0
$\begingroup$

Consider the following figure:enter image description here

There are two curves $p_1$ and $p_2$ given by $y= ax^2+bx+c$ and $y=dx^2+ex+f$ respectively. Also the curves $p_1$ and $p_2$ pass through $(x_1,y_1)$ and $(x_2,y_2)$ respectively.

Now I would like to draw a curve ($y=f(x)$) which connects the curves $p_1$ and $p_2$ with the following conditions:

  1. The connecting curve should pass through $(x_1,y_1)$ and $(x_2,y_2)$
  2. When the three curves are considered together, should have a second derivative at $(x_1,y_1)$ and $(x_2,y_2)$
  3. The connecting curve should be $C^2$ function (differentiable twice)

What I have done so far: From second condition, I have $g''(x1) = 2a$ and $g''(x2)=2d$.

I see that these two are different constants. So, Is it even possible to draw a curve with such a conditions? If so, how do I move from here?

$\endgroup$
  • 1
    $\begingroup$ May we assume $y_1=y_2$? $\endgroup$ – Michael Hoppe Nov 25 '16 at 9:37
  • $\begingroup$ @MichaelHoppe: yes. we can assume $y1= y2$. $\endgroup$ – Rhinocerotidae Nov 25 '16 at 9:38
0
$\begingroup$

EDIT: as the OP edited the conditions imposed on the curve, this suggestion is wrong as of now. I will update my answer ASAP

[Your curve could be the concatenation of both continuations. If the graphs would be prolonged, they would intersect and keep their derivatives at the desired points. Thus you can say your curve is the same as the left-hand side one until they meet, and then equal to the right-hand side one until the second point.]

$\endgroup$
  • $\begingroup$ Thanks for your suggestion. But I have missed an important condition of the connecting cure (that it should be doubly differentiable). See my edit. My apology for the mistake. $\endgroup$ – Rhinocerotidae Nov 25 '16 at 9:35
  • 1
    $\begingroup$ @SaravanaKumar on all points? $\endgroup$ – RGS Nov 25 '16 at 9:35
  • $\begingroup$ Yes. On all points. $\endgroup$ – Rhinocerotidae Nov 25 '16 at 9:36
0
$\begingroup$

Ok. This is the method I used to find the connecting curve (lets call the connecting curve as cc). I assumed the connecting curve to be a sixth degree polynomial.(Although this can be higher or lower order polynomial depending on the conditions to be met). $$c1x^6+c2x^5+c3x^4+c4x^3+c5x^2+c6x+c7=y$$

This has 7 unknowns ($c1, c2,...,c7$). Six of these can be found via the following conditions.

  1. cc must pass through $(x1,y1)$
  2. cc must pass through $(x2,y2)$
  3. $\Big(\frac{dy}{dx}\Big)_{(x1,y1)}$ of p1 = $\Big(\frac{dy}{dx}\Big)_{(x1,y1)}$ of cc
  4. $\Big(\frac{dy}{dx}\Big)_{(x2,y2)}$ of p2 = $\Big(\frac{dy}{dx}\Big)_{(x2,y2)}$ of cc
  5. $\Big(\frac{d^2y}{dx^2}\Big)_{(x1,y1)}$ of p1 = $\Big(\frac{d^2y}{dx^2}\Big)_{(x1,y1)}$ of cc
  6. $\Big(\frac{d^2y}{dx^2}\Big)_{(x2,y2)}$ of p2 = $\Big(\frac{d^2y}{dx^2}\Big)_{(x2,y2)}$ of cc

These six conditions will give six equations and six constants of the polynomial can be found by setting an arbitrary value to the seventh constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.