I'm having difficulty on the following problem of differential geometry:
$a)$ Verify that the curve $\alpha\colon (0,\infty)\to\mathbb R^3$ given by $\alpha(t)=\left(t,\dfrac{1+t}{t},\dfrac{1-t^2}{t}\right)$ is a plane curve.
$b)$ Verify that every regular curve of $\mathbb R^3$, whose coordinate functions are polynomials of degree less than or equal to $2$, is a plane curve.
So, for the first item I tried the following:
$$ \begin{array}{cc} \left\{ \begin{array}{ccc} x &=& t \\ y &=& \dfrac{1+t}{t} \\ z &=& \dfrac{1-t^2}{t} \end{array} \right. & \Rightarrow \left\{ \begin{array}{ccc} y = & \dfrac{1+x}{x} & & \\ z = & \dfrac{1-x^2}{x} & = & \dfrac{(1-x)(1+x)}{x} \end{array} \right. \end{array} $$
Hence, $$ z=(1-x)y=y-xy=y-(1+x)=y-1-x $$
Thus, $\alpha$ lies in the plane $-x+y-z=1$.
For the second item I took the curve
$$ \beta(t)=(a_0+a_1t+a_2t^2,b_0+b_1t+b_2t^2,c_0+c_1t+c_2t^2), a_i,b_i,c_i\in\mathbb R. $$
And wlog I assumed $\beta$ parameterized by the arc length. I tried something over the relation $|\beta'(t)|=1$, but I did not get anything because of the lack of control of the coefficients $a_i,b_i,c_i$.
The equation $E(s,t)=(a_0+a_1t+a_2s,b_0+b_1t+b_2s,c_0+c_1t+c_2s)$ parametrises a plane. You have $\beta(t)=E(t^2,t)$ so $\beta$ is a plane curve.