I have been given two lines and am asked to connect them with a parabola; the resulting shape needs to be continuous and differentiable (at either point where the parabola meets a line, the instantaneous rates of change need to be the same. The project itself is to "design a rollercoaster track" by getting the formula of the lines/parabola making up the track.)
The parabola is/needs to be in y = ax^2 + bx + c formula. I need to find a, b, and c.
So the line on the left, L1, has a slope of 0.7 and meets the parabola at point P, which is at (0, 0)
The line to the right of the parabola, L2, has a a slope of -1.5 and meets the parabola at point Q.
I know point P and point Q are 40 meters apart, so the x value of Q must be 40. I know by the wording of the question that P is higher than Q.
This is my rough sketch of the graph, just to put an image to my words. (it's rough so ignore the units of measurement) Link to image.
I know about derivatives but I know next to nothing about parabolas. I know my c = 0 because c is the Y intercept and the parabola starts at point P which is the origin. But… I know nothing else. I don't even know where to start. Can I get some hints or something? 🙁
EDIT: I think I graphed this right with the answer I got and the parabola & the second line don't ever intersect
Best Answer
Use the general form $y=ax^2+bx+c$, and its derivative, $y'=2ax+b$ with the $(x,y)$ coordinates and slopes at the required points. As stated, you know $c=0$. You must find $a$ and $b$.
Solution assuming that $Q(x_Q,y_Q)$ lies on the circle with radius 40 centered at $P(0,0)$:
$f(x)=ax^2+bx$
$f'(x)=2ax+b$
$P$ lies on the parabola and $m_P=0.7$ so $f'(0)=b=0.7$
Now we have
$f(x)=ax^2+0.7x$
$f'(x)=2ax+0.7$
Q lies on both the parabola and the circle. We also know $m_Q=-1.5$.
(1) $y_Q=ax_Q^2+0.7x_Q$
(2) $x_Q^2+y_Q^2=40^2$
(3) $-1.5=2ax_Q+0.7$
This is a system of three independent equations with three unknowns, $a$, $x_Q$, and $y_Q$. Use substitution to first eliminate $x_Q$ and then again to eliminate $y_Q$. Now you can solve for $a$.