Dottie Parra
2021-08-09
Answered

Graph the lines and conic sections $r=\frac{1}{(1+2\mathrm{sin}\theta )}$

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joshyoung05M

Answered 2021-08-10
Author has **97** answers

Step 1

Consider the given polar equation,

$r=\frac{1}{(1+2\mathrm{sin}\left(\theta \right))}$

Compare the given with standard equation of the hyperbola,

$r=\frac{ke}{1+2\mathrm{sin}\left(\theta \right)}$

we get,

$e=2$ and $k=\frac{1}{2}$

so the given equation is the equation of the hyperbola.

Step 2

So, the graph of the hyperbola is,

Consider the given polar equation,

Compare the given with standard equation of the hyperbola,

we get,

so the given equation is the equation of the hyperbola.

Step 2

So, the graph of the hyperbola is,

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I am approximating a solution to a first order LODE using Euler's method. I made two tables, one using a step size of .01 and another using .05 ( I had to start at x=0 and end at x=1). I am not understanding the directions for the second part of my assignment:

It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $C{h}^{n}$, where $C$ is a generally unknown constant and $n$ is the order. For Euler's method, plot the points:

$(0.1,\varphi (1)-{y}_{10}),$

$(.05,\varphi (1)-{y}_{20}),$

$(.025,\varphi (1)-{y}_{40}),$

$(.0125,\varphi (1)-{y}_{80}),$

$(.00625,\varphi (1)-{y}_{160})$

And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?

Any clarification is appreciated.

Thanks

Edit:

The LODE I am given is ${y}^{\prime}=x+2y,y(0)=1$ and the exact solution I found was $\varphi (x)=\frac{1}{4}(-2x+5{e}^{2x}-1)$.

It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $C{h}^{n}$, where $C$ is a generally unknown constant and $n$ is the order. For Euler's method, plot the points:

$(0.1,\varphi (1)-{y}_{10}),$

$(.05,\varphi (1)-{y}_{20}),$

$(.025,\varphi (1)-{y}_{40}),$

$(.0125,\varphi (1)-{y}_{80}),$

$(.00625,\varphi (1)-{y}_{160})$

And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?

Any clarification is appreciated.

Thanks

Edit:

The LODE I am given is ${y}^{\prime}=x+2y,y(0)=1$ and the exact solution I found was $\varphi (x)=\frac{1}{4}(-2x+5{e}^{2x}-1)$.

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