Let a > b > 0. Consider the ellipse formed by the points (x, y) = (a sin , b cos
ID: 2888337 • Letter: L
Question
Let a > b > 0. Consider the ellipse formed by the points (x, y) = (a sin , b cos ) as we vary the angle from 0 to 2.
(a) Consider the top half of the ellipse, that is, restrict to the points having y 0. Write a function f(x) whose graph coincide with the top half of the ellipse. Make sure to describe domain and range of f(x).
(b) Write a formula for the total length of the ellipse.
(c) Use the Simpson’s rule with N = 6 to approximate the length I (5, 4) of the ellipse having a = 5, b = 4. You do not need to simplify your answer.
Explanation / Answer
(x, y) = (a sin , b cos )
(a)
(x/a)2+(y/b)2 =1
=>(y/b)2 =1-(x/a)2
=>(y/b) =[1-(x/a)2]
=>y =b[1-(x/a)2]
=>y =(b/a)(a2-x2)
=>f(x) =(b/a)(a2-x2)
domain is [-a,a]
range is [0,b]
(b)
f(x) =(b/a)(a2-x2)
f '(x) =(b/a)(-x/(a2-x2))
total length of ellipse = 2*[-a to a][1+(f '(x))2] dx
total length of ellipse = 2*[-a to a][1+((b/a)(-x/(a2-x2)))2] dx
total length of ellipse = 2*[-a to a][1+((b2x2/(a4-a2x2))] dx
total length of ellipse = [-a to a]2((a4-a2x2+b2x2)/(a4-a2x2))dx
(c)
ellipse having a = 5, b = 4
total length of ellipse = [-5 to 5]2((54-52x2+42x2)/(54-52x2))dx
total length of ellipse = [-5 to 5]2((625-9x2)/(625-25x2))dx
x=(5-(-5))/6 =(5/3)
by simpson's rule
total length of ellipse = ((5/3)/3)[(2((625-9(-5)2)/(625-25(-5)2)))+4(2((625-9(-10/3)2)/(625-25(-10/3)2)))+2(2((625-9(-5/3)2)/(625-25(-5/3)2)))+4(2((625-9*02)/(625-25*02)))+2(2((625-9(5/3)2)/(625-25(5/3)2)))+4(2((625-9(10/3)2)/(625-25(10/3)2)))+(2((625-9*52)/(625-25*52)))]
total length of ellipse =22.4967 approximately
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