Find value of $t$ where slope of parametrically-defined curve $=4$ using multivariable calculus A recent problem I encountered gave me curve $C$ defined by the parametric equations $x(t)=2t^{2}+t-1$ and $y(t)=t^{2}-3t+1$, and asked me to find point $t$ where the slope of the tangent line to $C$ would equal $4$. Obviously, this is trivial using the Single-Variable equation 
$$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y/\text{d}t}{\text{d}x/\text{d}t}=4,$$
But I'm interested in interpreting $C$ as a literal curve using Multivariable Calculus. I first tried defining the vector-valued function
$$\mathbf{r} = <x(t),y(t)> = <2t^{2}+t-1,t^{2}-3t+1>,$$
but I quickly ran into problems regarding what would constitute the slope of a vector-valued function. One attempt I tried was to evaluate $||\mathbf{r'}||$ and match that to $4$ - that is, 
$$||\mathbf{r'}||=4,$$
but that quickly went nowhere, as one might imagine. 
Further research indicated to me that the best equivalent of a "slope" in Multivariable Calculus would be via a curve's gradient. However, that approach would require to define $t$ in terms of $x$ and $y$ such that an appropriate function $f(x,y)$ can be defined, a task that would be rather messy in the case of $\mathbf{r}$, even in the case of invoking the Quadratic Formula. This also ignores the lack of a guiding vector $\mathbf{u}$ which I would need to take the dot product of with $\nabla\mathbf{f}$ to get a magnitude resembling slope to match with $4$, or more specifically 
$$\nabla \mathbf{f}\cdot\mathbf{u}.$$
My question is thus what would be the best way to approach the problem using Multivariable Calculus, either via expressing $C$ as a vector-valued function or as a function of 2 variables. 

Hint: you will get the equation: $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t-3}{4t+1}=4$$. Solve this for $t$!
