1
$\begingroup$

Please see the schematic. I know the points (x1, y1) and (x2,y2) located on a line y1=c. There is another line (y2=mx+c2) which intersect y1 at a point which does not lie on or between(x1, y1) and (x2,y2). I would like to know the function y3 which takes a form of sine function (y3 = Bsin(xpi/d - x1 ) +y1 , where d is the distance between points (x1, y1) and (x2,y2)) and touches the line (y2=mx+c2) at some point but not intersect it. Let me know if this make sense or what part doesn't make sense. Any help/lead would do.

Older version: Please see the schematic. I know points (x1, y1) and (x2,y2) and equations y1 and y2 (m and c). I want to figure out the curve y3, which is a sine curve with amplitude B and touches y2 only at one point. How can I approach this problem. Any lead?

$\endgroup$
7
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Sep 7, 2021 at 8:42
  • $\begingroup$ Is $B$ given as well? And must $y_3$ be strictly of the form written? $\endgroup$
    – Ugo
    Commented Sep 7, 2021 at 8:59
  • $\begingroup$ No. B is unknown. in fact the question is: at what B, curve y3 touches the line y2? Is it clear now. And yes y3 takes that (sine) form (y3=Bsin(xpi/d-x1)+y1 ). $\endgroup$
    – D Dum
    Commented Sep 7, 2021 at 9:03
  • $\begingroup$ Have you tried finding where $y_2' = y_3'$? $\endgroup$
    – Ugo
    Commented Sep 7, 2021 at 9:17
  • $\begingroup$ yes. you get m= (B*pi/d)*cos(x_t pi/d - x1). This has both x_t and B unkown. x_s is the x coordinate of point (x_t, y_t) where the line y2 touches curve y3 (emphasized dot in fig) (which we dont know). $\endgroup$
    – D Dum
    Commented Sep 7, 2021 at 9:50

1 Answer 1

Reset to default
0
$\begingroup$

We can always move the origin to the intersection of the two lines so $c$ is irrelevant.

Call the points $A(a,0)$ and $B(b,0)$ with $b>a$. The green line is given by $f(x) = m x$ and the black line by $g(x) = C \sin \left( \frac{\pi}{b-a} (x-a) \right)$. We now want $f(x) = g(x)$ to have only one solution on the interval $(a,b)$. Unfortunately there is no closed-form solution for equations like $mx = \sin x$ so we use the Taylor series at $x = \frac{a+b}{2}$ as an approximation (the blue line):

$$C \sin \left( \frac{\pi}{b-a} (x-a) \right) \approx C \left(1 - \frac{\pi^2}{8} \left( \frac{a+b-2x}{a-b} \right)^2 \right)$$

Now $f(x) = g(x)$ is a quadratic equation which has only one solution if the discriminant is equal to $0$. We then rearrange and obtain

$$C \approx \frac{m}{4} \left( a+b + \sqrt{(a+b)^2 - \frac{8}{\pi^2} (a-b)^2} \right)$$

This approximation works best if $b \ll 2a$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .