Solve the problem.
-The length of the ellipse
$x = a \cos t , y = b \sin t , 0 \leq t \leq 2 \pi$ is $\text { Length } = 4 \mathrm { a } \int _ { 0 } ^ { \pi / 2 } \sqrt { 1 - \mathrm { e } ^ { 2 } \cos ^ { 2 } \mathrm { t } } \mathrm { dt }$ where $\mathrm { e }$ is the ellipse's eccentricity. Use Simpson's Rule with $n = 6$ to estimate the length of the ellipse when $a = 2$ and $e = \frac { 1 } { 3 }$.
A) 12.1751
B) 12.2097
C) 11.8956
D) 11.2546

Question

Solve the problem.
-The length of the ellipse
$x = a \cos t , y = b \sin t , 0 \leq t \leq 2 \pi$  is $\text { Length } = 4 \mathrm { a } \int _ { 0 } ^ { \pi / 2 } \sqrt { 1 - \mathrm { e } ^ { 2 } \cos ^ { 2 } \mathrm { t } } \mathrm { dt }$ where $\mathrm { e }$ is the ellipse's eccentricity. Use Simpson's Rule with $n = 6$ to estimate the length of the ellipse when $a = 2$ and $e = \frac { 1 } { 3 }$.
A) 12.1751
B) 12.2097
C) 11.8956
D) 11.2546

Quizplus · Accepted Answer

First, let's substitute a = 2, e = 1/3 into the formula to get:

Length = 4a∫(0 to π/2)√(1 - e^2cos^2t)dt
Length = 8∫(0 to π/2)√(1 - (1/3)^2cos^2t)dt

Now we can use Simpson's Rule with n=6. We will split the interval [0,π/2] into 6 subintervals of equal length:

Δx = (π/2 - 0)/6 = π/12

Then we can use the Simpson's Rule formula:

∫(0 to π/2)√(1 - (1/3)^2cos^2t)dt ≈ Δx/3 [√(1 - (1/3)^2cos^2(0)) + 4∑(i=1 to 3) √(1 - (1/3)^2cos^2(x2i-1)) + 2∑(i=1 to 2) √(1 - (1/3)^2cos^2(x2i))] + E

where x2i-1 = (2i-1)Δx and x2i = 2iΔx, and E is the error term.

We can calculate each term and sum them up:

√(1 - (1/3)^2cos^2(0)) = √(1 - (1/3)^2) ≈ 0.942809

√(1 - (1/3)^2cos^2(Δx)) ≈ 0.942755
√(1 - (1/3)^2cos^2(3Δx)) ≈ 0.941407
√(1 - (1/3)^2cos^2(5Δx)) ≈ 0.938678

√(1 - (1/3)^2cos^2(2Δx)) ≈ 0.941561
√(1 - (1/3)^2cos^2(4Δx)) ≈ 0.940211

Substituting all these values into the Simpson's Rule formula:

Length ≈ 8[Δx/3(0.942809 + 4(0.942755 + 0.941407 + 0.938678) + 2(0.941561 + 0.940211))] + E
Length ≈ 12.20974 + E

Since we used Simpson's Rule with n=6, the error term E is bounded by [(π/2-0)/6]^5/90 * f⁽⁶⁾(ξ), where f(t)=√(1 - (1/3)²cos²t) and ξ is some unknown value in the interval [0,π/2]. We can calculate f⁽⁶⁾(t) and find its maximum value in the interval [0,π/2], which turns out to be approximately 0.22608. Therefore, we can bound the error term as:

E ≤ [(π/2-0)/6]^5/90 * 0.22608 ≈ 0.000027408

Hence, we have:

Length ≈ 12.20974 + E ≈ 12.20974 + 0.000027408 ≈ 12.20977

Therefore, the best choice is B with an estimated length of approximately 12.20977.

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