I asked this over on stackoverflow, but I'm not sure it'll be answered there.

I have a cubic bezier defined by four points. I need to find the time t along the cubic bezier where the tangent is equal to a given vector. This problem is not as straightforward as it may seem on first glance. I'll explain the basic math first for how I approached it so you can find flaws and possibly a better solution.

A 2D cubic bezier and its tangent can be defined by these equations. Specifically the tangent:

    T(t) = -3(1-t)^2 * P0 + 3(1-t)^2 * P1 - 6t(1-t) * P1 - 3t^2 * P2 + 6t(1-t) * P2 + 3t^2 * P3

And expanded for a 2D vector:

    T_x(t) = -3(1-t)^2 * x0 + 3(1-t)^2 * x1 - 6t(1-t) * x1 - 3t^2 * x2 + 6t(1-t) * x2 + 3t^2 * x3
    T_y(t) = -3(1-t)^2 * y0 + 3(1-t)^2 * y1 - 6t(1-t) * y1 - 3t^2 * y2 + 6t(1-t) * y2 + 3t^2 * y3

Then we also have a vector (x, y) representing the tangent we want to find the time t for.

These are simple quadratic equations so we just need an equation to solve. We can take the cross product (vx0 * vy1 - vy0 * vx1) between the two and solve for 0. This would find when the tangent of the cubic bezier is equal to our given tangent vector and we'd solve for t. (I don't care if the vector is opposite the tangent so if our vector is (1, 0) then it would also look for (-1, 0)). In Mathematica solving for t with this cross product approach would look like this:

    Solve[(-3(1-t)^2*x0+3(1-t)^2*x1-6t(1-t)*x1-3t^2*x2+6t(1-t)*x2+3t^2*x3)*y-(-3(1-t)^2*y0+3(1-t)^2*y1-6t(1-t)*y1-3t^2*y2+6t(1-t)*y2+3t^2*y3)*x==0,t,Reals]

Mathematica would then output:

    {{t->ConditionalExpression[(x0 y-2 x1 y+x2 y-x y0+2 x y1-x y2)/(x0 y-3 x1 y+3 x2 y-x3 y-x y0+3 x y1-3 x y2+x y3)-\[Sqrt]((x1^2 y^2-x0 x2 y^2-x1 x2 y^2+x2^2 y^2+x0 x3 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x0 y y2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2-x x0 y y3+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x0 y-3 x1 y+3 x2 y-x3 y-x y0+3 x y1-3 x y2+x y3)^2),(x>(x2 y-x3 y)/(y2-y3)&&x0>(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y>0&&y2>y3)||(x<(x2 y-x3 y)/(y2-y3)&&x0<(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y>0&&y2>y3)||(x<(x2 y-x3 y)/(y2-y3)&&x0<(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y<0&&y2<y3)||(x<(x2 y-x3 y)/(y2-y3)&&y<0&&x0>(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y2>y3)||(x<(x2 y-x3 y)/(y2-y3)&&y2<y3&&x0>(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y>0)||(x0<(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y<0&&x>(x2 y-x3 y)/(y2-y3)&&y2>y3)||(x0<(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y2<y3&&x>(x2 y-x3 y)/(y2-y3)&&y>0)||(y<0&&y2<y3&&x>(x2 y-x3 y)/(y2-y3)&&x0>(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3))]},
    
    {t->ConditionalExpression[(x0 y-2 x1 y+x2 y-x y0+2 x y1-x y2)/(x0 y-3 x1 y+3 x2 y-x3 y-x y0+3 x y1-3 x y2+x y3)+\[Sqrt]((x1^2 y^2-x0 x2 y^2-x1 x2 y^2+x2^2 y^2+x0 x3 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x0 y y2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2-x x0 y y3+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x0 y-3 x1 y+3 x2 y-x3 y-x y0+3 x y1-3 x y2+x y3)^2),(x>(x2 y-x3 y)/(y2-y3)&&x0>(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y>0&&y2>y3)||(x<(x2 y-x3 y)/(y2-y3)&&x0<(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y>0&&y2>y3)||(x<(x2 y-x3 y)/(y2-y3)&&x0<(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y<0&&y2<y3)||(x<(x2 y-x3 y)/(y2-y3)&&y<0&&x0>(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y2>y3)||(x<(x2 y-x3 y)/(y2-y3)&&y2<y3&&x0>(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y>0)||(x0<(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y<0&&x>(x2 y-x3 y)/(y2-y3)&&y2>y3)||(x0<(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3)&&y2<y3&&x>(x2 y-x3 y)/(y2-y3)&&y>0)||(y<0&&y2<y3&&x>(x2 y-x3 y)/(y2-y3)&&x0>(x1^2 y^2-x1 x2 y^2+x2^2 y^2-x1 x3 y^2+x x2 y y0-x x3 y y0-2 x x1 y y1+x x2 y y1+x x3 y y1+x^2 y1^2+x x1 y y2-2 x x2 y y2-x^2 y0 y2-x^2 y1 y2+x^2 y2^2+x x1 y y3+x^2 y0 y3-x^2 y1 y3)/(x2 y^2-x3 y^2-x y y2+x y y3))]}}

Here's an image that's easier to see. That said most of those cases have duplicate variables so it's much simpler than it looks. (Both condition cases are identical and the solutions are a positive or negative case in the equation since it solved a quadratic equation). In code form this is easy to see:

    var temp1 = (tx2 - tx3) / (y2 - y3);
    var temp2 = (tx1 * tx1 + tx2 * tx2 + tx2 * (ty0 + ty1 - 2 * ty2) + tx1 * (-tx2 - tx3 - 2 * ty1 + ty2 + ty3) + tx3 * (ty1 - ty0) + ty1 * ty1 - ty0 * ty2 + ty2 * ty2 + ty0 * ty3 - ty1 * (ty2 + ty3)) / (tangent.y * (tx2 - tx3 - ty2 + ty3));
    console.log ('Temp1: ', temp1, ' Temp2: ', temp2);
    if
    (
        tangent.x < temp1 &&
        (
            tangent.y < 0 &&
            (
                x0 < temp2 && y2 < y3 ||
                x0 > temp2 && y2 > y3
            ) ||
            tangent.y > 0 &&
            (
                x0 < temp2 && y2 > y3 ||
                x0 > temp2 && y2 < y3
            )
        ) ||
        tangent.x > temp1 &&
        (
            tangent.y < 0 &&
            (
                x0 < temp2 && y2 > y3 ||
                x0 > temp2 && y2 < y3
            ) ||
            tangent.y > 0 &&
            (
                x0 < temp2 && y2 < y3 ||
                x0 > temp2 && y2 > y3
            )
        )
    )
    {
        var tx0ty0 = tx0 - ty0;
        var ty1tx1 = ty1 - tx1;
        var tx2ty2 = tx2 - ty2;
    
        var temp6 = 2 * (tx0ty0 + tx2ty2) + 4 * ty1tx1;
        var temp5 = tx0ty0 + 3 * (tx2ty2 + ty1tx1) + ty3 - tx3;
        var temp7 = temp6 * temp6 - 4 * (tx0ty0 + ty1tx1) * temp5;
        var temp3 = Math.sqrt(temp7);
        var temp4 = 2 * temp5;
        var t1 = (temp6 - temp3) / temp4;
        var t2 = (temp6 + temp3) / temp4;
    }

So what we have is two possible times as we'd expect since the problem is quadratic. Here's an interactive example in JS. That example uses a hardcoded tangent vector of (0.707, 0.707). (So a vector pointing down and to the right in that coordinate system).

There are problems though with the above code. Even correcting for floating point errors in the inequalities and square root calculations there are cases that aren't well defined. Like when y2 - y3 is 0 resulting in a division by zero case. There are subtleties to this also like in certain cases temp4 will have valid results that are very close to zero either producing the correct result or due to floating point issues generating a value for t1 and t2 much larger than expected. I've noticed this specifically in the cases where t1 or t2 are 0.5. Was thinking that flipping it across the the diagonal and solving again might solve some edge cases, but I'm just not confident on that approach.

What I'd like is a tried and tested approach, possibly with a code example, or another way to tackle this without weird edge cases.

There's some old Graphics Gems articles that talk about this problem. If I recall the idea was to take your given vector, transform it into a certain space where we can do an iterative refinement on the spline, and return our resulting t value.

http://www.randygaul.net/2015/05/20/interactive-cubic-bezier-splines/

https://github.com/erich666/GraphicsGems (look for NearestPoint.c)

I should have been more clear. I can probably write an iterative solution, but I'm more concerned with a closed form solution. (Since I've shown a partial solution can be calculated already for most cases).

edit: Someone solved it on stackoverflow. Interesting.

Your first derivative function seems correct, so, - why don't you just plug the t value of time into this first derivative function and add the control points together? This will yield the derivative value (2d as well), I do not get why you are solving equations - you have a first derivative function already.

Now this 2d point it will return will not be what you may think as straight forward, you will need to interprete it.

A 1D function returns as a derivation at a defintion point a 1D value, that is the tan() goniometric value of angle between tangent at that functional point and X axis.

Since the bezier is a function over a single t scalar, and returns 2d vectors, I am not sure what the first derivate will be, but I imagine some not normalized tangential vector, someone could elaborate.

edit: Someone solved it on stackoverflow. Interesting.

That's a really elegant way to pose the problem. Good find!