Find the length of the loop for the given curve:
x = 12t - 4t^(3) , y = 12t^(2)

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- anonymous

Find the length of the loop for the given curve:
x = 12t - 4t^(3) , y = 12t^(2)

- katieb

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- 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

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

points**

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## More answers

- 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

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

I squred the derivatives and collected liked terms inside the radical

- 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

the square root in the integral will disappear if you spot how to factorize this.

- myininaya

sorry i had a thing
i had my limits backwards lol

- myininaya

ok so did you guys find out how to find the intersections?
i can show you how dedesigns

- asnaseer

also @DEdesigns57 - your limits are not right. should be from minus square root 3 to plus square root 3.

- myininaya

well he could double it instead

- asnaseer

remember:\[x^2=3\implies x=\pm\sqrt{3}\]

- anonymous

why? it seems the (0,0) is the other intersection point.

- asnaseer

we are trying to find the start and end limits of 't' - not x.

- asnaseer

you should have ended up with something like:\[t^2=3\implies t=\pm\sqrt{3}\]

- anonymous

if t = 0 than (x=0,y=0) though...

- 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

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

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

and i'm late
you guys already said this

- asnaseer

thanks for your assistance @myininaya - always appreciated :-)

- asnaseer

ok @DEdesigns57 - are you following the explanations above?

- anonymous

so its not from t=0 to t=sqrt(3)?

- asnaseer

let me draw it to try and help

- asnaseer

|dw:1327172842200:dw|

- 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

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

Awesome! thank you so much, wish my teacher took the time to explain things like this.

- 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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