November | 2017 | Mathematical Poetry

The beta function is yet another function that can be represented as an integral:

$\displaystyle B(a, b)=\int\limits _0^1 t^{a-1}(1-t)^{b-1}\,dt$

Now you may think that this function is too complicated to deal with, but luckily for us, it is intimately connected with the Gamma function, which is easier to calculate. To see this, remember the definition of the Gamma function,

$\displaystyle \Gamma(a)=\int\limits_0^{\infty} x^{a-1}e^{-x}\,dx$

From which it follows that

$\displaystyle \Gamma(a)\Gamma(b)=\int\limits_0^{\infty} x^{a-1}e^{-x}\,dx \cdot \int\limits_0^{\infty} y^{b-1}e^{-y}\,dy=\int\limits_0^{\infty}\int\limits_0^{\infty} x^{a-1}y^{b-1}e^{-x-y}dA$

whereupon making the change of variable, $x=zt$ , $y=z(1-t)$ , we see that

$\displaystyle \Gamma(a)\Gamma(b)=\Gamma(a+b)B(a,b)$

$\displaystyle B(a, b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$

This allows us to relatively easily compute various values of the Beta function, especially if its inputs are integers, since

$\displaystyle \Gamma(n)=(n-1)!$

Our previous result can then be transformed into

$\displaystyle B(a,b)=\frac{(a-1)!(b-a)!}{(a+b-1)!}$

If we now make the substitution $t=\sin^2(\theta)$ , the integral transforms into

$\displaystyle B(a, b)=\int_0^{\pi/2} 2\sin^{2a-1}(\theta)\cos^{2b-1}(\theta)\,d\theta$

This form allows us to calculate trigonometric integrals which would otherwise be very hard to calculate. For example,

$\displaystyle \int_0^{\pi/2} 2\sin^{11}(\theta)\cos^{19}(\theta)\,d\theta=B(6,10)=\frac{(5!)(9!)}{13!}$

Let’s finish with a rather beautiful result:

$\displaystyle I=\int_0^{\pi/2}\sqrt{\sin(\theta)}\,d\theta=\frac{1}{2}\int_0^{\pi/2}2\sin^{2(3/4)-1}(\theta)\cos^{2(1/2)-1}(\theta)\,d\theta=\frac{1}{2}B(3/4,1/2)$

$\displaystyle I=\frac{\sqrt{\pi}}{2}\left(\frac{\Gamma(3/4)}{\Gamma(5/4)}\right)\approx 1.198$

In a previous post, I went over the equations of motion for a particle in free fall experiencing drag, with $F_d\propto v^2$ . In this post, I will do the same, but with $F_d\propto v$ . First let’s choose our coordinate system: the origin will be at the point where the object is dropped, and down will taken to be positive. Thus, our equation of motion is

$\displaystyle mg-bv=ma$

$\displaystyle mg-bv=\frac{dv}{dt}$

We will solve this differential equation by separation for the velocity:

$\displaystyle dt=\frac{dv}{mg-bv}$

We can now integrate it.

$\displaystyle \int dt=\int\,\frac{dv}{mg-bv}$

$\displaystyle t+C=-\frac{1}{b}\ln(mg-bv)$

We can now solve for C, demanding that $v(0)=v_0$ . We find that the constant is equal to

$\displaystyle C=\frac{1}{b}\ln(mg-bv_0)$

After solving for the $v$ , we find that the velocity asa function of time is

$\displaystyle v(t)=\frac{mg}{b}\left(1-e^{-bt}\right)+v_0e^{-bt}$

After letting $t\to\infty$ , we find the so called terminal velocity is given by

$\displaystyle v_t=\frac{mg}{b}$

I find it also interesting that the initial velocity doesn’t impact what the final velocity is, since that term goes to 0 as $t\to\infty$ . Now, we can easily find the function $y(t)&fg=000000$ of position of our object by integrating our velocity function.

$\displaystyle y(t)=\int \frac{mg}{b}\left(1-e^{-bt}\right)+v_0e^{-bt}\,dt$

$\displaystyle y(t)=\frac{mg}{b^2}\left(bt+e^{-bt}\right)-\frac{v_0}{b}e^{-bt}+A$

Where $A$ is a constant. Since we chose our origin at the point of fall, then we have that $y(0)=0$ , allowing use to solve for our constant. Using these conditions, we can find $A$ to be

$\displaystyle A=\frac{v_0}{b}-\frac{mg}{b^2}$

Our equation is then

$\displaystyle y(t)=\frac{mg}{b^2}\left(bt+e^{-bt}-1\right)+\frac{v_0}{b}\left(1-e^{-bt}\right)$

However, this equation is hard to deal with, as there are exponential and linear terms. But if $b$ and/or $t$ is large, the exponential term will be very small. We can then get a good approximation of our function by treating it as zero; this also makes the calculations easier.

$\displaystyle y(t)\approx \frac{mg}{b}t+\frac{v_{0}b-mg}{b^2}$

Let’s see what we can do with this equation. If we drop an object at a height $H$ , then we can find the approximate time it takes to hit the ground under the influence of gravity and drag. Since we chose the origin as our starting point, and positive as the downward, direction, we set $y(t)=H$ .

$\displaystyle \frac{mg}{b}t+\frac{v_{0}b-mg}{b^2}=H$

$\displaystyle t=\frac{Hb^3 -v_{0}b^2+mgb}{mgb^2}$

If you haven’t read my previous post, you might no know that $b$ stands for. Well $b=\frac{1}{2}\rho C_dA$ where $\rho$ is the density of the object, $C_d$ is the drag coefficient, and $A$ is the cross-sectional area of the object.