We begin by considering two lines. Namely, \(f(t): x = 1 + 2t, y = 2 - 3t, z = 3 + 4t\) and \(h(s): x = -1 + 3s, y = 3-s, z = -5 + 5s\). I now plot these two lines in 3 space in order to justify their skewness i.e. They do not intersect and are not parallel.

I now introduce a new function \(D_l\) and this is defined as the distance between the two lines i.e. \(|(f(t)-h(s)|\).  We now work with this equation to derive a general method for calculating the distance between two skew lines.

Before we begin, recall that \(|{(a,b,c)}| = \sqrt{a^2+b^2+c^2}\)

Now,

\[D_l = |{(1,2,3)-(-1,3,-5)+t(2,-3,4)-s(3,-1,5)}| \\= |{(2 + 2t - 3s, -1 -3t + s, 8 + 4t -5s)}|  \\= \sqrt{(2+2t-3s)^2+(-1 -3t+s)^2+(8+4t-5s)^2}\\=\sqrt{35s^2 - 58st - 94s + 29t^2 + 78t + 69}\]

I now bring in some Calculus. We use the fact that minimizing a function is the same as minimizing the square of that function (does not always hold but it holds here because we are dealing with a distance function that is non-negative and monotonic). Hence, we do the following:

\[(D_l)^2 =  35s^2 - 58st - 94s + 29t^2 + 78t + 69\]

We now take the partial derivatives \((D_l)^2\) with respect to s and t and we set it equal to zero. This is as follows.

\[\frac{\partial (D_l)^2}{\partial s} = 70s -58t -94 = 0\]

\[\frac{\partial (D_l)^2}{\partial t} = -58s+58t+78 = 0\]

Solving the system of linear equations we arrive at \(s=\frac{4}{3}\) and \(t=\frac{-1}{87}\).…

I recently received a message from a friend and the heading of the post perfectly describes what was said to me. Thereafter, an interesting integration question was sent to me. It read as follows:

I must admit, it does look quiet scary. My immediate thought was that some sort of substitution was required but I really had no idea as to where and how this should be done. Two pages and a headache later, I thought to myself why don’t I get a rough idea of what the answer should look like. Once again, let’s start approximating things (as it turns out, my approximation gets me the absolutely correct answer).

I looked at the integration bounds and noted that the point x = 3 was in fact the midpoint. I then decided to construct a Taylor polynomial for the above function around the point x = 3. It was done as follows:

You might be wondering why I didn’t bother taking anymore derivatives.…

Why \(\pi\)? I assume this is the question on everyone’s mind. (Whether you’re a Math lover or not)

The simple answer would be that we all love pie, now don’t we?

Before I begin discussing any technicalities, I’d like to acknowledge that it is possible for some of us to find the concepts easy whilst others might struggle with them. This is the reason why I’m choosing to speak in a very simple and understandable manner. (I’m baby proofing my post!)

Firstly, let us have a look at the Maclaurin series of \(\arctan(x)\)

Aside: A Maclaurin series is a polynomial which approximates a function around the point x = 0. The level of accuracy decreases as you move further away from x = 0. The only way to get an exact answer and not an approximate is to let the sum go to infinity.

Below is a table of the first few derivatives of \(\arctan(x)\).…