anonymous
  • anonymous
The following parametric equations trace out a loop. x=3-(2/2)t^2 y=(-2/6)t^3+2t+2 Find the t values at which the curve intersects itself: t = +/- . What is the total area inside the loop? Area = .
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
blockcolder
  • blockcolder
First, you need to find \(t_1\) and \(t_2\) such that \(x(t_1)=x(t_2)\) and \(t_1\neq t_2\): A little clarification: is \(x= 3-\frac{2}{2}t^2\) or \(x=3-\frac{2}{2t^2}\)?
blockcolder
  • blockcolder
Well, whichever it is, the t-values you're looking for ate \(\large t=\pm\frac{\sqrt{6}}{2}\).
anonymous
  • anonymous
i think its 2/2, it is very unclear, So I have to equate the x equation?

Looking for something else?

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

More answers

blockcolder
  • blockcolder
Yeah, so you'll end up with \(3-t_1^2=3-t_2^2 \Rightarrow\) \(t_1^2=t_2^2 \Rightarrow\) \(|t_1|=|t_2| \). WLOG, let \(t_2=-t_1\), since we don't want \(t_1=t_2\). Now, you have to equate \(y(t_1)=y(t_2)\) and use the fact that t2=-t1 to find out the value of t1. Let me know what you got.
anonymous
  • anonymous
aaahhhhh!!!! I think I got it, after equating both x and finding out that for t_2 to be equal to t_1, one of them has to be the opposite. Then when I equated y(t1)=y(t2) I can just do -(yt2)=(yt2) and then solve for t, getting t=0 and t= sqrt(6). However the one thing I don't understand is how we're allowed to equate both x.

Looking for something else?

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