3. [W] Consider the following data from NASA about the density of the sun. • When you are 10% of the way out from the center of the sun (use x = 0.1), the density is approximately 82.9619 g/cm3 . • When you are 20% of the way out of the sun, the density is approximately 38.7504 g/cm3 . • When you are 70% of the way out of the sun, the density is approximately 1.7819 g/cm3 . • When you are 80% of the way out of the sun, the density is approximately 1.9824 g/cm3 . • When you are 90% of the way out of the sun, the density is approximately 0.7859 g/cm3 . We want to use this data to find a quartic (4th degree) interpolating polynomial for the density of the sun throughout. y = Ax4 + Bx3 + Cx2 + Dx + E (a) Set up this problem as a system of linear equations. (b) Find the quartic polynomial that goes through these data points. (c) Use this polynomial to approximate the density 50% of the way out of the sun.

a)

A(0.1)^4 + B(0.1)^3 + C(0.1)^2 + D(0.1) + E = 82.9619  1

A(0.2)^4 + B(0.2)^3 + C(0.2)^2 + D(0.2) + E = 38.7504 2

A(0.7)^4 + B(0.7)^3 + C(0.7)^2 + D(0.7) + E = 1.7819   3

A(0.8)^4 + B(0.8)^3 + C(0.8)^2 + D(0.8) + E = 1.9824 4

A(0.9)^4 + B(0.9)^3 + C(0.9)^2 + D(0.9) + E = 0.7859 5

b)  y = 519x^4 - 1630x^3 + 1844x^2 - 889x + 155

c)  0.1875

Step-by-step explanation:

a)

Let y = Ax^4 + Bx^3 + Cx^2 + Dx + E

Setting up the problem as a system of linear equations

when x=(10%)=0.1, y = 82.9619

therefore

A(0.1)^4 + B(0.1)^3 + C(0.1)^2 + D(0.1) + E = 82.9619  1

when x=(20%)=0.2, y = 38.7504

A(0.2)^4 + B(0.2)^3 + C(0.2)^2 + D(0.2) + E = 38.7504 2

when x=(70%)=0.7, y = 1.7819

A(0.7)^4 + B(0.7)^3 + C(0.7)^2 + D(0.7) + E = 1.7819   3

when x=(80%)=0.8, y = 1.9824

A(0.8)^4 + B(0.8)^3 + C(0.8)^2 + D(0.8) + E = 1.9824 4

when x=(90%)=0.9, y = 0.7859

A(0.9)^4 + B(0.9)^3 + C(0.9)^2 + D(0.9) + E = 0.7859 5

b)

Find the quartic polynomial that goes through these data points

Now to find the values ( A,B,C,D,E)

[MATRIX]

║A║         ║   (0.1)^4  (0.1)^3    (0.1)^2   (0.1)    1    ║^-1    ║ 82.9619  ║

║B║         ║     (0.2)^4  (0.2)^3  (0.2)^2   (0.2) 1   ║         ║ 38.7504 ║

║C║     =  ║    (0.7)^4  (0.7)^3  (0.7)^2   (0.7)   1   ║         ║ 1.7819     ║

║D║         ║    (0.8)^4  (0.8)^3  (0.8)^2  (0.8)   1   ║         ║ 1.9824    ║

║E║         ║    (0.9)^4  (0.9)^3  (0.9)^2  (0.9)   1   ║         ║ 0.7859    ║

║A║      ║  29.7619  -47.619  166.667   -238.095   89.2857 ║ ║ 82.9619║

║B║      ║  -77.381  119.048  -333.333   452.381 -160.714      ║ ║ 38.7504║

║C║  =  ║   71.131  -102.381  208.333  -269.048 91.9643      ║ ║ 1.7819    ║

║D║      ║  -26.369 33.0952  -41.6667  52.619  -17.6786     ║ ║ 1.9824   ║

║E║       ║   3.00      -2.4         -2.4        -3.00          -1.00       ║  ║ 0.7859  ║

║A║       ║ 519   ║

║B║       ║ -1630║

║C║  =   ║ 1844 ║ ,         y = 519x^4 - 1630x^3 + 1844x^2 - 889x + 155

║D║       ║ -889 ║

║E║       ║ 155    ║

c)

approximate the density 50% of the way out of the sun

so at 50% (0.5) x=0.5

we substitute

y = 519(0.5)^4 - 1630(0.5)^3 + 1844(0.5)^2 - 889(0.5) + 155

= 0.1875

correct answers below:

need more information

step-by-step explanation: