Length of a parametric curve

Question

I need a hint or some help regarding a parametric curve length;

The equation is $\gamma=(t^2(t-1),t^2(t+1))$, with $t \in [-1,1]$!

I started with $$\text{length} = \int\sqrt{x'(t)^2 + y'(t)^2}dt$$ but I always get $0$,

where am i going wrong?

Answer 1

You should take $t^2$ out as $|t|$ (abs value). Your integral may by symmetry be evaluated as twice the integral from 0 to 1.

Answer 2

After the correction suggest by H. H. Rugh, you should find $$ \int_{-1}^1\sqrt{x'(t)^2 + y'(t)^2}dt=\int_{-1}^1\sqrt{(3t^2-2t)^2 + (3t^2+2t)^2}dt=\int_{-1}^1\sqrt{18t^4+8t^2}dt\\=\int_{-1}^1|t|\sqrt{18t^2+8}dt=2\int_{0}^1t\sqrt{18t^2+8}dt=\frac{1}{18}\int_{0}^1\sqrt{18t^2+8}d(18t^2+8)=\frac{1}{18}\left[\frac{2}{3}(18t^2+8)^{3/2}\right]_{0}^1=\frac{26^{3/2}-8^{3/2}}{27}.$$

Answer 3

We have $$L(\gamma) = \int_{-1}^1\sqrt{(3t^2 -2t)^2 + (3t^2 + 2t)^2}dt = \int_{-1}^1\sqrt{9t^4 - 12t^3 + 4t^2 + 9t^4 + 12t^3 + 4t^2}dt \\ = 2\int_{0}^1\sqrt{18t^4 + 8t^2}dt = 2\int_{0}^1t\sqrt{18t^2 + 8}dt = \frac{1}{18}\int_{8}^{26}\sqrt{u}dt = \frac{1}{27} \left[ u^{3/2}\right]_8^{26} = \frac{1}{27}(26^{3/2} - 8^{3/2})$$