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

1. 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?

2. 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.

3. anonymous

points**

4. 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

5. anonymous

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

6. anonymous

I squred the derivatives and collected liked terms inside the radical

7. 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.

8. asnaseer

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

9. myininaya

10. myininaya

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

11. asnaseer

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

12. myininaya

well he could double it instead

13. asnaseer

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

14. anonymous

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

15. asnaseer

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

16. asnaseer

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

17. anonymous

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

18. 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.

19. 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?

20. 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

21. myininaya

and i'm late you guys already said this

22. asnaseer

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

23. asnaseer

ok @DEdesigns57 - are you following the explanations above?

24. anonymous

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

25. asnaseer

let me draw it to try and help

26. asnaseer

|dw:1327172842200:dw|

27. 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 >?

28. 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 :-)

29. anonymous

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

30. asnaseer

you're more than welcome. I am glad I was able to explain it well enough for you to understand.