mathkaveli

In our previous posts, we explored how a particle traces out an ellipse and what happens when its speed changes. But motion isn’t just about where and how fast something goes — it’s also about energy. Today, we’ll explore the kinetic and potential energy of a particle moving along an elliptical path. We'll use physics to understand the motion and simulate the energies using Octave. --- The Position and Velocity Revisited Let’s return to our position vector: r(t) = < a*cos(ωt), b*sin(ωt) > From this, the velocity is: v(t) = < -a*ω*sin(ωt), b*ω*cos(ωt) > This gives us the magnitude of velocity (speed): |v(t)| = ω \* sqrt( a²*sin²(ωt) + b²*cos²(ωt) ) --- Kinetic Energy (KE) Kinetic energy is given by the formula: KE = (1/2) \* m \* |v(t)|² So for our particle of mass m , we get: KE(t) = (1/2) \* m \* ω² \* ( a²*sin²(ωt) + b²*cos²(ωt) ) This energy varies over time as the particle speeds up and slows down along the ellipse. --- Potential Energy (PE) ...

In our last post, we explored how a particle moves along an elliptical path, using a parametric position vector: r(t) = \ That model assumed a constant angular speed, ω (omega). But real life isn’t always so smooth. What happens when ω is not constant? In this post, we’ll explore how changing speed affects the motion of a particle. We’ll go from elegant ellipses to beautiful spirals, and we’ll simulate them using **Octave**, a free programming tool for doing math and science with code. --- Motion with Changing Angular Speed In our original setup, we had: x(t) = a*cos(ωt) y(t) = b*sin(ωt) If ω is constant, the motion traces out an ellipse over and over again. But what if we let the angular speed change over time — for example: ω(t) = ω₀ + k\*t This means that as time increases, the angle grows **faster** (if k > 0) or **slower** (if k r(t) = \ where θ(t) = ∫₀ᵗ ω(s) ds = ω₀\*t + (1/2)*k*t² Now we’re dealing with **non-uniform angular motion**, and the particle no ...

Have you ever watched a satellite orbit a planet or marveled at the way planets orbit the sun? Their paths often trace out more than just perfect circles — they move in ellipses. But how do we describe such motion mathematically? And how can we measure how far an object has moved if it's not traveling in a straight line? Let’s take a look at a mathematical model for this type of motion, break it down, and even compute how far the particle travels using some Octave code (a free alternative to MATLAB). --- The Position Vector: A Mathematical Story In physics, the position of a moving particle in 2D can be described with a position vector. Here's a special one that traces out an ellipse: r(t) = < a*cos(ωt), b*sin(ωt) > This vector has: a : the horizontal radius of the ellipse b : the vertical radius ω (omega): the angular speed of the particle t : time --- What Shape Is This? Let’s prove it's an ellipse. We define: x(t) = a*cos(ωt) y(t...

Have you ever come across a second-order differential equation that looks like this? x²y″ − 3xy′ + 4y = 0 This is what's called a Cauchy-Euler equation (also known as an equidimensional equation). These are special because they match the pattern: x²y″ + axy′ + by = 0 and we can solve them with a neat trick— try solutions of the form y = xʳ . But what happens when that method gives us a repeated root ? Let's walk through it together! Step 1: Assume a Power Function Suppose we guess a solution: y = xʳ Then its derivatives are: y′ = r·x r−1 y″ = r(r−1)·x r−2 Plug these into the equation: x²y″ − 3xy′ + 4y = x²·r(r−1)x r−2 − 3x·r x r−1 + 4 x r Simplify: xʳ [r(r−1) − 3r + 4] = 0 This gives the characteristic equation: r² − 4r + 4 = 0 → (r − 2)² = 0 We have a repeated root! r = 2 Wait... Isn't That Only One Solution? Yes! The solution we get is: y₁(x) = x² But second-order differential equations need two linearly independent...

Both constructions aim to "complete" the rationals Q \mathbb{Q} , meaning to create a number system where every "approaching" sequence or gap is fully realized as a number. Feature Dedekind Cuts Cauchy Sequences Basic object A downward-closed set of rationals without a greatest element A sequence ( q n ) (q_n) of rationals that becomes arbitrarily close together Intuition A "cut" describes everything less than a real number A sequence approximates a real number Completeness The set of cuts fills in all gaps The set of limits of Cauchy sequences fills in all gaps Key property Each cut represents a real number, possibly irrational Each equivalence class of Cauchy sequences represents a real number Real number A specific subset of Q \mathbb{Q} An equivalence class of Cauchy sequences Least Upper Bound Property The union of a set of cuts is a cut Every bounded increasing sequence of Cauchy classes converges Construc...

Ever wondered how mathematicians spot patterns, predict outcomes, or even build formulas that show up in everything from genetics to video games? One of the most powerful tools for this is something called the binomial theorem —a formula you might have seen in class, but probably never realized how deep (and useful) it actually is. Let’s dive into what it is, why it works , and how it opens the door to something even cooler: infinite series . What Is a Binomial? First things first: a binomial is just a simple expression with two terms, like: ( a + b ) , ( 1 + x ) , ( 2 − y ) (a + b), \quad (1 + x), \quad (2 - y) When we raise it to a power, we’re multiplying it by itself several times: ( 1 + x ) 3 = ( 1 + x ) ( 1 + x ) ( 1 + x ) (1 + x)^3 = (1 + x)(1 + x)(1 + x) Multiplying all that out might look like a mess… but there's a beautiful pattern hidden inside. The Binomial Theorem: A Pattern of Powers and Coefficients Here’s the general formula: ( a + b ) n = ∑ k = 0 n ( n k ) a...

In this blog post, we'll solve a partial differential equation (PDE) using a numerical method called the Finite Element Method (FEM). Don’t worry if these terms seem intimidating; we'll break down every step! Problem Statement We're looking at this PDE: ∂ 2 u ( x , y ) ∂ x 2 + ∂ 2 u ( x , y ) ∂ y 2 − 12.5 π 2 u ( x , y ) = − 25 π 2 sin ⁡ ( 5 π x 2 ) sin ⁡ ( 5 π y 2 ) , 0 < x , y < 0.4 \frac{\partial^2 u(x,y)}{\partial x^2}+\frac{\partial^2 u(x,y)}{\partial y^2}-12.5\pi^2 u(x,y)=-25\pi^2\sin\left(\frac{5\pi x}{2}\right)\sin\left(\frac{5\pi y}{2}\right),\quad 0<x,y<0.4 with boundary condition: u ( x , y ) = 0 at the boundary of the region u(x,y)=0\quad \text{at the boundary of the region} The exact solution to compare with our numerical result is: u ( x , y ) = sin ⁡ ( 5 π x 2 ) sin ⁡ ( 5 π y 2 ) u(x,y)=\sin\left(\frac{5\pi x}{2}\right)\sin\left(\frac{5\pi y}{2}\right) We will approximate this numerically. Numerical Method: FEM Explained Simpl...