Using Polynomial equations in C#
In this C# polynomial equation tutorial, you'll learn how to model and animate a moving object along a cubic curve using mathematical formulas. This hands-on project demonstrates how to solve polynomial equations and translate them into visual motion-ideal for senior secondary students learning both math and coding.
Understanding Polynomial and Cubic Equations | Maths Explanation for C# Kids
                        A polynomial equation expresses a relationship involving powers of a variable. 
                        For a cubic curve, the general form is:
                        y = ax3 + bx2 + cx + d;
                        Here, a, b, c, and d are constants. Every third-degree polynomial equation 
                        has both a maximum and a minimum point. 
                        These turning points are useful in generating smooth motion when graphing or animating curves with C#.
                    
Deriving the Equation of a Cubic Curve | Maths Explanation for C# Kids
To generate a cubic equation, all we will need are the maximum and minimum points of the curve.
                        y = ax3 + bx2 + cx + d  ----- (eqn 0)
                    
By differentiating y = ax³ + bx² + cx + d, we get dy/dx = 3ax² + 2bx + c. Setting the derivative equal to zero at both the maximum and minimum points allows us to calculate a, b, c, and d.
                            dy/dx = yI = 3ax2 + 2bx + c
                            At maximum point, yI = 0
                            yI|(x = xmax) = 0
                            3axmax2 + 2bxmax + c = 0  ----- (eqn 1)
                            At minimum point, yI = 0
                            yI|(x = xmin) = 0  ----- (eqn 2)
                            3axmin2 + 2bxmin + c = 0
                            Subtracting both derived equations
                            yI|(x = xmax) -
                            yI|(x = xmin)
                            ⇒
                            3a(xmax2 - xmin2) 
                            + 2b(xmax - xmin) = 0
                            2b(xmax - xmin) = 
                            -3a(xmax2 - xmin2)
                        
| b = | -3a(xmax - xmin)(xmax + xmin) | 
| 2(xmax - xmin) | 
                            b = -3/2a(xmax + xmin)
                            
                            Substituting b in (eqn 1)
                            3axmax2 + 2bxmax + c = 0
                            3axmax2 + 
                            2(-3a/2)(xmax + xmin)xmax 
                            + c = 0
                            3axmax2 - 
                            3axmax(xmax + xmin)
                            + c = 0
                            3axmax2 - 3axmax2
                            - 3axmaxxmin + c = 0
                            c = 3axmaxxmin
                            
                            From the general equation(eqn 0)
                            y = ax3 + bx2 + cx + d
                            ymax = axmax3 + 
                            bxmax2 + cxmax + d
                            
                            Substituting for b & c
                            ⇒ ymax = axmax3 - 
                            3/2a(xmax + xmin)xmax2
                            + 3axmaxxminxmax + d
                            
                            ymax = axmax3 - 
                            3/2axmax3 - 
                            3/2axmax2xmin
                            + 3axmax2xmin + d
                            
                            ymax = 1/2[2axmax3
                            - 3axmax3 - 3axmax2xmin
                            + 6axmax2xmin + 2d]
                            
                            ymax = 1/2[
                            -axmax3 +
                            3axmax2xmin + 2d]
                            
                            2ymax = -a(xmax - 
                            3axmin)xmax2 + 2d
                            
                            2d = 2ymax + a(xmax - 
                            3axmin)xmax2 
                            
                            d = ymax + a/2(xmax - 
                                3axmin)xmax2 
                            
                            
                            
                            From the general equation(eqn 0)
                            y = ax3 + bx2 + cx + d
                            
                            ymax = axmax3 + 
                            bxmax2 + cxmax + d
                            
                            ymin = axmin3 + 
                            bxmin2 + cxmin + d
                            
                            Subtracting both derived equations
                            ymax - ymin =
                            a(xmax3 - xmin3)
                            + b(xmax2 - xmin2)
                            + c(xmax - xmin)
                            
                            ymax - ymin = 
                            (xmax - xmin)[a(xmax2 
                            + xmaxxmin + xmin2)
                            + b(xmax + xmin) + c]
                            
                            Substituting for b & c
                            
                            ymax - ymin = 
                            (xmax - xmin)[a(xmax2 
                            + xmaxxmin + xmin2)
                            - 3a/2(xmax + xmin)2
                            + 3axmaxxmin]
                            
                            ymax - ymin = 
                            a(xmax - xmin)[xmax2 
                            + xmaxxmin + xmin2
                            - 3/2(xmax2 + 
                            2xmaxxmin + xmin2)
                            + 3xmaxxmin]
                            
                            2(ymax - ymin) = 
                            a(xmax - xmin)[2xmax2 
                            + 2xmaxxmin + 2xmin2
                            - 3(xmax2 + 2xmaxxmin 
                            + xmin2) + 6xmaxxmin]
                            
                            2(ymax - ymin) = 
                            a(xmax - xmin)(2xmax2 
                            + 2xmaxxmin + 2xmin2
                            - 3xmax2 - 6xmaxxmin 
                            - 6xmin2 + 6xmaxxmin)
                            
                            2(ymax - ymin) = 
                            a(xmax - xmin)(-xmax2 
                            + 2xmaxxmin - xmin2)
                            
                            2(ymax - ymin) = 
                            -a(xmax - xmin)(xmax2 
                            - 2xmaxxmin + xmin2)
                            
                            2(ymax - ymin) = 
                            -a(xmax - xmin)(xmax 
                            -  xmin)2
                            
                            2(ymax - ymin) = 
                            -a(xmax - xmin)3
                            
                        
| a = | -2(ymax - ymin) | 
| (xmax - xmin)3 | 
                        b = -3/2a(xmax + xmin)
                        
                        c = 3axmaxxmin
                        
        &
                        d = ymax + a/2(xmax - 
                            3axmin)xmax2
                    
These formulas form the mathematical basis of our C# polynomial solver.
Generating and Animating along a Cubic Polynomial Curve in C#
Once we determine the constants, we can implement a C# cubic equation solver to animate motion along the curve. The following example shows how to code a polynomial equation in C# using simple variables and C# windows form graphics.
To animate an object along a polynomial curve, increment x continuously and compute its corresponding y value using the cubic polynomial equation.
This C# code allows you to visualize the trajectory of a polynomial equation by plotting the curve dynamically on a C# windows form. The roots of the polynomial equation and the coefficients determine the shape and symmetry of the curve.
                        
                            Create a new C# Windows Forms Application
 project
                        ;
                        call it Dymetric_CS.
                        Create 2 new C# classes;
                        Call them Dymetric and CubicPath.
                        Type out the adjoining C# code for animating an image body through 
                        the path of a cubic / polynomial curve.
                    
Key Takeaways on Cubic Path Animation in C#
In this tutorial, you learned how to:
- Understand and derive cubic polynomial equations
 - Find coefficients from maximum and minimum points
 - Implement a polynomial equation solver using C#
 - Animate an object along a polynomial curve
 
By combining algebraic reasoning with code, senior secondary students can see how mathematics powers real-world applications like animation, computer graphics, and game design.
Applications of Polynomial Equations C# Programming and STEM Education
Polynomial equations are used in:
- Data modeling and curve fitting
 - Graphics programming for drawing smooth curves
 - Physics simulations and motion paths
 - Machine learning and optimization problems
 
Learning how to solve polynomial equations in C# provides a strong foundation for both mathematics and computational thinking.
Summary: Visualizing Polynomial Equations in C#
Polynomial equations are powerful tools for generating smooth, curved motion in graphics and animations. In this tutorial, you've learnt how to solve polynomial equations in C#, understand the mathematics of cubic curves, and create a simple animation that moves an image body along a polynomial equation path.
This interactive C# polynomial solver visually demonstrates how mathematical equations can be represented as real motion on a graph. It's a simple yet powerful example of combining coding and mathematics for educational purposes.
So! C# Fun Practice Exercise - Animate along Cubic Path
                        As a fun practice exercise, try modifying the values of xmax, xmin, ymax, 
                        and ymin to observe how they affect the polynomial equation graph. You can also:
                    
- Write a function to calculate the roots of the polynomial.
 - Compare your results with a quadratic equation solver.
 - Build a reusable polynomial equation solver in C#.
 
C# Cubic Path Window Display Code Stub
C# Cubic Path Code for Dymetric Class
namespace Dymetric
{
class Dymetric
{
private CubicPath cube_curve;
private bool do_simulation;
public Dymetric(int screen_width, int screen_height)
{
cube_curve = new CubicPath(screen_width, screen_height);
do_simulation = false;
}
// decide what course of action to take
public void decideAction(PaintEventArgs e, bool click_check)
{
if (do_simulation && click_check)
{
// do animation
cube_curve.inPlay(e);
do_simulation = false;
}
else
{
// Put ball on screen
cube_curve.clearAndDraw(e);
do_simulation = true;
}
}
}
}
C# Animation Code for Cubic Path Class
using System.Threading;
using System.Drawing;
using System.Windows.Forms;
namespace Dymetric
{
class CubicPath
{
private int x_start, x_max, y_max, x_min, y_min, x, y;
private double a, b, c, d;
private const int dotDIAMETER = 10;
// we'll be drawing to and from a bitmap image
private Bitmap offscreen_bitmap;
Graphics offscreen_g;
private Brush dot_colour, bg_colour;
public CubicPath(int screen_width, int screen_height)
{
dot_colour = new SolidBrush(Color.Yellow);
bg_colour = new SolidBrush(Color.LightGray);
// Create bitmap image
offscreen_bitmap = new Bitmap(screen_width, screen_height - 55,
System.Drawing.Imaging.PixelFormat.Format24bppRgb);
// point graphic object to bitmap image
offscreen_g = Graphics.FromImage(offscreen_bitmap);
// Set background of bitmap graphic
offscreen_g.Clear(Color.LightGray);
x_start = x = 20;
x_max = offscreen_bitmap.Width / 4 + 10;
y_max = 20;
x_min = 3 * offscreen_bitmap.Width / 4 - 10;
y_min = offscreen_bitmap.Height - 70;
// constants
a = (-2 * (y_max - y_min)) / Math.Pow((x_max - x_min), 3);
b = -((double)3 / 2) * a * (x_max + x_min);
c = 3 * a * x_max * x_min;
d = y_max + (a / 2) * (x_max - 3 * x_min) * Math.Pow(x_max, 2);
y = (int)Math.Round(a * Math.Pow(x, 3) + b * Math.Pow(x, 2) + c * x + d);
}
// draw first appearance of dot on the screen
public void clearAndDraw(PaintEventArgs e)
{
/*
* draw to offscreen bitmap
*/
// clear entire bitmap
offscreen_g.Clear(Color.LightGray);
// draw dot
offscreen_g.FillEllipse(dot_colour, x, y, dotDIAMETER, dotDIAMETER);
// draw to screen
Graphics gr = e.Graphics;
gr.DrawImage(offscreen_bitmap, 0, 55, offscreen_bitmap.Width, offscreen_bitmap.Height);
}
// repetitively clear and draw dot on the screen - Simulate motion
public void inPlay(PaintEventArgs e)
{
Graphics gr = e.Graphics;
// condition for continuing motion
while (x < offscreen_bitmap.Width - dotDIAMETER && y >= y_max)
{
// redraw dot
offscreen_g.FillEllipse(dot_colour, x, y, dotDIAMETER, dotDIAMETER);
// draw to screen
gr.DrawImage(offscreen_bitmap, 0, 55, offscreen_bitmap.Width, offscreen_bitmap.Height);
x += 20;
y = (int)Math.Round(a * Math.Pow(x, 3) + b * Math.Pow(x, 2) + c * x + d);
// take a time pause
Thread.Sleep(50);
}
x = x_start;
y = (int)Math.Round(a * Math.Pow(x, 3) + b * Math.Pow(x, 2) + c * x + d);
}
}
}