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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
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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