anonymous
  • anonymous
Find the length of the loop for the given curve: x = 12t - 4t^(3) , y = 12t^(2)
Mathematics
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anonymous
  • anonymous
Find the length of the loop for the given curve: x = 12t - 4t^(3) , y = 12t^(2)
Mathematics
katieb
  • katieb
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myininaya
  • myininaya
\[L=\int\limits_{upperintersection}^{lowerintersection}\sqrt{(x')^2+(y')^2} dt\] I believe that is what I remember do you know how to find the intersections?
anonymous
  • anonymous
no I dont. To be exact my main problem is finding the values of t that I should use for my limits of integration which I guess are just the intersection point.
anonymous
  • anonymous
points**

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asnaseer
  • asnaseer
yes - the loop in question begins and ends at x=0 (see here: http://www.wolframalpha.com/input/?i=x%3D12t-4t%5E3%2C+y%3D12t%5E2) so find the values for t for which x=0 - they will form the lower and upper limits of the integral referred to by @myininaya
anonymous
  • anonymous
after plugging in all my stuff I get some thing like this: \[\int\limits_{t=0}^{t=\sqrt{3}}\sqrt{144t ^{4}+288t ^{2}+144}dt\]
anonymous
  • anonymous
I squred the derivatives and collected liked terms inside the radical
asnaseer
  • asnaseer
yup that looks correct - you should be able to spot that 144 can be factored out. that should leave you with a simpler equation that should also be factorable.
asnaseer
  • asnaseer
the square root in the integral will disappear if you spot how to factorize this.
myininaya
  • myininaya
sorry i had a thing i had my limits backwards lol
myininaya
  • myininaya
ok so did you guys find out how to find the intersections? i can show you how dedesigns
asnaseer
  • asnaseer
also @DEdesigns57 - your limits are not right. should be from minus square root 3 to plus square root 3.
myininaya
  • myininaya
well he could double it instead
asnaseer
  • asnaseer
remember:\[x^2=3\implies x=\pm\sqrt{3}\]
anonymous
  • anonymous
why? it seems the (0,0) is the other intersection point.
asnaseer
  • asnaseer
we are trying to find the start and end limits of 't' - not x.
asnaseer
  • asnaseer
you should have ended up with something like:\[t^2=3\implies t=\pm\sqrt{3}\]
anonymous
  • anonymous
if t = 0 than (x=0,y=0) though...
asnaseer
  • asnaseer
that only gives you half the curve - so as @myininaya said, you could do 0 to square root 3 and then double the answer.
anonymous
  • anonymous
and when I look at the graph (0,0) and (0,36) are the intersection points. so I just found the values of t that made that happen. What am I doing wrong?
myininaya
  • myininaya
ok if we look at the graph we will see two intersection clearly x is 0 when both intersection happen so x=12t-4t^3 => we have 0=12t-4t^3 at the intersections 0=4t(3-t^2) => t=0, pm sqrt(3) and if we plug these into y we would get y=0 and y=36 And our graph confirms these are the right points of intersection: (0,0) and (0, 36) --------------- based on your graphs symmetry I belive you could just double what you have
myininaya
  • myininaya
and i'm late you guys already said this
asnaseer
  • asnaseer
thanks for your assistance @myininaya - always appreciated :-)
asnaseer
  • asnaseer
ok @DEdesigns57 - are you following the explanations above?
anonymous
  • anonymous
so its not from t=0 to t=sqrt(3)?
asnaseer
  • asnaseer
let me draw it to try and help
asnaseer
  • asnaseer
|dw:1327172842200:dw|
anonymous
  • anonymous
ah so your saying the from t=-sqrt(3) to t = sqrt(3) is one revolution around the curve, and that if I were to use from t=0 to t=sqrt(3) I would have to multiply my integral by 2 since it would only be half of it >?
asnaseer
  • asnaseer
so length of loop can be found either by integrating from \(-\sqrt{3}\) to \(\sqrt{3}\), or by doubling the integral from 0 to \(\sqrt{3}\). yes - I think you have it now - well done :-)
anonymous
  • anonymous
Awesome! thank you so much, wish my teacher took the time to explain things like this.
asnaseer
  • asnaseer
you're more than welcome. I am glad I was able to explain it well enough for you to understand.

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