Let $\textbf{t}$ be a continuous vector field which is parallel to the unit tangent vector at each point of a smooth curve C.
Prove that,
$$
\int_{C} \textbf{t} \cdot d\textbf{r}=\int_{C} ||\textbf{t}|| d\textbf{s}\space.
$$
$C$ is parametrised by $<u,\frac{u^2}{2},\frac{u^3}{6}>$ and I have been given the points $(0,0,0)$ and $(6,18,36)$
Assuming we're working with three dimensions here, any curve can be parameterized in one variable (called a parameter), with some bounds on the parameter. (Usually the bounds are specified in an interval, e.g. $a \leq s \leq b$.) Let us assume that $\mathbf{r}(s)$ is a parameterization of $C$, and define $\mathbf{r}'(s)$ to be the derivative (tangent) vector at the point $\mathbf{r}(s)$ (the point on the curve corresponding to when the parameter is equal to $s$). What the left-hand side denotes is then is the vector line integral, which is by definition $$\int_a^b \mathbf{t}(\mathbf{r}(s)) \cdot \mathbf{r}'(s) ds$$ However, it is also given that for any point $(x, y, z) = \mathbf{r}(s)$ which lies on the curve $C$, $\mathbf{t}(x, y, z)$ returns a vector parallel to the tangent vector at $\mathbf{r}(s)$, meaning that $ \mathbf{t}(\mathbf{r}(s))$ is actually always parallel to $\mathbf{r}'(s)$. We know that the dot product of two parallel vectors is just the product of their magnitudes, so it would follow that $$\mathbf{t}(\mathbf{r}(s)) \cdot \mathbf{r}'(s) = \|\mathbf{t}(\mathbf{r}(s))\| \cdot \| \mathbf{r}'(s) \|$$ (The dot on the right-hand side is just normal multiplication.) Thus, the integral we wrote down is equivalent to $$\int_a^b \|\mathbf{t}(\mathbf{r}(\mathbf{s}))\| \cdot \|\mathbf{r}'(s)\| ds$$ which is exactly the definition of the scalar line integral $$\int_C \|\mathbf{t}\| ds$$ on the right-hand side of the identity you wanted to verify.
One key takeaway here: when evaluating line integrals (scalar or vector), you will most likely need to find a parameterization of the curve you are evaluating across.