Parametric Equations

This one will be a short post, it's mostly intuition as opposed to a definite proof.

So I was musing over finding the area under a parametric curve. And I considered how we'd usually find the area under a curve, integration. Which in 2D with an x and y axis can be expressed generally as:
\[ \int_a^b y \ \mathrm{d}x\]
Parametric equations define a curve in terms of a parameter, usually t. So let's consider an arbitrary set of parametric equations:
\[\begin{align*} x=x(t) \\ y = y(t) \end{align*} \]
So to change the dx to a dt to allow integration one must differentiate x with respect to t and solve for dx.
\[ \begin{align*} x &= x(t) \\ \frac{\mathrm{d}x}{\mathrm{d}t} &=x'(t)\\ \mathrm{d}x &= x'(t) \mathrm{d}t \end{align*}\]
The value for dx must then be substituted back into the original expression.
\[\int_a^b y(t)x'(t) \ \mathrm{d}t\]
And that's it, how to find the area under a curve defined by a parametric equation.

Kinetic Energy

So I started writing this a while ago but never got around to finishing it, this post is fairly calculus heavy so be warned.

Another equation we so often use in physics is $E = \frac{1}{2}mv^2$, this is the equation for the kinetic energy of an object. Kinetic energy is also used pretty vaguely at GCSE and A level. A more formal definition is "The energy possessed by an object because of its motion" or "The energy required to accelerate a mass to a velocity". In an ideal system, i.e. one with no external forces acting the accelerated body will possess this energy and associated velocity until collision.

Clearly kinetic energy is related to $E = \int F \mathrm{d}s$_[1] as this is work equation and is the very definition of mechanical work in physics. The unit of energy we use in physics is the joule. When considering a mechanical situation such as this one, 1 joule is defined as the amount of energy used in applying a force of 1 newton over 1 metre.

However this knowledge alone is not enough to derive the expression for kinetic energy, we also need to know what force is. Force is defined mathematically in Newton's second law as $F=\frac{\mathrm{d}p}{\mathrm{d}t}$ or more simply in the case of constant mass as $F=m\frac{\mathrm{d}v}{dt}=ma$

From Newton's second law and the work equation the equation for kinetic energy can easily be found using some calculus.

First we substitute $F=m\frac{\mathrm{d}v}{dt}$ into the work equation: \[F\mathrm{d}s = m \frac{\mathrm{d}v}{\mathrm{d}t}\mathrm{d}s \] Second we need to rewrite the work equation into a different form to remove the ds. s, is just distance and can be expressed as velocity multiplied by time, $\mathrm{d}s = v \ \mathrm{d}t$. So using this knowledge the work equation can be rewritten as: \[F\mathrm{d}s = m \frac{\mathrm{d}v}{\mathrm{d}t}v \ \mathrm{d}t \] Cancelling the dt's and simplifying leaves: \[F\mathrm{d}s = mv \ \mathrm{d}v \] This can now be substituted back into the work equation: \[E = \int \! mv \ \mathrm{d}v \] The equation for kinetic energy assumes that mass is constant, so this can be factored out of the integral as a constant. \[E = m \int v \ \mathrm{d}v \] Integrating this yields: \[E = m \left[ \frac{v^2}{2} + C\right] \] Putting limits on this integral will eliminate the +C and provide the energy associated with a particular change in velocity. Assuming the particle was always accelerated from 0 to a velocity v this simplifies to: \[E = m \left[ \frac{v^2}{2} + C\right]_{0}^{v} = \frac{1}{2}mv^2\] Which is the equation for kinetic energy!

[1] strictly speaking the equation is $E = \int F\cdot \mathrm{d}s$, this includes use of the dot product. In this post have assumed that the force is always acting in the same direction as the motion, however this is often not the case and hence the dot product is required.