This document discusses linear equations and curve fitting. It provides 18 examples of using a linear system to solve for the coefficients of linear, quadratic, and cubic polynomials that fit given data points. It also provides examples of using a linear system to solve for the coefficients of circle and central conic equations that fit given points. The linear systems are set up and solved, providing the resulting equations that fit the data in each example.

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.