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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread We'd need to see Question 2) for this, but basically if they gave you a formula in terms of n for T_n, then you'd plot the graph as though it's a function but only at discrete points (e.g. n = 1,2,3,...). If they only gave the...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Inspection. $\noindent (a) $T_n = T_{n-1} + 5$$ $\noindent (b) $T_n = 2T_{n-1}$$ $\noindent (c) $T_n = T_{n-1} -7$$ $\noindent (d) $T_n = -T_{n-1}$$

Well, you asked for uncommon. Anyway, don't think there's 'out of syllabus' for this.

Many can be found here for example: https://en.wikipedia.org/wiki/Glossary_of_literary_terms.

What's the main thing you hate about it?

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent The answer should be an integer. Note that the coefficients of $x,x^2$ and $x^3$ are $n,\frac{n\left(n-1\right)}{2}$ and $\frac{n(n-1)(n-2)}{3!}$ (i.e. binomial coefficients). Since this is an A.P., we have...

$\noindent The confusion here has arisen because there are two possible main conventions. One is where `$n$' refers to the number of parabolas used in the approximation, and the other is where `$n$' refers to the number of \textsl{sub-intervals} used. Note that one of these is just twice the...

The payoff P is 0 if and only if 50 – X ≤ 0, i.e. iff X ≥ 50. So find Pr(X ≥ 50) (this'll be the answer). The reason is that P = max(50 – X, 0). So for P to be 0, we need and want the maximum of those two to be 0. This'll happen if and only if the 50 – X is 0 or less.

$\noindent Note $\sinh t >t$ for all $t > 0$ (in fact equality iff $t =0$). One way to see this is via differentiation. Another way is via integration: recall $\cosh u >1$ for all $u>0$. So $\int _{0}^{t} \cosh u\text{ d}u > \int _0 ^ t 1 \text{ d}t\Rightarrow \int _{0}^{t} \cosh u\text{ d}u >...

$\noindent One thing to note is that if we write $\theta = t$, then squaring and adding, $x^2 + y^2 = \sin^2 2\theta \Rightarrow r \equiv f(\theta)= \left |\sin 2\theta\right |$. This may be a more familiar (polar) curve for you. Anyway, note $f(\theta) = f(-\theta)$, so the curve is symmetric...

What have you gotten up to so far? The first part of doing part (a) (just setting up the Riemann sum) should be rather routine. What did you get for the Riemann sum?

Well it's because the notations are both referring to the partial derivative of ƒ wrt its first designated variable. As seanieg89 said: .

Yeah that's valid (remember to say we let t = 1). But do you know why it's valid, i.e. why ∂ƒ/∂x and ∂ƒ/∂u are the same (well I guess this already got answered as leehuan raised this question)? (Also note here dƒ/dt means dz/dt, where z = ƒ(tx,ty) and x and y are considered fixed.)

Well they're just two notations for the same thing. If you get the answer out in the subscript notation form and get penalised, it's kind of like penalising someone for writing y' instead of dy/dx (assuming that you are allowed to use ƒ1 notation, which should be the case at least if you write a...

What can't you get away with? Writing something like ƒ1?

E.g. Say ƒ(u,v) = u^3 + v^3. So ƒ1(x,y) = 3u^2 (i.e. "3 times the square of the first variable in f). Then ƒ(tx,ty) = (tx)^3 + (ty)^3 = t^3 x^3 + t^3 y^3 = t^3 (x^3 + y^3) (incidentally, this is an example of a function ƒ that is homogeneous of degree 3). Now, ƒ1(tx,ty) = 3(tx)^2. We wouldn't...

I think the confusion was more of a notational confusion rather than a mathematical confusion.

OK I think I get what you're doing now. When you're saying the derivative of ƒ wrt t would be ∂ƒ/∂x ∂x/∂t + etc., by ∂ƒ/∂x, you mean "partial of f wrt its first variable", right? And then ∂x/∂t, you mean "partial of the thing in the first variable's position wrt t", right (which is x, yeah)...

Well you wrote something like ∂x/∂t, but x and t are independent of each other (so we wouldn't have ∂x/∂t = x. Anyway, if ∂x/∂t equalled x, that'd be like the population growth formulas y' = ky, so ∂x/∂t = x would basically imply something like x = e^t. Like usually when we take partial...

Well u and v instead of x and y, with u = tx, and v = ty. And then note that when you're saying ∂ƒ/∂u, what you're referring to is ƒ1, etc. It might help if you look at a concrete example of a function ƒ. E.g. ƒ(x,y) = x3y + sin(x + y).