = 11 + 11 + 11 + 11 + 11 = 55

Since 55 is not divisible by 9. So 254 254 254 254 254 is also not divisible by 9
b)
1356 modulo 19
1318 ≡ 1(mod 19)
1356=132 1354=169 (13¿¿ 18)3=169 ≡17(mod 19) ¿
Question 3
a)i)
27 v ≡1( mod 70)
27 v=1−70 w
27 v+70 w=1
70=2× 27+16
27=1× 16+11
16=1× 11+ 5
11=2 ×5+1
1=11−(2 ×5)
1=11− ( 2× ( 16−1×11 ) ) =3 ×11−2×16
1=3 × ( 27−1 ×16 ) −2 ×16=3× 27−5× 16
1=3 × 27−5 × (70−2 ×27 )=13 ×27−5 ×70
V = 13,
W = -5
27 ×13 ≡1 (mod 70)
27 x ≡ 10(mod 70)
13 ×27 x ≡ 13 ×10(mod 70)
x ≡ 13 0(mod 70)≡ 6 0(mod 70)
ii)
21 x=6+70 k
21 x−70 k=6

7 ( 3 x−10 k ) =6
Since ( 3 x−10 k ) is an integer and the equation won’t hold therefore we can say that no solution
exists
iii)
7 x ≡ 2(mod 10)
10=1×7 +3
7=2× 3+1
1=7−(2× 3)
1=7− ( 2 × ( 10−1× 7 ) ) =3× 7−2×10
V = 3,
W = -2
7 x ≡ 2(mod 10)
3 ×7 x ≡ 6( mod 10)
x ≡ 6( mod10)
b)
The answer is not coming.
Question 4
a)
i)
16 x2− y2=25
16 x2
25 − y2
25 =1
b=5∧a= 5
4
Vertex is (0, 0)
y=± b
a x =± 4 x
ii)

iii)
c2=a2+b2= 25 ×17
16
c= 5
4 √17
Foci is,
( ± c ,0 ) =(± 5
4 √ 17 , 0)
Ecentricity is,
e= √1+ b2
a2 = √1+ 42= √17
Directrix is,
x= a2
c = 5
4 √17
iv)
x (t )= 5
4 sec t
y ( t ) =5 tan t
90 °<t<180°
b)
i)
From the equation we can write,
B = -3, A = 5, C = 4.
Since B2−4 AC=−71<0 ,

We can say that it is a equation of the ellipse
ii)
Do not have the software to do the plot
Question 5
a)
x=3 t−4
y=t−4=¿ t= y +4
Putting t in x,
x=3 ( y +4 ) −4=3 y +8
y= 1
3 x −8
3
b)
d2= ( 3 t−5−4 t−2 ) 2+ ( t−4+ 2t−7 ) 2
d2= (−t−7 ) 2 + ( 3 t−11 ) 2
d2=t2 +49+14 t+9 t2 +121−66 t2=10 t2 −52t +170
c)
The mistakes are there in the last line and the second last line
d2=10 t2−52 t +170=10 ( t2−5.2t )+170
d2=10 ( t−2.6 ) 2−10 ( 6.76 ) +170=10 ( t−2.6 ) 2−67.6+170
d2=10 ( t −2.6 )2+102.4
The minimum occurs at t = 2.6 and the minimum value is
d2=102.4=¿ d =10.12
Question 6

F=μ N
N=(mg cos 15 °−T sin 20 °)
F=μ(mg cos 15 °−T sin 20°)
Force acting opposite to friction is given by,
( mg sin 15°−T cos 20 ° )=μ (mg cos 15° −T sin 20 °)
Solving for T we get,
T =81.205 Newton
Question 7
1)
F = 2N
So if we break in to the components
−i− √ 3 j
2)
F = 10 N
So if we break in to the components
−10 cos 20 °i−10 sin20 ° j=−9.397 i−3.42 j
G = 6 N
So if we break in to the components
6 cos 35 ° i+ 6 sin35 ° j=4.915 i+3.441 j
H is
−4.46i+0.02 j

Magnitude of H = 4.5 N and it makes 0 degree with horizontal towards left
3)
F = 85 N
So if we break in to the components
−85 cos 45 ° i+85 sin 45 ° j=−60.1i+60.1 j
G = G N
So if we break in to the components
−G i
H = H N
So if we break in to the components
H cos 30 °i−H sin 30 ° j=−60.1 i+60.1 j
Solving for the directions
We get,
H = 120.21 N
G = 44 N
4)
T1 is left and T2 is right
T 1 cos 45=T 2 cos 60
T 1=T 2
√2
In the vertical direction the force equation is,
T 1 sin 45+T 2sin 60=50 ×9.8
T 1
√ 2 + √3 T 2
2 =490
T 2
2 + √3 T 2
2 =490
T 2=358.7 N
T 1=253.64 N
5)
T is the tension
T sin 50=mg cos 35

T 1=T 2
√2
m=33.4 kg
6)
F=mg sin 20−¿ T cos 60=377.223 N ∈the opposite ¿ the assumed direction
7)
μ ¿
T =4937.34
8)
μ ¿
T =36.58=m1 g
m1=3.73