514_James Stewart Calculus 5 Edition Answers

514_James Stewart Calculus 5 Edition Answers - Stewart...

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Unformatted text preview: Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 5 4 3 2 42. (a) p(x)=x (2+r )x +(1+2r )x (1 r )x +2(1 r )x+r 1 / 4 3 2 p (x)=5x 4(2+r )x +3(1+2r )x 2(1 r )x+2(1 r ) . So we use 5 4 3 2 n n n n x (2+r )x +(1+2r )x (1 r )x +2(1 r )x +r 1 x =x 4 n 3 2 n n+1 n n n . We substitute in the value r 3.04042 10 6 5x 4(2+r )x +3(1+2r )x 2(1 r )x +2(1 r ) n in order to evaluate the approximations numerically. The libration point L is slightly less than 1 AU 1 from the Sun, so we take x =0.95 as our first approximation, and get x 1 x 0.98451 , x 4 5 2 0.98830 , x 0.98976 , x 6 0.98998 , x 7 0.96682 , x 3 0.97770 , 0.98999 x . 8 9 So, to five decimal places, L is located 0.98999 AU from the Sun (or 0.01001 AU from Earth). 1 (b) In this case we use Newton’s method with the function 2 5 4 3 2 p(x) 2rx =x (2+r )x +(1+2r )x (1+r )x +2(1 r )x+r 1 2 p(x) 2rx / 4 3 2 =5x 4(2+r )x +3(1+2r )x 2(1+r )x+2(1 r ) . So 5 4 3 2 n n n n x (2+r )x +(1+2r )x (1+r )x +2(1 r )x +r 1 x =x n 4 3 2 n n+1 n n n . Again, we substitute r 3.04042 10 5x 4(2+r )x +3(1+2r )x 2(1+r )x +2(1 r ) 6 . L is 2 n slightly more than 1 AU from the Sun and, judging from the result of part (a), probably less than 0.02 AU from Earth. So we take x =1.02 and get x 1.01422 , x 1.01118 , x 1.01018 , 1 x 5 2 3 4 1.01008 x . So, to five decimal places, L is located 1.01008 AU from the Sun (or 0.01008 AU 6 2 from Earth). 18 ...
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