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

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1\[L=\int\limits_{upperintersection}^{lowerintersection}\sqrt{(x')^2+(y')^2} dt\] I believe that is what I remember do you know how to find the intersections?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no 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.

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1yes  the loop in question begins and ends at x=0 (see here: http://www.wolframalpha.com/input/?i=x%3D12t4t%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
 5 years ago
Best ResponseYou've already chosen the best response.0after 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
 5 years ago
Best ResponseYou've already chosen the best response.0I squred the derivatives and collected liked terms inside the radical

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1yup 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
 5 years ago
Best ResponseYou've already chosen the best response.1the square root in the integral will disappear if you spot how to factorize this.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1sorry i had a thing i had my limits backwards lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1ok so did you guys find out how to find the intersections? i can show you how dedesigns

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1also @DEdesigns57  your limits are not right. should be from minus square root 3 to plus square root 3.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1well he could double it instead

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1remember:\[x^2=3\implies x=\pm\sqrt{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why? it seems the (0,0) is the other intersection point.

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1we are trying to find the start and end limits of 't'  not x.

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1you should have ended up with something like:\[t^2=3\implies t=\pm\sqrt{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if t = 0 than (x=0,y=0) though...

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1that only gives you half the curve  so as @myininaya said, you could do 0 to square root 3 and then double the answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 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
 5 years ago
Best ResponseYou've already chosen the best response.1ok if we look at the graph we will see two intersection clearly x is 0 when both intersection happen so x=12t4t^3 => we have 0=12t4t^3 at the intersections 0=4t(3t^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
 5 years ago
Best ResponseYou've already chosen the best response.1and i'm late you guys already said this

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1thanks for your assistance @myininaya  always appreciated :)

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1ok @DEdesigns57  are you following the explanations above?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its not from t=0 to t=sqrt(3)?

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1let me draw it to try and help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah 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
 5 years ago
Best ResponseYou've already chosen the best response.1so 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
 5 years ago
Best ResponseYou've already chosen the best response.0Awesome! thank you so much, wish my teacher took the time to explain things like this.

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.1you're more than welcome. I am glad I was able to explain it well enough for you to understand.
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