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

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation:

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent