Determine unit tangent vectors to curve $C$ at $(x_0, y_0)$, where: $C : y=ax^2 + bx + c, a \neq 0$.
I know that unit tangent vector at point $P(x(t), y(t))$ is given by $\mathbf{T}(t)=\frac{\mathbf{r}^\prime(t)}{|\mathbf{r}^\prime(t)|}$, if $\mathbf{r}(t)$ is a parametric smooth curve, but what is the parametric equation of parabola in such case?
This is my first time determining parametric form, so I am confused.
Note that every point on $C$ is defined by $(t, at^2 + bt + c)$. Thus, we define $\bar{r}(t) = \langle t, at^2 + bt + c \rangle$, a parametric function.
Additionally, you need not put it into parametric form. The derivative of the function which defines $C$ is given by $2at + b$ (by the power rule), which must be the slope of the tangent line. We know slope is change in $y$ divided by change in $x$, so we have that the unit tangent vector must be in the form
$$T(t) = \left\langle n, n(2at+b)\right\rangle $$
in order for the slope of $T(x)$ to be $2at + b$. Thus, simply solve for the $n$ such that $T(t)$ is a unit vector, and you have your unit tangent vector. In this case, we get
$$\sqrt{(n(2at+b))^2 + n^2} = 1$$ $$\sqrt{n^2 \cdot (2at+b)^2 + n^2} = 1$$ $$\sqrt{n^2((2at+b)^2 + 1 )} = 1$$ $$n \sqrt{(2at+b)^2 + 1} = 1$$ $$n = \frac{1}{\sqrt{(2at+b)^2 + 1}}$$