08448380779 Call Girls In Civil Lines Women Seeking Men
Sect3 7
1. SECTION 3.7
LINEAR EQUATIONS AND CURVE FITTING
In Problems 1-10 we first set up the linear system in the coefficients a , b, that we get by
substituting each given point ( xi , yi ) into the desired interpolating polynomial equation
y = a + bx + . Then we give the polynomial that results from solution of this linear system.
1. y ( x ) = a + bx
1 1 a 1
1 3 b = 7 ⇒ a = − 2, b = 3 so y ( x) = − 2 + 3 x
2. y ( x ) = a + bx
1 −1 a 11
1 2 b = −10 ⇒ a = 4, b = −7 so y ( x) = 4 − 7 x
3. y ( x ) = a + bx + cx 2
1 0 0 a 3
1 1 1 b = 1 ⇒ a = 3, b = 0, c = −2 so y ( x) = 3 − 2 x 2
1 2 4 c
−5
4. y ( x ) = a + bx + cx 2
1 −1 1 a 1
1 1 1 b = 5 ⇒ a = 0, b = 2, c = 3 so y ( x) = 2 x + 3 x 2
1 2 4 c
16
5. y ( x ) = a + bx + cx 2
1 1 1 a 3
1 2 4 b = 3 ⇒ a = 5, b = −3, c = 1 so y ( x) = 5 − 3 x + x 2
1 3 9 c
5
6. y ( x ) = a + bx + cx 2
1 −1 1 a −1
1 3 9 b = −13
1 5 25 c
5
2. ⇒ a = − 10, b = −7, c = 2 so y ( x) = − 10 − 7 x + 2 x 2
7. y ( x ) = a + bx + cx 2 + dx 3
1 −1 1 −1 a 1
1 0 b
0 0 0
=
1 1 1 1 c 1
1 2 4 8 d −4
4
⇒ a = 0, b = , c = 1, d = −
3
4
3
so y( x) =
1
3
( 4 x + 3x 2 − 4 x3 )
8. y ( x ) = a + bx + cx 2 + dx 3
1 −1 1 −1 a 3
1 0 b
0 0 5
=
1 1 1 1 c 7
1 2 4 8 d 3
⇒ a = 5, b = 3, c = 0, d = −1 so y ( x) = 5 + 3x − x3
9. y ( x ) = a + bx + cx 2 + dx 3
1 −2 4 −8 a −2
1 −1 1 −1 b
= 2
1 1 1 1 c 10
1 2 4 8 d 26
⇒ a = 4, b = 3, c = 2, d = 1 so y ( x) = 4 + 3 x + 2 x 2 + x 3
10. y ( x ) = a + bx + cx 2 + dx 3
1 −1 1 −1 a 17
1 1 b
1 1 −5
=
1 2 4 8 c 3
1 3 9 27 d −2
⇒ a = 17, b = −5, c = 3, d = −2 so y ( x) = 17 − 5 x + 3 x 2 − 2 x 3
In Problems 11-14 we first set up the linear system in the coefficients A, B, C that we get by
substituting each given point ( xi , yi ) into the circle equation Ax + By + C = − x 2 − y 2 (see
Eq. (9) in the text). Then we give the circle that results from solution of this linear system.
3. 11. Ax + By + C = − x 2 − y 2
−1 −1 1 A −2
6 6 1 B = −72 ⇒ A = −6, B = −4, C = −12
7 5 1 C
−74
x 2 + y 2 − 6 x − 4 y − 12 = 0
( x − 3)2 + ( y − 2)2 = 25 center (3, 2) and radius 5
12. Ax + By + C = − x 2 − y 2
3 −4 1 A −25
5 10 1 B = −125 ⇒ A = 6, B = −8, C = −75
−9 12 1 C
−225
x 2 + y 2 + 6 x − 8 y − 75 = 0
( x + 3)2 + ( y − 4)2 = 100 center (–3, 4) and radius 10
13. Ax + By + C = − x 2 − y 2
1 0 1 A −1
0 −5 1 B = −25 ⇒ A = 4, B = 4, C = −5
−5 −4 1 C
−41
x2 + y 2 + 4 x + 4 y − 5 = 0
( x + 2)2 + ( y + 2)2 = 13 center (–3, –2) and radius 13
14. Ax + By + C = − x 2 − y 2
0 0 1 A 0
10 0 1 B = −100 ⇒ A = −10, B = −24, C = 0
−7 7 1 C
−98
x 2 + y 2 − 10 x − 24 y = 0
( x − 5)2 + ( y − 12)2 = 169 center (5, 12) and radius 13
In Problems 15-18 we first set up the linear system in the coefficients A, B, C that we get by
substituting each given point ( xi , yi ) into the central conic equation Ax 2 + Bxy + Cy 2 = 1 (see
Eq. (10) in the text). Then we give the equation that results from solution of this linear system.
4. 15. Ax 2 + Bxy + Cy 2 = 1
0 0 25 A 1
25 0 0 B = 1 ⇒ A=
1 1
, B=− , C=
1
25 25 25
25 25 25 C
1
x 2 − xy + y 2 = 25
16. Ax 2 + Bxy + Cy 2 = 1
0 0 25 A 1
25 0 B = 1
0 ⇒ A=
1
, B=−
7
, C=
1
25 100 25
100 100 100 C
1
4 x 2 − 7 xy + 4 y 2 = 100
17. Ax 2 + Bxy + Cy 2 = 1
0 0 1 A 1
1 0 0 B = 1 ⇒ A = 1, B = −
199
, C =1
100
100 100 100 C
1
100 x 2 − 199 xy + 100 y 2 = 100
18. Ax 2 + Bxy + Cy 2 = 1
0 0 16 A 1
9 0 0 B = 1 ⇒
1
A= , B =−
481
, C=
1
9 3600 16
25 25 25 C
1
400 x 2 − 481xy + 225 y 2 = 3600
B
19. We substitute each of the two given points into the equation y = A + .
x
1 1
1 A = 5 ⇒ A = 3, B = 2 so y = 3 +
2
1 B 4
2 x
5. B C
20. We substitute each of the three given points into the equation y = Ax + + .
x x2
1 1 1
A 2
1 1
B = 20
2 8 16
⇒ A = 10, B = 8, C = −16 so y = 10 x + −
2 4 x x2
C 41
1 1
4
4 16
In Problems 21 and 22 we fit the sphere equation ( x − h )2 + ( y − k )2 + ( z − l ) 2 = r 2 in the expanded
form Ax + By + Cz + D = − x 2 − y 2 − z 2 that is analogous to Eq. (9) in the text (for a circle).
21. Ax + By + Cz + D = − x 2 − y 2 − z 2
4 6 15 1 A −277
13 5 7 1 B
= −243 ⇒ A = −2, B = −4, C = −6, D = −155
5 14 6 1 C −257
5 5 −9 1 D −131
x 2 + y 2 + z 2 − 2 x − 4 y − 6 z − 155 = 0
( x − 1)2 + ( y − 2)2 + ( z − 3)2 = 169 center (1, 2, 3) and radius 13
22. Ax + By + Cz + D = − x 2 − y 2 − z 2
11 17 17 1 A −699
29 B
1 15 1 −1067
= ⇒ A = −10, B = 14, C = −18, D = −521
13 −1 33 1 C −1259
−19 −13 1 1 D −531
x 2 + y 2 + z 2 − 10 x + 14 y − 18 z − 521 = 0
( x − 5) 2 + ( y + 7)2 + ( z − 9) 2 = 676 center (5, –7, 9) and radius 26
In Problems 23-26 we first take t = 0 in 1970 to fit a quadratic polynomial P(t ) = a + bt + ct 2 .
Then we write the quadratic polynomial Q(T ) = P(T − 1970) that expresses the predicted
population in terms of the actual calendar year T.
6. 23. P(t ) = a + bt + ct 2
1 0 0 a 49.061
1 10 100 b = 49.137
1 20 400 c
50.809
P(t ) = 49.061 − 0.0722 t + 0.00798 t 2
Q(T ) = 31160.9 − 31.5134 T + 0.00798 T 2
24. P(t ) = a + bt + ct 2
1 0 0 a 56.590
1 10 100 b = 58.867
1 20 400 c
59.669
P(t ) = 56.590 + 0.30145 t − 0.007375 t 2
Q(T ) = − 29158.9 + 29.3589 T − 0.007375 T 2
25. P(t ) = a + bt + ct 2
1 0 0 a 62.813
1 10 100 b = 75.367
1 20 400 c
85.446
P(t ) = 62.813 + 1.37915 t − 0.012375 t 2
Q(T ) = − 50680.3 + 50.1367 T − 0.012375 T 2
26. P(t ) = a + bt + ct 2
1 0 0 a 34.838
1 10 100 b = 43.171
1 20 400 c
52.786
P(t ) = 34.838 + 0.7692 t + 0.00641t 2
Q(T ) = 23396.1 − 24.4862 T + 0.00641T 2
In Problems 27-30 we first take t = 0 in 1960 to fit a cubic polynomial P(t ) = a + bt + ct 2 + dt 3 .
Then we write the cubic polynomial Q(T ) = P(T − 1960) that expresses the predicted population
in terms of the actual calendar year T.
7. 27. P(t ) = a + bt + ct 2 + dt 3
1 0 0 0 a 44.678
1 10 100 1000 b
= 49.061
1 20 400 8000 c 49.137
1 30 900 27000 d 50.809
P(t ) = 44.678 + 0.850417 t − 0.05105 t 2 + 0.000983833 t 3
Q(T ) = − 7.60554 × 106 + 11539.4 T − 5.83599 T 2 + 0.000983833 T 3
28. P(t ) = a + bt + ct 2 + dt 3
1 0 0 0 a 51.619
1 10 100 1000 b
= 56.590
1 20 400 8000 c 58.867
1 30 900 27000 d 59.669
P(t ) = 51.619 + 0.672433 t − 0.019565 t 2 + 0.000203167 t 3
Q(T ) = − 1.60618 × 106 + 2418.82 T − 1.21419 T 2 + 0.000203167 T 3
29. P(t ) = a + bt + ct 2 + dt 3
1 0 0 0 a 54.973
1 10 100 1000 b
= 62.813
1 20 400 8000 c 75.367
1 30 900 27000 d 85.446
P(t ) = 54.973 + 0.308667 t + 0.059515 t 2 − 0.00119817 t 3
Q(T ) = 9.24972 ×106 − 14041.6 T + 7.10474 T 2 − 0.00119817 T 3
30. P(t ) = a + bt + ct 2 + dt 3
1 0 0 0 a 28.053
1 10 100 1000 b
= 34.838
1 20 400 8000 c 43.171
1 30 900 27000 d 52.786
P(t ) = 28.053 + 0.592233 t + 0.00907 t 2 − 0.0000443333 t 3
Q(T ) = 367520 − 545.895 T + 0.26975 T 2 − 0.0000443333T 3
8. In Problems 31-34 we take t = 0 in 1950 to fit a quartic polynomial P(t ) = a + bt + ct 2 + dt 3 + et 4 .
Then we write the quartic polynomial Q(T ) = P(T − 1950) that expresses the predicted
population in terms of the actual calendar year T.
31. P(t ) = a + bt + ct 2 + dt 3 + et 4 .
1 0 0 0 0 a 39.478
1 10 100 1000 b
10000 44.678
1 20 400 8000 160000 c = 49.061
1 30 900 27000 810000 d 49.137
1
40 1600 64000 2560000 e
50.809
P(t ) = 39.478 + 0.209692 t + 0.0564163 t 2 − 0.00292992 t 3 + 0.0000391375 t 4
Q(T ) = 5.87828 × 108 − 1.19444 ×106 T + 910.118 T 2 − 0.308202 T 3 + 0.0000391375 T 4
32. P(t ) = a + bt + ct 2 + dt 3 + et 4 .
1 0 0 0 0 a 44.461
1 10 100 1000 b
10000 51.619
1 20 400 8000 160000 c = 56.590
1 30 900 27000 810000 d 58.867
1
40 1600 64000 2560000 e
59.669
P(t ) = 44.461 + 0.7651t − 0.000489167 t 2 − 0.000516 t 3 + 7.19167 × 10−6 t 4
Q(T ) = 1.07807 × 108 − 219185 T + 167.096 T 2 − 0.056611T 3 + 7.19167 ×10−6 T 4
33. P(t ) = a + bt + ct 2 + dt 3 + et 4 .
1 0 0 0 0 a 47.197
1 10 100 1000 b
10000 54.973
1 20 400 8000 160000 c = 62.813
1 30 900 27000 810000 d 75.367
1
40 1600 64000 2560000 e
85.446
P(t ) = 47.197 + 1.22537 t − 0.0771921t 2 + 0.00373475 t 3 − 0.0000493292 t 4
Q(T ) = − 7.41239 × 108 + 1.50598 × 106 T − 1147.37 T 2 + 0.388502 T 3 − 0.0000493292 T 4
9. 34. P(t ) = a + bt + ct 2 + dt 3 + et 4 .
1 0 0 0 0 a 20.190
1 10 100 1000 b
10000 28.053
1 20 400 8000 160000 c = 34.838
1 30 900 27000 810000 d 43.171
1
40 1600 64000 2560000 e
52.786
P(t ) = 20.190 + 1.00003 t − 0.031775 t 2 + 0.00116067 t 3 − 0.00001205 t 4
Q(T ) = − 1.8296 ×108 + 370762 T − 281.742 T 2 + 0.0951507 T 3 − 0.00001205 T 4
35. Expansion of the determinant along the first row gives an equation of the form
ay + bx 2 + cx + d = 0 that can be solved for y = Ax 2 + Bx + C. If the coordinates of any
one of the three given points ( x1 , y1 ), ( x2 , y2 ), ( x3 , y3 ) are substituted in the first row, then
the determinant has two identical rows and therefore vanishes.
36. Expansion of the determinant along the first row gives
y x2 x 1
1 1 1 3 1 1 3 1 1 3 1 1
3 1 1 1
= y 4 2 1−x 3 2 1+x 3 4 1− 3 4 2 =
2
3 4 2 1
9 3 1 7 3 1 7 9 1 7 9 3
7 9 3 1
−2 y + 4 x 2 − 12 x + 14 = 0 .
Hence y = 2 x 2 − 6 x + 7 is the parabola that interpolates the three given points.
37. Expansion of the determinant along the first row gives an equation of the form
a( x 2 + y 2 ) + bx + cy + d = 0, and we get the desired form of the equation of a circle upon
division by a. If the coordinates of any one of the three given points ( x1 , y1 ), ( x2 , y2 ), and
( x3 , y3 ) are substituted in the first row, then the determinant has two identical rows and
therefore vanishes.
38. Expansion of the determinant along the first row gives
x2 + y 2 x y 1
25 3 −4 1
=
125 5 10 1
225 −9 12 1
10. 3 −4 1 25 −4 1 25 3 1 25 3 −4
= ( x + y ) 5 10 1 − x 125 10 1 + y 125 5 1 − 125 5 10
2 2
−9 12 1 225 12 1 225 −9 1 225 −9 12
= 200( x 2 + y 2 ) + 1200 x − 1600 y − 15000 = 0.
Division by 200 and completion of squares gives ( x + 3)2 + ( y − 4)2 = 100, so the circle has
center (–3, 4) and radius 10.
39. Expansion of the determinant along the first row gives an equation of the form
ax 2 + bxy + cy 2 + d = 0, which can be written in the central conic form
Ax 2 + Bxy + Cy 2 = 1 upon division by –d. If the coordinates of any one of the three given
points ( x1 , y1 ), ( x2 , y2 ), and ( x3 , y3 ) are substituted in the first row, then the determinant
has two identical rows and therefore vanishes.
40. Expansion of the determinant along the first row gives
x2 y2 1
xy
0 0 16 1
=
9 0 0 1
25 25 25 1
0 16 1 0 16 1 0 0 1 0 0 16
= x 0 0 1 − xy 9 0 1 + y 9 0 1 − 9 0 0
2
25 25 1 25 25 1 25 25 1 25 25 25
= 400 x 2 − 481xy + 225 y 2 − 3600 = 0.
