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 t1 and t2 such that [ x(t1) y(t1) ]=[ x(t2) y(t2) ] , so:

[ 3t1−t13 3t12 ]=[ 3t2−t23 3t22 ]

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

t1=−

√

3

and t2=

√

3

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

L=∫

t2

t1

√

(

dy

dt

)2+(

dx

dt

)2

dt

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

=12

√

3