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(a) Sketch a graph of the following parametric curve

$\:C:\left\{x=3+2cos\theta \right\}\left\{y=5+2sin\theta \right\}$

HINT: use the fact that $\left(2cos\theta \right)^2+\left(2sin\theta \right)^2\:=\:4$

i started with the fact that $x^2+y^2=$$\left(3+2cos\theta \right)^2+\left(5+2sin\theta \right)^2$ and then i kept the HINT in mind and only squared the $\:3$ and $5$ to get

$\left(2cos\theta \right)^2+\left(2sin\theta \right)^2$$+$$\:9\:+\:25$

so i switch $\left(2cos\theta \right)^2+\left(2sin\theta \right)^2\:=\:4$

and got $33+4\:=37$.

so, $x^2+y^2=36$

so when i sketch this it's just a circle going counterclockwise with radius $=36$ ??

(b) Find the length of the curve C in part (a) for $\:0\le \theta \le \frac{5\pi }{4}$

so the formula is

$L=\int _a^b\:\sqrt{\left(\frac{dx}{dy}\right)^2+\left(\frac{dy}{dx}\right)^2}dx$

$\frac{d\theta }{dx}$=-2sin$\theta $

$\frac{d\theta }{dy}$=2cos$\theta $

$\:L=\int \:_0^{\frac{5\pi }{4}}\:\sqrt{4sin^2\theta +4cos^2\theta }d\theta $

$\:L=\int \:_0^{\frac{5\pi }{4}}\:\sqrt{4}\sqrt{sin^2\theta \:+cos^2\theta \:}d\theta $

$sin^2\theta +cos^2\theta =1\:$

$2\int \:_0^{\frac{5\pi }{4}}\:1\:d\theta $

$L=2\left[\theta \right]\frac{\frac{5\pi }{4}}{0}$

$2\left[\frac{5\pi }{4}-0\right]$

= $\frac{5\pi }{2}$

does all this look correct?

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  • $\begingroup$ You can't leave out the cross term in the square, just because it makes things! You'e missing $12\cos\theta$ and $20\sin\theta$ $\endgroup$ – DJohnM Oct 25 '14 at 2:45
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$\bullet\;\;\; (x-3)=2\cos \phi\Rightarrow (x-3)^2 = 4\cos^2 \phi$

$\bullet\;\;\; (y-5)=2\sin \phi\Rightarrow (y-5)^2 = 4\sin^2 \phi$

So $(x-3)^2+(y-5)^2 = 4$

So It Represent a Circle whose Center is at $(3,5)$ and Radius is $=2$

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  • $\begingroup$ okay, so i would just bring over the 3 and 5 right away and i also forgot that 4 would be r^2 not r. thank you $\endgroup$ – Charlene Oct 25 '14 at 2:11
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The formula $x^2 + y^2 = r^2$ is the general formula for a circle whose center is at the origin. Your circle is not centered at the origin.

In part (b), notice that actually $\frac{dx}{dy} \neq \frac{dx}{d\theta} = -2\cos\theta.$ The fact that if you multiply together the values you got for $\frac{dx}{dy}$ and $\frac{dy}{dx},$ and the product is not $1,$ should be a tip-off that these are incorrect derivatives. However, the second integral you wrote is much better. It isn't really at all the same as the first integral, but it is equal to what you would get for the first integral if you wrote $d\theta$ (instead of $dx$ or $dy$) in all the places you should, but kept $dx$ and $dy$ in the places where they belong.

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