Convert arc into two arcs to get the correct final tangent

Question

I have an arc which is governed by a start tangent, a start point and an end point.

Now i want to add a tangent to the final point so the path created between the two points are correct. Here is a visual of the issue.

So the two desired tangents are the same colour as the arc. But as you can see the final tangent point (in pink) does not match the blue tangent that i am aiming for.

So i need some way to calculate a new path between the two points which will create arcs that have the desired tangents.

How would i go about calculating this ?

Things i know about my arc:

1. Its origin and radius
2. Start and end points (and start and end angles)
3. The start tangent (blue) top left and final tangent (red) bottom right.
4. The final desired tangent (blue) bottom right.

Answer 1

The shape of the curve you are looking for is given by the equation $-\mathrm{ln}(x) + 0.2x+0.609$ on the interval $[0,5]$. If we treat the origin as the point that is located three intersections below the initial pink ball then,

$$y'(0) = U$$ $$y'(5) = 0$$ $$y''(0-5) > 0$$

Based on this information, we can construct an equation for the derivative of the function which governs the shape of your curve. Which is

$$\frac{dy}{dx} = \frac{1}{x} - 0.2$$

To solve this differential equation, we multiply the integrand $dx$ to the other side of the equation and then integrate! $$\int dy = \int \frac{1}{x} - 0.2 dx $$ Following some basic integration rules which can be found here, the general solution to the differential equation is: $$y = \mathrm{ln}(x) - 0.2x + C$$ However, to satisfy the positive concavity requirements we must negate the above equation to yield the following: $$y = -\mathrm{ln}(x) + 0.2x + 0.609$$ The constant of integration is 0.609 to shift $y(5)$ up to $0$. I hope this helps! :)