Question

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

Answer 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?

Answer 2

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.

Answer 3

points**

Answer 4

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

Answer 5

after plugging in all my stuff I get some thing like this: \[\int\limits_{t=0}^{t=\sqrt{3}}\sqrt{144t ^{4}+288t ^{2}+144}dt\]

Answer 6

I squred the derivatives and collected liked terms inside the radical

Answer 7

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.

Answer 8

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

Answer 9

sorry i had a thing
i had my limits backwards lol

Answer 10

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

Answer 11

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

Answer 12

well he could double it instead

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

that only gives you half the curve - so as @myininaya said, you could do 0 to square root 3 and then double the answer.

Answer 19

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?

Answer 20

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)


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based on your graphs symmetry I belive you could just double what you have

Answer 21

and i'm late
you guys already said this

Answer 22

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

Answer 23

ok @DEdesigns57 - are you following the explanations above?

Answer 24

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

Answer 25

let me draw it to try and help