Sect3 7

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 
 

⇒ 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.

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.

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
 

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.

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.

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

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

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.

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.


x2 + y 2 x y 1
25 3 −4 1
=
125 5 10 1
225 −9 12 1

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.

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.


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.

Sect3 7

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Sect3 7