Question

Find the length of the loop of the curve x=3t-t^3 y=3t^2

Answer 1

If the path described makes a loop, then there will be 2 values t1t1 and t2t2 such that [x(t1)y(t1)]=[x(t2)y(t2)][x(t1)y(t1)]=[x(t2)y(t2)] , so:

[3t1−t133t12]=[3t2−t233t22]

[3t1−t133t12]=[3t2−t233t22]

From 3t12=3t223t12=3t22, we get t1=−t2t1=−t2, and from 3t1−t13=−3t1+t133t1−t13=−3t1+t13, we get that

t1=−3–√ and t2=3–√

t1=−3 and t2=3

Now the formula for parametrized path length is similar to the pythagorean formula, explicitly:

L=∫t2t1(dydt)2+(dxdt)2−−−−−−−−−−−−−−−√dt

L=∫t1t2(dydt)2+(dxdt)2dt

And filling in:

L=∫3√−3√(6t)2+(3−3t2)2−−−−−−−−−−−−−−√dt=∫3√−3√9t4+18t2+9−−−−−−−−−−−√dt=∫3√−3√∣∣3t2+3∣∣dt=t3+3t ∣∣∣t=3√t=−3√=123