A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Find the length of the loop for the given curve:
x = 12t  4t^(3) , y = 12t^(2)
anonymous
 4 years ago
Find the length of the loop for the given curve: x = 12t  4t^(3) , y = 12t^(2)

This Question is Closed

myininaya
 4 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
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0I squred the derivatives and collected liked terms inside the radical

asnaseer
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1sorry i had a thing i had my limits backwards lol

myininaya
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1well he could double it instead

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

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

asnaseer
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0if t = 0 than (x=0,y=0) though...

asnaseer
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1and i'm late you guys already said this

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

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

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

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

anonymous
 4 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
 4 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
 4 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
 4 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.