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In-s [12.5K]
4 years ago
7

Determine the graph of the polar equation r=4/(1-1/2cosx(theta))

Mathematics
1 answer:
crimeas [40]4 years ago
6 0

Answer:

  see below

Step-by-step explanation:

We assume you want the graph of ...

  r=\dfrac{4}{1-\frac{1}{2}\cos{(\theta)}}

A graphing calculator or spreadsheet is useful for this.

__

You know cos(θ) = cos(-θ), so the graph is symmetrical about the x-axis. You can evaluate the function at a few points to find the general outline.

  r at 0° = 8

  r at 30° ≈ 7.05

  r at 45° ≈ 6.19

  r at 60° ≈ 5.33

  r at 90° = 4

  r at 120° = 3.2

  r at 135° ≈ 2.96

  r at 150° ≈ 2.79

  r at 180° ≈ 2.67

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We're looking for a solution of the form

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\displaystyle\sum_{n\ge0}\left(\bigg(2(n+2)(n+1)a_{n+2}-4a_n\bigg)x^{n+2}+2(n+1)a_{n+1}x^{n+1}+(n+2)(n+1)a_{n+2}x^n\right)=0

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2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0

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\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of y(0)=y'(0)=1, a degree-8 approximation of

y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).

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Answer:

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Hey ya didnt get the question here

so I creat a equation that ya can use

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Hope this help ya

<3

Red

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