   Chapter 10.5, Problem 60E

Chapter
Section
Textbook Problem

The dwarf planet Pluto travels in an elliptical orbit around the sun (at one focus). The length of the major axis is 1.18 × 1010 km and the length of the minor axis is 1.14 × 1010 km. Use Simpson’s Rule with n = 10 to estimate the distance traveled by the planet during one complete orbit around the sun.

To determine

To Find: The distance traveled by the planet during one complete orbit around the sun.

Explanation

Given:

The equation of the ellipse is x2a2+y2b2=1.

Formula used:

The circumference of the ellipse is L=ab[dxdt]2+[dydt]2dt.

Calculation:

The parametric representation for the given ellipse is as below,

(x,y)=(acost,bsint)0t2π

Then, the value of 2a is 1.18×1010km and 2b is 1.14×1010km.

Find the derivatives as follows,

x=acostdxdt=asint

y=bsintdydt=bcost

Substitute the asint for [dxdt], bcost for [dydt] and limit is 0 to 2π in the above mentioned formula.

L=02π(asint)2+(bcost)2dtL=02πa2sin2t+b2cos2tdt

Substitute the value 12(1.18×1010)km for a and 12(1.14×1010)km for b.

L=02π(12(1.18×1010))2sin2t+(12(1.14×1010))2cos2tdtL=1010202π(1.18)2sin2t+(1.14)2cos2tdtL=101020πf(t)dt

Compute the value of Δt.

We know that n is 10.

Δt=ban

Substitute the value 0 for a and 2π for b.

Δt=2π010Δt=2π10Δt=π5

Compute the function values.

Consider the function f(0).

f(0)=f(t)

Substitute the value of 0 for t.

f(0)=(1.18)2sin20+(1.14)2cos20f(0)=1.14

Consider the function f(Δt).

f(Δt)=f(t)f(π5)=f(t)

Substitute the value of π5 for t.

f(π5)=(1.18)2sin2(π5)+(1.14)2cos2(π5)f(π5)=1.154

Consider the function f(2Δt).

f(2Δt)=f(t)f(2×π5)=f(t)

Substitute the value of 2π5 for t.

f(2π5)=(1.18)2sin2(2π5)+(1.14)2cos2(2π5)f(2π5)=1.176

Consider the function f(3Δt).

f(3Δt)=f(t)f(3×π5)=f(t)

Substitute the value of 3π5 for t.

f(3π5)=(1.18)2sin2(3π5)+(1.14)2cos2(3π5)f(3π5)=1.176

Consider the function f(4Δt).

f(4Δt)=f(t)f(4×π5)=f(t)

Substitute the value of 4π5 for t.

f(4π5)=(1.18)2sin2(4π5)+(1

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