Search results

There's actually quite a bit of nice physics behind the didgeridoo, which one can find out about by searching online (e.g. here is one page that came up: https://newt.phys.unsw.edu.au/jw/dij/dij.html ).

Re: MATH1231/1241/1251 SOS Thread $\noindent Multiply through by $x^p y^q$ to obtain $\underbrace{\left(2x^p y^{2+q} + 4x^{2+p}y^{1+q}\right)}_{A(x,y)}\, \mathrm{d}x +\underbrace{\left(4x^{1+p}y^{1+q} + 3x^{3+p}y^q\right)}_{B(x,y)}\, \mathrm{d}y = 0$.$ $\noindent Now, $\frac{\partial...

Re: MATH1231/1241/1251 SOS Thread Note the LHS is d/dx (y/x).

Re: HSC 2016 2U Marathon If that's all it said, then 3,3,3 satisfies those conditions (AP with common difference 0 and GP with common ratio 1). Were there some other conditions on the GP or anything?

Re: HSC 2016 2U Marathon What was the Q. asking to prove/do?

Re: Discrete Maths Sem 2 2016 There's some discussion about this question at this page: http://mathforum.org/library/drmath/view/62543.html .

Re: MATH1231/1241/1251 SOS Thread $\noindent At any time $t$, the change in $B$ comes from the instantaneous deposit rate of $C+Dt$, and from the continuous compounding, which means instantaneous rate of increase in $B$ due to interest added as $RB(t)$. So the balance satisfies $B^\prime (t) =...

Find expressions involving x and y for the slopes (gradients) of PA and PB using the gradient formula. Then, equate the expression for the slope of PA to 2 times the expression for the slope of PB, and rearrange and simplify.

That orbital speed formula may be needed, but it's literally just speed = distance/time (using the formula for circumference of a circle for distance), so I wouldn't worry about memorising that one.

Re: MATH1231/1241/1251 SOS Thread We can use either for the normal, because those two vectors you've written are just negatives of each other, which means they are both valid normals (remember, two vectors that are just multiples of each other will both be valid direction vectors for a line).

Re: MATH1231/1241/1251 SOS Thread Yeah.

Re: MATH1231/1241/1251 SOS Thread Correct!

Re: MATH1231/1241/1251 SOS Thread I think you made a simplification error in getting the second last line.

Here are some hints for a method: • For any given i, find an expression for Pr(X = i), i.e. The probability that the randomly chosen family is a family with i children. This expression is very simple and easy to find once you understand the notation used. • Thus, write down an expression...

Don't worry, I think I've figured out what the Q. is asking. If you haven't made a typo, then the Q. itself essentially has a typo I think, because i is meant to be the thing we're summing over, yet it appears as the lower limit of a sum, which cannot happen.

I think there are typos here because the sums don't really make sense. (I presume i was meant to be a summation index, but it appears in the limits of summation too, which a summation index should never do. Also, r hasn't been defined I think.) I guess the first sum should just say that the...

$\noindent Let $p\equiv \mathbb{P}\left(X=1\right)$. Then $X$ is a $\text{Bernoulli}\left(p\right)$ random variable. Using standard formulas for mean and variance of a Bernoulli r.v., we have $\mu_X \equiv \mathbb{E}\left[X\right] = p$ and $\sigma^2 _X \equiv \mathrm{Var}\left[X\right] = pq$...

$\noindent We only need regular differentiation. Recall that for these ODE's, $y$ is just a(n unknown) function of $x$, so it's a \emph{dependent} variable, so there's only one independent variable here (namely $x$), so we just differentiate normally with respect to $x$. To help make this...

$\noindent To obtain the solution provided by the answers, let $u= y-x$, so $y= u+x \Rightarrow y^\prime = u^\prime + 1$. Substituting into the ODE, we obtain$ $$\begin{align*}u^\prime + 1 &= u^3 \\ \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} &= u^3 -1 = \left(u-1\right) \left(u^2 + u +...

How did you get your solution? Note we can't just integrate the RHS of the given ODE as though we're just integrating a square, because there's an unknown function y(x) in there.