anonymous
  • anonymous
Find a Cartesian equation relating x and y corresponding to the parametric equations x=5sin(5 t) y=9cos(5 t) Write your answer in the form P(x,y)=0 where P(x,y) is a polynomial in x and y such that the coefficient of y^2 is 25.
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!
anonymous
  • anonymous
x^2+(25y^2/81)-1 this is the answer I got :/
amistre64
  • amistre64
it does seem rather elliptic doesnt it ...
anonymous
  • anonymous
Thats what I did, I used the ellipse formula but I don't know where I'm going wrong.

Looking for something else?

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

More answers

amistre64
  • amistre64
r cos(t) = 5 sin(5t) r sin(t) = 9 cos(5t) \[r^2 = \sqrt{25sin^2(5t)+81cos^2(5t)}\] hmmm
amistre64
  • amistre64
got a spurious ^2 on that last r
anonymous
  • anonymous
wait where did you get the r, I thought I just had to get "rid" of my t variable by squaring both side and using the unit circle formula,
amistre64
  • amistre64
just trying to recall a few things is all
amistre64
  • amistre64
if we unparameterize it, we get x = 5 sin(5t) arcsin(x/5)/5 = t y = 9 cos(5 arcsin(x/5)/5) y = 9 cos(arcsin(x/5))
amistre64
  • amistre64
|dw:1378409548824:dw|
amistre64
  • amistre64
\[y = 9\sqrt{25-x^2}\] \[y^2 = 81(25-x^2)\]
anonymous
  • anonymous
okay so I can't use the unit circle in this case? I thought it would be easier to get rid of the t that way.
amistre64
  • amistre64
since the x and y to start with are ... not the same amplitude, one is moving at a different rate than the other which throws the circle outta whack
amistre64
  • amistre64
but even them, i seem to have made a wrong turn somewhere
anonymous
  • anonymous
ahhh alright, so the value for x and y should be the same a value? Also, you got the x and 5 from the equation right? I mean cosx/5 thought shouldn't be 5 being the adjacent side?
amistre64
  • amistre64
x=5 sin(5t) , divide by 5 and arcsin it arcsin(x/5)= 5t , divide by 5 to solve for t arcsin(x/5)/5 = t plugging into y = 9 cos(5t) y = 9 cos(5*arcsin(x/5)/5), the 5s cancel y = 9 cos(arcsin(x/5)), the rest is to determine some angle whose sin is x/5
amistre64
  • amistre64
i drew the triangle, cos = sqrt(5^2-x^2) / 5 forgot the /5
anonymous
  • anonymous
Ahhhh I get it, then do find the get the tangent i only need dy/dx= dy/dt / dx/dt and solve for pi/5.
amistre64
  • amistre64
y^2 = 81/25 (25-x^2) 25y^2 = 81(25) -81x^2 81x^2 + 25y^2 = 81(25)
anonymous
  • anonymous
yup makes sense, thanks a lot for your help! Haven't done things like that inever
amistre64
  • amistre64
lol, been awhile for me so this was a nice refresher

Looking for something else?

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