XP 18

Posted: August 27th, 2021

Student’s Name

Instructor’s Name

Course

Date

XP 18

Given the equation

Question 1

Using your calculator/technology to help, sketch the graph of this function on a set of Polar axes over the interval [0, 2Pi].  Mark the orientation of the graph; mark the intercepts and identify them with their Polar coordinates. Also, identify the angle in this interval for which r is undefined.

Polar plot =>

Polar Plots

https://www4f.wolframalpha.com/Calculate/MSP/MSP437914990h21488836ib00000hieg6dbf92g80b2?MSPStoreType=image/gif&s=43&w=430.&h=71.&cdf=Animation
https://www4f.wolframalpha.com/Calculate/MSP/MSP438214990h21488836ib000062gc09cc06721eie?MSPStoreType=image/gif&s=43&w=430.&h=164.&cdf=Animation
https://www4f.wolframalpha.com/Calculate/MSP/MSP438514990h21488836ib000049gi80ghb79i5f87?MSPStoreType=image/gif&s=43&w=430.&h=289.&cdf=Animation
https://www4f.wolframalpha.com/Calculate/MSP/MSP438814990h21488836ib00003a9c8f4ifi1ffe6i?MSPStoreType=image/gif&s=43&w=430.&h=360.&cdf=Animation

Question 2

A). Let R be the region swept out by r over the theta-interval, where.   Set up the Polar-form integral whose value represents the Area of R. 

Integration; definite integral

 integral_0^pi 1\/2 (8\/(1 + cos(k)))^2 dx = (32 pi)\/(cos(k) + 1)^2

Series expansion at k=0

Indefinite integral

 integral 1\/2 (8\/(1 + cos(k)))^2 dx = (32 x)\/(cos(k) + 1)^2 + constant

B). Use the formula

                                                                        (CRC Handbook, #366)

            to help write another formula, using variable k, that would represent the Area of    region R, as defined in part A.

New formula;

Thus,

 integral 1\/2 (8\/(1 + cos(k)))^2 dx = (32 x)\/(cos(k) + 1)^2 + constant

C). Use your formula from part B to find the Area of R when k =

            I)                                     II)                                    III)     

Make small sketches illustrating what R looks like in each case; report the areas rounded to the nearest tenth.

At k =π/4 area bound is;

 integral_0^(pi\/4) 1\/2 (8\/(1 + cos(k)))^2 dx = (8 pi)\/(cos(k) + 1)^2

  however, π= 3.14 and k = π/4

Area =  = 25.10

https://www4f.wolframalpha.com/Calculate/MSP/MSP1176171c8b93691806f000000d21aa34bd3dc5a2?MSPStoreType=image/gif&s=10&w=407.&h=176.

Figure 1: Sketch varying k from 0 to π/4

At k =π/2 area bound is;

 integral_0^(pi\/2) 1\/2 (8\/(1 + cos(k)))^2 dx = (16 pi)\/(cos(k) + 1)^2

  however, π= 3.14 and k = π/2

Area =  = 54.20

https://www4f.wolframalpha.com/Calculate/MSP/MSP12131afc26d77dhg8di400002ef3cc9g41gbai86?MSPStoreType=image/gif&s=33&w=407.&h=176.

Figure 2: Sketch varying k from 0 to π/2

At k =3π/4 area bound is;

 integral_0^((3 pi)\/4) 1\/2 (8\/(1 + cos(k)))^2 dx = (24 pi)\/(cos(k) + 1)^2

  however, π= 3.14 and k = 3π/4

Area =  = 75.43

https://www4f.wolframalpha.com/Calculate/MSP/MSP204417h4f85b79fafd2b00001510d878f4b87ch0?MSPStoreType=image/gif&s=44&w=407.&h=178.

Figure 3:Sketch varying k from 0 to 3π/4

Expert paper writers are just a few clicks away

Place an order in 3 easy steps. Takes less than 5 mins.

Calculate the price of your order

You will get a personal manager and a discount.
We'll send you the first draft for approval by at
Total price:
$0.00