Babynini
  • Babynini
Graph and convert this equation to rectangular coordinates polar equation: r=8cos(theta)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
dan815
  • dan815
|dw:1434650063164:dw|
dan815
  • dan815
|dw:1434650153992:dw|
dan815
  • dan815
what do u think an expression for r is? in terms of x and y

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Babynini
  • Babynini
How did you graph that? o.0
dan815
  • dan815
u click on draw button
Babynini
  • Babynini
em, well I tried multiplying both sides by r and got r^2=8rcos(theta) (x^2+y^2)=8x
dan815
  • dan815
yep that is right
Babynini
  • Babynini
haha i know that. But i meant how did you come up with that graph.
Babynini
  • Babynini
but i think it needs to be simplified further than that. right?
dan815
  • dan815
thats the relationship between polar and cartesian
Babynini
  • Babynini
(x^2+y^2)-8x = 0 now what?
dan815
  • dan815
u can put it in circle form
dan815
  • dan815
x^2+y^2=8x rewrite it as this first x^2-8x + y^2=0 ^------------ complete the square and put the constant on the right side then
Babynini
  • Babynini
I don't know how to complete squares :/ that is where I got stuck.
dan815
  • dan815
um
dan815
  • dan815
watch youtube video, its faster
dan815
  • dan815
its going to look like this
dan815
  • dan815
(x-4)^2-16+y^2=0 (x-4)^2+y^2=16 (x-4)^2+y^2=4^2 radius 4, shift in x by 4
dan815
  • dan815
|dw:1434650531548:dw|
Babynini
  • Babynini
hrm ok. I tried looking for a good youtube video but just got all confused =.=
imqwerty
  • imqwerty
we have the basic formulas - x = rcos(theta) y = rsin(theta) If x^2 + y^2 = r^2 and r^2 = 8rcos(theta) = 8(x) then you have been given that: x^2 + y^2 = 8x x^2 + 8x + y^2 = 0 Convert to the center-vertex form for circles.
dan815
  • dan815
um completing square therse nothing to it
Babynini
  • Babynini
haha sigh, i know it should be super easy. But not for me xD
dan815
  • dan815
Any quadratic is in the form ax^2+bx+c, where a b and c are anything
dan815
  • dan815
okay no lets change this... to something simpler first
Babynini
  • Babynini
k
dan815
  • dan815
x^2+bx+c lets say u have an equation of this form
dan815
  • dan815
to put in square form, okay so u need a bracket with the variable + a constant squares
dan815
  • dan815
(x+d)^2 okay when u square this what happens (x+d)(x+d)=x*(x+d) +d*(x+d) = x^2+dx+dx+d^2 = x^2+2dx+d^2 therefore if you want to rewrite x^2+bx+c in (x+d)^2 form then you know that the 2dx has add up to bx 2dx=bx d=b/2 so we see that it has to be in this form (x+b/2)^2 now the constant part will be d^2 or (b/2)^2, this we subtract away as it is in cluded when u square it
dan815
  • dan815
x^2+bx+c ---> becomes (x+b/2)^2-(b/2)^2+c
dan815
  • dan815
in your case we have the equation x^2+bx+c x^2-8x+0 which means b=-8 and c=0 okay so (x+b/2)^2 -(b/2)^2+c (x+(-8/2))^2 - (-8/2)^2 +0
Babynini
  • Babynini
k..i'm trying to follow hah
Babynini
  • Babynini
and then we just put the y^2 back in there?

Looking for something else?

Not the answer you are looking for? Search for more explanations.