Recent content by ElectronicTeaCup

Oh yes, I meant the latter. Sorry for the confusion that it may have lead to.

It is a tough call, since now I will have to again restart if I am to pick a new course. Other than MIT OCW, I haven't been able to find courses freely available in such an organized manner (notes/text + tests + exams). The MIT ocw course was very tough, and I was very demotivated by how much I...

Yes, Kline's book only covers single variate calculus. It is supposed to serve as my first step into learning more math.

Thanks for the reply. The current model we use consists of a system of differential equations, thus the direct motivation towards calculus. Plus, like you said, it is the prerequisite/foundation for other advanced mathematics. But my question is more about my particular method, is it best to...

Hello everyone! I finished a masters in integrative neuroscience about a year back, which was supposed to have a very strong mathematics tilt. Despite this, and the two semesters of mathematics, I feel that it did not help me out much. I ended up doing my masters thesis in a lab of physicists...

I'm not sure how to simplify this without spending a lot of time on it. Is there a pattern that I need to weed out?

Yes, thank you for letting me know. I had issues with a previous thread where I did not give enough information (where I thought I had). Also, since you mention it, I do have a lot of difficulty with identities. I just went back through my notes and realized that I had derived this formula...

It appears that I needed to use $$ \begin{array}{l} \cos ^{2}(\theta)=\frac{1+\cos (2 \theta)}{2} \\ \sin ^{2}(\theta)=\frac{1-\cos (2 \theta)}{2} \end{array} $$ To get the values of cos and sin in the solution. I was not familiar with this formula :nb).

Thank you for your replies. It seems that in trying to post only the relevant parts of the question, I am missing possibly essential information (that I am not picking up myself). The question in its entirety is: Reduce to standard form and graph the curve whose equation is ##x^{2}+4 x y+4...

So I get that: $$ \sin 2 \theta=-\frac{4}{5} $$ $$ \cos 2 \theta=-\frac{3}{5} $$ But what is the next step?

Oh right, I wasn't even thinking about infinity, I was just thinking of it as "undefined" Also, is this also correct?##\begin{array}{l} \cot 2 \theta=0 \\ \frac{\cos 2 \theta}{\sin 2 \theta}=0 \\ \cos 2 \theta=0 \\ 2 \theta=90 \end{array}##

One of my solutions had this in one part. Why is this the case?

There wasn't a mistake, just one more step was needed: Is there a method to do this division, and how do you get the intuition to divide it anyway? o_O

Oh I see! Using the answers from WolframAlpha gives me the right results. Now to find that mistake! Thank you all!

Oh yes, the ##x_{1}## is supposed to be squared. I tried averaging them but it became very cumbersome