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## Homework Statement

1.

put in polar form

x

^{2}+y

^{2}-3x+4y=0

my work:

x

^{2}+y

^{2}=3x-4y

r

^{2}=3rcos[tex]\theta[/tex]-4rsin[tex]\theta[/tex]

r=3cos[tex]\theta[/tex]-4sin[tex]\theta[/tex]

2.

put in cartesian form

r

^{2}=tan[tex]\theta[/tex]

r

^{2}=y/x

r=sqrt(y/x)

3.

find slope at [tex]\theta[/tex]=[tex]\pi[/tex]/2

then find points where tangent line is horizontal

3/(2+2cos[tex]\theta[/tex])

using the long formula which I don't feel like typing out, I get -1 as the slope when I plug in [tex]\pi[/tex]/2 for theta.

for the part with the horizontal tangent line, I set the numerator of the derivative to 0, but couldnt get an answer. the graph of the original function seems to be like the graph of +-sqrt(x) reflected about the y axis, like the graph of a backwards c. because the graph continuously increased/decreased as i moved left, there were no places where the slope was 0

4.

find the area of one loop of the graph

2sin(5[tex]\theta[/tex])

first off, i have one question. what does one loop mean? does it mean simply the graph i get when I graph the graph from 0 to 2 pi? or is one loop just one petal of the graph? 2sin5[tex]\theta[/tex] gives me a graph of a rose curve with 5 petals, so to find the area of one loop would I find the area of 5 petals or just one?

now that that's out of the way, I'm having a little trouble finding the limits of integration. no idea where to start on this one. do I set the original equation to 0 or something?

5.

find the area in common b/w both graphs.

-6cos[tex]\theta[/tex]

2-2cos[tex]\theta[/tex]

the 2 graphs where a circle and a limacon.

kinda hard to really explain the graph but here's my integral

i worked with only the part above the x axis so everything is mnultipled by 2

2[[tex]\int[/tex][[tex]\pi[/tex]/2 to [tex]\pi[/tex]] (1/2)(-6cos[tex]\theta[/tex])

^{2}d[tex]\theta[/tex] - [tex]\int[/tex][2[tex]\pi[/tex]/3 to [tex]\pi[/tex]] (1/2)(2-2cos[tex]\theta[/tex])

^{2}d[tex]\theta[/tex] - [tex]\int[/tex][0 to 2[tex]\pi[/tex]/3] (1/2)(-6cos[tex]\theta[/tex])

^{2}d[tex]\theta[/tex]

dunno if that will come out correctly. basically, if i did it right, there should be 3 integrals there. the first integral gives you the area of the cirlce. the second integral subtracts...fuuu....

alright, actually, while i am just typing this i realized I possible mistake. I won't erase what i just typed just in case its really correct or w/e. here's my new integral

2([tex]\int[/tex][2[tex]\pi[/tex]/3 to [tex]\pi[/tex]] (1/2)(2-2cos[tex]\theta[/tex])

^{2}d[tex]\theta[/tex]+[tex]\int[/tex][0 to 2[tex]\pi[/tex]/3](1/2)(-6cos[tex]\theta[/tex])

^{2}) d[tex]\theta[/tex])

here, I find the area of most of the common area using the area of the limacon , and add the area of a small sliver that is missed by using the integral of the circle. i probably didnt explain this well, but w/e.

anyways, for the first integral i attempted (the one i decided had a mistake), i got -5[tex]\pi[/tex]. should my answer even be negative? for the second integral i get 14 pi....does that seem right?

I know its a lot of problems, but any help would be appreciated. thanks.

## Homework Equations

I did my work in part 1 b/c it was just easier that way.

## The Attempt at a Solution

work in part 1...

EDIT: integrals came out poorly. how do I do the limits of integration in latex again? also, the 2s in fron tof the integrals should be distributed to all the integrals, as in its 2 times all the integrals, which should have been bracketed