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Consider the curve $a\subset \mathbb{R}^3 $ defined by $\alpha((t)=(2t, t^2,\frac{t^3}{3})$ where $t \in \mathbb{R}^3$. Verify that {$\alpha', \alpha''$} is a linearly independent set for each t

Find $\alpha'=(2,2t,t^2)$ and $\alpha''=(0,2,2t)$

To test linear dependence I am used to finding the determinant. I was looking into proofs using cross-product and came up with the following but am unsure if its correct.

$\alpha'$x $\alpha'' = (2t^2, -4t, 4)$

Would I know do $2t^2-4t+4$ and factorise to get:

$\frac{4 \pm \sqrt{16-32}}{4}= \frac{4 \pm 2i}{4}$ which obviously isn't in the real numbers and therefore it can not equal zero proving linear independence?

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    Using determinants is only for figuring out whether a set of $n$ vectors in $\Bbb R^n$ is linearly independent. You have $2$ vectors in $\Bbb R^3$, so that doesn't work. Think, instead, what linear independence actually _means_ and calculate more directly. – 2017-01-10

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You computed $\alpha'\times\alpha''$, and by inspection you can verify that this vector is $\ne0$ for all $t$. $\qquad\square$

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Two nonzero vectors are linearly dependent if and only if one is a nonzero multiple of the other. Given $t \in \mathbb{R}$, suppose $\alpha'(t)$ and $\alpha''(t)$ were linearly dependent. Then there would exist a nonzero scalar $\lambda$ such that $$ (0,2,2t) = \lambda (2,2t,t^2) $$ By comparing the first components, we see $\lambda = 0$, a contradiction.