Author Topic: 4U Maths Question Thread (Read 665269 times) Tweet Share

envisagator · « **Reply #2025 on:** October 07, 2018, 11:21:17 am »

Hi Rui, can you help me with part (c) of the question. Thanks in advance

clovvy · « **Reply #2026 on:** October 07, 2018, 04:15:53 pm »

Quote from: envisagator on October 07, 2018, 11:21:17 am

Hi Rui, can you help me with part (c) of the question. Thanks in advance

Use the equaion from part (i)
$\text{Let the equations}\, F_1=\frac{m\cos\alpha}{r}(V_1^2-V_0^2)\, \text{and a similar expression for}\, F_2\\ \text{now divide both expressions to get}\, \frac{F_1}{F_2}=\frac{V_1^2-V_0^2}{V_2^2-V_0^2}\\ \text{Then rearrange to get}\, F_1(V_2^2-V_0^2)=F_2(V_1^2-V_0^2)\\ \text{then rearrange for}\, V_0^2\, \text{to get the required result}$

Also I don't know how to do part (ii) so if anyone can help there that would be great..

RuiAce · « **Reply #2027 on:** October 07, 2018, 06:37:49 pm »

Quote from: envisagator on October 07, 2018, 11:21:17 am

Hi Rui, can you help me with part (c) of the question. Thanks in advance

Yeah pretty much see above. All that it really wanted you to interpret is that when $v = V_1$, $F = F_1$, and also when $v = V_2$, $F = F_2$. Then we just do the best thing we can (i.e. cancellation) to get rid of the $m$, $r$ and $\cos \alpha$.

Quote from: clovvy on October 07, 2018, 04:15:53 pm

Use the equaion from part (i)
$\text{Let the equations}\, F_1=\frac{m\cos\alpha}{r}(V_1^2-V_0^2)\, \text{and a similar expression for}\, F_2\\ \text{now divide both expressions to get}\, \frac{F_1}{F_2}=\frac{V_1^2-V_0^2}{V_2^2-V_0^2}\\ \text{Then rearrange to get}\, F_1(V_2^2-V_0^2)=F_2(V_1^2-V_0^2)\\ \text{then rearrange for}\, V_0^2\, \text{to get the required result}$

Also I don't know how to do part (ii) so if anyone can help there that would be great..

Then the friction is reversed and starts going upwards instead. Which makes sense because if the velocity is too small, the particle will start to slide down the banked track, so you wanna push it back up there.

wlam · « **Reply #2028 on:** October 09, 2018, 03:39:56 pm »

Hi could someone pls help me w this question thanks

clovvy · « **Reply #2029 on:** October 09, 2018, 03:44:23 pm »

Quote from: wlam on October 09, 2018, 03:39:56 pm

Hi could someone pls help me w this question thanks

which parts do you need help with specifically?
Anyway, I'll start you off with part (i), expand the expression (cisx)^5 manually
$\cos5\theta+i\sin5\theta=\cos^5\theta+5cos^4\theta i\sin\theta-10\cos^3\theta\sin^2\theta-10\cos^2\theta i\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^5\theta\\ \text{Now equate real and imaginary parts i.e.}\\ \cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\,\text{and}\, \sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta\\ \text{Now divide the expressions for}\, \cos5\theta\, \text{with the expression for}\, \sin5\theta\, \text{then divide top and bottom with}\, \cos^4\theta\\ \therefore \tan5\theta=\frac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}\div\frac{\cos^5\theta}{\cos^5\theta}\\=\frac{\tan\theta-10\tan^3\theta+5\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$

wlam · « **Reply #2030 on:** October 09, 2018, 05:02:54 pm »

Quote from: clovvy on October 09, 2018, 03:44:23 pm

which parts do you need help with specifically?
Anyway, I'll start you off with part (i), expand the expression (cisx)^5 manually
$\cos5\theta+i\sin5\theta=\cos^5\theta+5cos^4\theta i\sin\theta-10\cos^3\theta\sin^2\theta-10\cos^2\theta i\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^5\theta\\ \text{Now equate real and imaginary parts i.e.}\\ \cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\,\text{and}\, \sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta\\ \text{Now divide the expressions for}\, \cos5\theta\, \text{with the expression for}\, \sin5\theta\, \text{then divide top and bottom with}\, \cos^4\theta\\ \therefore \tan5\theta=\frac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}\div\frac{\cos^5\theta}{\cos^5\theta}\\=\frac{\tan\theta-10\tan^3\theta+5\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$

hi thanks!

i need help with part b

RuiAce · « **Reply #2031 on:** October 09, 2018, 08:09:26 pm »

Quote from: wlam on October 09, 2018, 03:39:56 pm

Hi could someone pls help me w this question thanks

$\text{Note that multiplying by }x\text{ gives the equation}\\ x^5-10x^3 + 5x = 0.$
$\text{Comparing with part i), it makes sense to make the substitution }x = \tan \theta.\\ \text{This then gives }\tan^5 \theta - 10\tan^3 \theta + 5 \tan \theta = 0.$
$\text{Which upon dividing }1 - 10\tan^2\theta + 5\tan^4\theta\text{ becomes }\boxed{ \frac{\tan^5\theta - 10\tan^3\theta + 5\tan\theta}{1-10\tan^2\theta+5\tan^4\theta} = 0}\\ \text{and by part i) becomes }\boxed{\tan 5 \theta = 0}.$
$\text{The general solution is }5\theta = k\pi \implies \theta = \frac{k\pi}{5}\\ \text{so taking the five consecutive numbers }k=-2,-1,0,1,2\text{ we have}\\ \theta = \pm \frac{2\pi}{5}, \pm \frac{\pi}{5}, 0.$
$\text{So the solutions to }x^5 - 10x^3 + 5x = 0\text{ are }x = \pm \tan \frac{2\pi}{5}, \pm \tan \frac{\pi}{5}, 0.\\ \text{Of course, }x=0\text{ is the solution to }x=0\text{, so after}\\ \text{some factorising, }x^4-10x^2+5=0\text{ solves to give }x = \pm \tan \frac{2\pi}{5}, \pm \tan \frac{\pi}{5}$

wlam · « **Reply #2032 on:** October 10, 2018, 07:33:24 pm »

Quote from: RuiAce on October 09, 2018, 08:09:26 pm

$\text{Note that multiplying by }x\text{ gives the equation}\\ x^5-10x^3 + 5x = 0.$
$\text{Comparing with part i), it makes sense to make the substitution }x = \tan \theta.\\ \text{This then gives }\tan^5 \theta - 10\tan^3 \theta + 5 \tan \theta = 0.$
$\text{Which upon dividing }1 - 10\tan^2\theta + 5\tan^4\theta\text{ becomes }\boxed{ \frac{\tan^5\theta - 10\tan^3\theta + 5\tan\theta}{1-10\tan^2\theta+5\tan^4\theta} = 0}\\ \text{and by part i) becomes }\boxed{\tan 5 \theta = 0}.$
$\text{The general solution is }5\theta = k\pi \implies \theta = \frac{k\pi}{5}\\ \text{so taking the five consecutive numbers }k=-2,-1,0,1,2\text{ we have}\\ \theta = \pm \frac{2\pi}{5}, \pm \frac{\pi}{5}, 0.$
$\text{So the solutions to }x^5 - 10x^3 + 5x = 0\text{ are }x = \pm \tan \frac{2\pi}{5}, \pm \tan \frac{\pi}{5}, 0.\\ \text{Of course, }x=0\text{ is the solution to }x=0\text{, so after}\\ \text{some factorising, }x^4-10x^2+5=0\text{ solves to give }x = \pm \tan \frac{2\pi}{5}, \pm \tan \frac{\pi}{5}$

thanks!!

A Selective Student · « **Reply #2033 on:** October 12, 2018, 10:24:09 am »

Hey Rui can i get help with these 2 questions. One is circle geo and one is complex. Would need some help on the working out.

RuiAce · « **Reply #2034 on:** October 12, 2018, 02:23:58 pm »

Quote from: A Selective Student on October 12, 2018, 10:24:09 am

Hey Rui can i get help with these 2 questions. One is circle geo and one is complex. Would need some help on the working out.

$\text{The complex numbers question can be done by brute force.}$
$\text{Let }z = x+iy.\text{ Then the equation becomes}\\ \boxed{\left(x+\sqrt{x^2+y^2} + iy = 2 + 8i \right)}.$
$\text{Equating imaginary parts gives }y=8.\\ \text{Equating real parts then gives}\\ \begin{align*} x + \sqrt{x^2+64} &= 2\\ \sqrt{x^2+64} &= 2-x\\ x^2+64&=4-4x+x^2\\ 4x &= -60\\ x &= -15\end{align*}$
$\text{So }|z|^2 = 225 + 64 = 289.$
There is a geometric way around this question as well, and it's quite interesting. But I found I had to do a tiny bit more work than this algebraic method.

envisagator · « **Reply #2035 on:** October 12, 2018, 03:18:30 pm »

Hi Rui, can I have help with this induction q: I was think assume n=k and n=k-1, but the answer's did n=k and n=k+1 just not sure.
$\text{ A sequence is defined by: } a_1=5, a_2=13, a_{n+2}=5a_{n+1}-6a_n \text{ for all natural numbers n} \\ \text{ Use mathematical induction to prove } a_{n}=2^n+3^n$

RuiAce · « **Reply #2036 on:** October 12, 2018, 03:35:30 pm »

Quote from: envisagator on October 12, 2018, 03:18:30 pm

Hi Rui, can I have help with this induction q: I was think assume n=k and n=k-1, but the answer's did n=k and n=k+1 just not sure.
$\text{ A sequence is defined by: } a_1=5, a_2=13, a_{n+2}=5a_{n+1}-6a_n \text{ for all natural numbers n} \\ \text{ Use mathematical induction to prove } a_{n}=2^n+3^n$

If it's just the issue of {assume n=k, n=k-1 and prove n=k+1} v.s. {assume n=k+1, n=k and prove n=k+2} either works. Perhaps the reason why they did the latter is because by doing the latter you let $k\geq 1$, in accordance with what the question provides. Assuming the former makes the working out heaps tidier, just that you need to let $k \geq 2$ instead.

Did you need help with the computations?

envisagator · « **Reply #2037 on:** October 12, 2018, 03:53:40 pm »

Quote from: RuiAce on October 12, 2018, 03:35:30 pm

If it's just the issue of {assume n=k, n=k-1 and prove n=k+1} v.s. {assume n=k+1, n=k and prove n=k+2} either works. Perhaps the reason why they did the latter is because by doing the latter you let $k\geq 1$, in accordance with what the question provides. Assuming the former makes the working out heaps tidier, just that you need to let $k \geq 2$ instead.

Did you need help with the computations?

Thats all I needed to know that you!!.

A Selective Student · « **Reply #2038 on:** October 12, 2018, 05:06:13 pm »

Quote from: RuiAce on October 12, 2018, 02:23:58 pm

(Image removed from quote.)
$\text{The complex numbers question can be done by brute force.}$
$\text{Let }z = x+iy.\text{ Then the equation becomes}\\ \boxed{\left(x+\sqrt{x^2+y^2} + iy = 2 + 8i \right)}.$
$\text{Equating imaginary parts gives }y=8.\\ \text{Equating real parts then gives}\\ \begin{align*} x + \sqrt{x^2+64} &= 2\\ \sqrt{x^2+64} &= 2-x\\ x^2+64&=4-4x+x^2\\ 4x &= -60\\ x &= -15\end{align*}$
$\text{So }|z|^2 = 225 + 64 = 289.$
There is a geometric way around this question as well, and it's quite interesting. But I found I had to do a tiny bit more work than this algebraic method.

Wow can't believe i didn't see those methods for both question o.O Thanks so much!

3.14159265359 · « **Reply #2039 on:** October 20, 2018, 04:31:01 pm »

hello

can someone help pleaseee because Im not getting the right answer.

thank you

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