Question

Find the length of the loop of the curve x = 3t - t^3 and y = 3t².

Answer 1

What i mainly need help is finding the limits of integration without using a graphing calculator. I know you can just use the equation for length afterwards you got the limits of integrationl

Answer 2

oh i see are we using \[L=\int\limits_{a}^{b} \sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}\]

Answer 3

Yepp correct, but how do we find a, b without knowing the graph?

Answer 4

you should have bounds for t such as ........0<t<2pi or an interval 4>t>1

Answer 5

I dont see any bounds given in the question, but they tell us a loop is formed, can we do anything with that information?

Answer 6

\[L=\int\limits_{0}^{2\pi} \sqrt{(3-3t^2)^2+(6t)^2}=L=\int\limits_{0}^{2\pi} \sqrt{9t^{4}+18t^{2}+9}\]

Answer 7

\[ L=\int\limits_{0}^{2\pi} \sqrt{(3t^{2}+3)^{2}}=L=\frac{3}{4}t^{3}+3t \]

Answer 8

let me double check bounds to make sure we don't play with the equations first

Answer 9

Where do we get the bounds from? and ye sure take your time thank you :)

Answer 10

bounds are t=0 and sqrt(3) haha

Answer 11

L=21/4sqrt(3)

Answer 12

Can you show me how you got those bounds plz? sqrt 3?

Answer 13

sorry for loops they start at x=0 so sub this into your parametric equation and solve for t such that 0=3t-t^3

Answer 14

ohhh but how do we know the loop starts at x = 0?

Answer 15

your parametric equation if i sub t=0 into we get x=0 and y=0 the loop is at the origin

Answer 16

ohhh got it! and wouldnt the counts be t = - root(3) to root(3)?