Question

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.

Answer 1

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

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