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. 
 A: You already have parametric equations for the curve, which you’re often going to want to derive from an implicit Cartesian equation, anyway, when working with tangents to the curve, computing line integrals, &c. However, we can go in the other direction as well.  
Start by converting the parametric equations into an implicit one. There are many ways to do this. An almost entirely mechanical way to do this is to use the method of resultants. Form the Sylvester matrix $$\begin{bmatrix}t^2&-3t&1-y&0\\0&t^2&-3t&1-y\\2t^2&t&-1-x&0\\0&2t^2&t&-1-x\end{bmatrix}$$ of the two polynomials and compute its determinant, which factors into $t^4(x^2-4xy+4y^2-15x-19y-5)$. This must vanish for all $t$, therefore an implicit equation for the curve is $$f(x,y) = x^2-4xy+4y^2-15x-19y-5=0.$$ This happens to be a parabola, which has well-defined tangents everywhere. With this in hand you can compute its gradient at any point and from that, using the fact that the gradient is orthogonal to the tangent, the slope of the tangent line at the point.  
For your specific problem, then, you need to find points on the curve for which $\nabla f = (2x-4y-15,8y-4x-19)$ is a multiple of $(4,-1)$, which becomes a matter of finding the intersections of a certain line with the curve. This all seems like a lot more work to me than computing $y'/x'$ from the parametric equations, setting it equal to $4$ and solving for $t$.
A: Hint: you will get the equation: $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t-3}{4t+1}=4$$. Solve this for $t$!
