Recent content by Drongoski

Yes, I still take students, but only a few.

3. \frac {dy}{dx} = e^{2y}\\ \\ \therefore \frac {dy}{e^{2y}} = dx \\ \\ \therefore \int dx = \int e^{-2y} dy \\ \\ \therefore x = -\frac {1}{2}e^{-2y} + C

2. \frac {dy}{dx} = 1 - \frac {y}{3} = \frac {3 - y}{3}\\ \\ \therefore \frac {dy}{3-y} = \frac {dx}{3}\\ \\ \therefore \int \cdots dy = \int \cdots dx \\ \\ \therefore -ln |3-y| = \frac {x}{3} + C \\ \\ x = 2 \implies y = 4 \implies -ln|3 - 4| = \frac {2}{3} + C \implies C = -\frac {2}{3} \\...

Question too bloody long!

You can use the "Terry Lee" method to find the square root of -60-32i. Here's how: \sqrt{-32i - 60} = 2\sqrt{-8i -15}\\ \\ \sqrt{-8i - 15} = \sqrt{1 - 8i -16} = \sqrt{1^2 + 2\times 1 \times ({-4i}) +(-4i)^2} \\ \\ \\ $Remember: $ (a + b)^2 = a^2 + 2 \times a \times b + b^2 = (a + b)^2 \\ \\...

A^4 + 4B^4 = A^4 + 4B^4 + 4A^2B^2 - 4A^2B^2 \\ \\ = (A^2 + 2B^2)^2 - (2ab)^2 = (A^2 + 2B^2 +2AB)\times(A^2 + 2B^2 - 2AB)

I don't think you can get an easy answer here. If you can afford it, get a good tutor to check you out or a friend who is good in maths to help you.

Maybe I'll show you how to do RCR later.

\int ^1 _0 \frac {\sqrt x}{1 + x} dx = \int ^1 _0 \frac {\sqrt x}{1+(\sqrt x)^2} \times 2\sqrt x d\sqrt x\\ \\ =2\int^1 _0 (1 - \frac {1}{1 + (\sqrt x)^2} )d \sqrt x \\ \\ =2 \left[\sqrt x - tan^{-1} \sqrt x \right ]^1 _0 \\ \\ = 2[(1 - \frac {\pi}{4} ) -(0 - 0)] = 2 - \frac {\pi}{2} \\ \\...

No.

Integrands which are derivatives(or constant multiples thereof) of composite functions are the ones usually handled by "integration by substitution". They often refer to reverse chain rule - but I've seldom seen how this reversal is achieved. Using my method, I can do the reversal of the chain...

Motion, in general, does not need to be in a straight line. In your Adv & Ext 1 Maths, you are usually dealing with motion in a straight line, because this is the easiest to deal with. Motion in a straight line is often referred to as "rectilinear motion".

\frac {dy}{dx} = -3\frac {d}{dx}(x^4-2x^2-1)^{-1} = -3\frac{d(x^4-2x^2-1)^{-1}}{d(x^4-2x^2-1)} \times \frac {d(x^4-2x^2-1)}{dx}\\ \\ = -3\times {-1}(x^4-2x^2-1)^{-2} \times (4x^3-4x) \\ \\ = \frac{3\times 4x(x-1)(x+1)}{(x^4-x^2-1)^2}\\ \\ = 0 $ at x $=0, -1,1\\ \\ y(-1) = \frac {-3}{1-2-1} =...

Again, if not required to use the substitution: \int ^0 _{-3} \frac {x}{\sqrt {1-x} }dx = -\int ^0 _{-3}\frac {1-x - 1}{\sqrt {1-x} }dx\\ \\ = \int ^0 _{-3} (1-x)^{0.5} d(1-x)-\int ^0 _{-3}(1-x)^{-0.5}d(1-x)\\ \\ = \left[ \frac{2}{3}(1-x)^{1.5} -2(1-x)^{0.5} \right]^0 _{-3} \\ \\ =-(\frac...

Avoiding formal substitution: \int ^1 _0 x\sqrt{1-x^2} dx = -\frac {1}{2}\int ^1 _0 (1-x^2)^{0.5} d(1-x^2) \\ \\ =-\frac{1}{2} \left[\frac {(1-x^2)^{1.5}}{1.5} \right]^1 _0 = -\frac {1}{2}\left [ (\frac{2}{3} \times(1-1)^{1.5}) -(\frac {2}{3}(1-0)^{1.5})\right ]\\ \\ = -\frac{1}{2} \times...