Author Topic: Mathematics Question Thread (Read 1626430 times) Tweet Share

pikachu975 · « **Reply #1920 on:** May 28, 2017, 09:07:44 pm »

Quote from: Fahim486 on May 28, 2017, 09:02:26 pm

Thanks for your help but may I ask where you got h/2 = 1/4 from?

Basically h is the difference between each function value.

The difference between each is 0.5, so 0.5/2 = 1/4!

Fahim486 · « **Reply #1921 on:** May 28, 2017, 09:43:50 pm »

Quote from: pikachu975 on May 28, 2017, 09:07:44 pm

Basically h is the difference between each function value.

The difference between each is 0.5, so 0.5/2 = 1/4!

ooohhhhh okay thats makes so much sense. Thanks for your help!!!!!!!!!!!

laurenf58 · « **Reply #1922 on:** May 29, 2017, 06:33:19 pm »

Can I have some help with these questions please?

Thanks

Fahim486 · « **Reply #1923 on:** May 29, 2017, 07:46:17 pm »

Hi really need help with this question. Thanks!!!

RuiAce · « **Reply #1924 on:** May 29, 2017, 08:32:49 pm »

Quote from: Fahim486 on May 29, 2017, 07:46:17 pm

Hi really need help with this question. Thanks!!!

$\text{Apply the relevant formulae}$
$A=\frac12r^2\theta = 4\pi \implies \boxed{r^2\theta=8\pi}\\ \ell = r\theta \implies \boxed{r\theta=\frac{\pi}{8}}$
$\text{This is now a pair of simultaneous equations}\\ \text{Upon division we have }r = \frac{8\pi}{\pi/8}=64$
$\text{You should be able to finish it off from here}$

RuiAce · « **Reply #1925 on:** May 29, 2017, 08:37:13 pm »

Quote from: laurenf58 on May 29, 2017, 06:33:19 pm

Can I have some help with these questions please?

Thanks

$y=x^3+2x \implies \frac{dy}{dx}=3x^2+2\\ \text{Clearly }\frac{dy}{dx} > 0\text{ so the curve is monotonic increasing}$
$\text{Note that we may more formally verify that }\frac{dy}{dx}>0\text{ for all }x\\ \text{The quadratic }3x^2+2\text{ is }\textit{positive definite}\\ \text{(You may want to revisit prelim to see how to prove it, and check what it means)}$
_________________
$\text{If the graph is monotonic increasing, then it should be easy to sketch it roughly}\\ \text{It will look something like this}\\ \textit{Note that the graph is only a visual aid - it need not be precise.}$

For convenience, y=5 was also plotted
$\text{Just reading off the graph, there is only 1 point of intersection}$
$\textbf{However, you should understand why in GENERAL}\\ \text{if the curve is monotonic, it intersects a straight line only once}$
_________________

This is just 1. That equation is formed when you try to find the point of intersection between the two graphs, and you already know from the previous part that there is only one.

Fahim486 · « **Reply #1926 on:** May 29, 2017, 09:33:27 pm »

Thanks for the help RuiAce. I've got another question I also need help with and I can't figure out why I keep getting the wrong answer. Thanks!!!

RuiAce · « **Reply #1927 on:** May 29, 2017, 09:37:58 pm »

Quote from: Fahim486 on May 29, 2017, 09:33:27 pm

Thanks for the help RuiAce. I've got another question I also need help with and I can't figure out why I keep getting the wrong answer. Thanks!!!

$y=e^{x+\ln x}\implies \frac{dy}{dx}=\left(1+\frac1x\right) e^{x+\ln x}$
$\text{Substituting }x=1\text{ gives }\frac{dy}{dx}=2e$

Thebarman · « **Reply #1928 on:** May 30, 2017, 07:28:33 pm »

Does anyone have some tips on how to solve questions like these? I'm basically using guess and check in my calculator at this point, but there must be a better way to solve it. Any help would be appreciated

y= 3cos(x) - cos(2x)

RuiAce · « **Reply #1929 on:** May 30, 2017, 08:43:31 pm »

Quote from: Thebarman on May 30, 2017, 07:28:33 pm

Does anyone have some tips on how to solve questions like these? I'm basically using guess and check in my calculator at this point, but there must be a better way to solve it. Any help would be appreciated

y= 3cos(x) - cos(2x)

What are you trying to solve? All you've specified is just an expression.

Thebarman · « **Reply #1930 on:** May 31, 2017, 12:13:25 am »

I'm trying to graph and find the the intercepts of the two curves.

RuiAce · « **Reply #1931 on:** May 31, 2017, 12:15:30 am »

As far as the HSC goes, the only way to sketch that is to first sketch y=3cos(x) and y=-cos(2x) separately (AND to scale), and then add the y-coordinates together. This is essentially how superposition of waves work.

My honest opinion - it's such an annoying process.

If you want to find the intercepts of 3cos(x) and cos(2x), upon equating you get 3cos(x) - cos(2x) = 0 so you just look at where all your x-intercepts are.

1937jk · « **Reply #1932 on:** May 31, 2017, 03:58:37 pm »

hi just stuck on this question,
A loan of $6000 over 5 years at 15%, charged monthly, is paid back in 5 annual instalments.
a) what is the amount of each instalment?
b) how much is paid back altogether?

RuiAce · « **Reply #1933 on:** May 31, 2017, 04:18:11 pm »

Quote from: 1937jk on May 31, 2017, 03:58:37 pm

hi just stuck on this question,
A loan of $6000 over 5 years at 15%, charged monthly, is paid back in 5 annual instalments.
a) what is the amount of each instalment?
b) how much is paid back altogether?

$15\text{ percent per annum }\implies 1.25\text{ percent per monthly.}\\ \textbf{This question differs from the norm in that the compounding frequency}\\ \textbf{does not match the frequency of the payments}$
$\text{Let }A_n\text{ be the amount after }n\text{ years}\\\text{and let }M\text{ be the annual payment} \\\text{Then, carefully note that }A_1 = 6000(1.0125)^{12} - M$
$\text{Doing the usual stuff}\\ A_5 = 6000(1.0125)^{60}-M(1.0125+1.0125^{12}+\dots+1.0125{48}) = 6000(1.0125)^{60}-M\left(\frac{1.0125^{60}-1}{1.0125^{12}-1}\right)\\ \text{Equate this to 0 to find the annual payment}$
______________________
$\text{The total amount paid off will just be }5M.$

K9810 · « **Reply #1934 on:** May 31, 2017, 09:48:47 pm »

Hey!
How can I differentiate this?

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