fx-3650P and TI-68: Quadratic Equation and Arc Length between Roots of a Quadratic Curve
Two approaches using two well-liked, classic calculators. The fx-3650P uses Basic like language while the TI-68 handles formulas.
Quadratic Equation
Solve for A x^2 + B x + C = 0, with the discriminant D = B^2 – 4 * A * C.
We know the solutions: X = (-B ± √(B² – 4 * A * C)) / (2 * A)
fx-3650P Program
? → A : ? → B : ? → C :
B² – 4 A C → D ◢
-B ÷ ( 2 A ) → M :
√ ( √ ( D² ) ) ÷ ( 2 A ) → Y :
D ≥ 0 ⇒ Goto 1 : M ◢ Y ◢ Goto 2 :
Lbl 1 : M + Y → X ◢ M – Y → Y ◢ Lbl 2
D: discriminant
If D<0; roots are in the form of M ± Yi
Else, the roots are real and are stored in X, Y
TI-68 Formula
X = 0 × A + (-B + √(B² – 4 × C × A) × J) ÷ (2 × A)
The 0 × A is added to force A to be prompted for first.
J = -1 for one root, J = 1 for the other
TI-68 takes care of both real and complex roots, no worries.
The coefficients can be complex!
Examples
A |
B |
C |
D |
Roots |
2 |
-3 |
-9 |
81 |
3, -1.5 |
1 |
0 |
25 |
-100 |
5i, -5i |
-48 |
64 |
28 |
9472 |
-0.347127088, 1.680464022 |
Arc Length of a Quadratic Equation between its Real Roots
Give roots X, Y: (t – X) * (t – Y) = t^2 – (X + Y )* t + X * Y
f(t) = t^2 – (X + Y) * t + X * Y
f’(t) = 2 * t – (X + Y)
arc = ∫( √(1 + f’(t)^2) dt
TI-68 will set up for the outside integral function, while the fx-3650P can use the integral function inside of the program.
We are going to assume that X < Y.
fx-3650P Program
? → A : ? → B : ∫ ( √ (1 + (2 X – A – B) ² ), A, B)
This is the direct approach.
TI-68 Formula
ARC = √(1 + (2 × X – A – B)²)
for X use the integral function (dx)
[ 3rd ] [ Σ+ ] (dx) [ = ]
Enter low, high, and the number of intervals.
The more intervals, generally, the more accurate the integral is.
Examples
I compared results against the fx-991CW.
A |
B |
TI-68, intv = 16 |
fx-3650P |
fx-991 CW |
2 |
9 |
26.070832160 |
26.070800000 |
26.070797720 |
0 |
5 |
13.903768900 |
13.904000000 |
13.903767950 |
-2 |
2 |
9.293567375 |
9.293568000 |
9.293567525 |
Eddie
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