Author Topic: 4U Maths Question Thread (Read 665725 times) Tweet Share

RuiAce · « **Reply #285 on:** July 03, 2016, 09:35:15 am »

Quote from: amandali on July 03, 2016, 02:18:32 am

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need help with these inequality ques thanks

$\textbf{Remember: A lot of steps are allowed because }a, b>0$
$\text{Attempt to work backwards as nothing is obvious.} \\ \begin{align*} \frac{1}{a}+\frac{1}{b} &\ge \frac{4}{a+b}\\ \iff \frac{a+b}{ab}&\ge \frac{4}{a+b} \\ \iff (a+b)^2 &\ge 4ab \end{align*}\\ \text{This is now very obvious. Start from }(a-b)^2 \ge 0\text{ and work forwards}$
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$\text{I spent so long on this question and could not find any shortcut} \\ \text{that utilises the previous part for some reason.}\\ \text{This is a very awkward proof that full on, works backwards.}$
$\begin{align*}\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}&\ge\frac{8}{{(a+b)}^{2}}\\\iff\left({a}^{2}+2ab+{b}^{2}\right)\left({a}^{2}+{b}^{2}\right)&\ge8{a}^{2}{b}^{2}\\ \iff{\left({a}^{2}+{b}^{2}\right)}^{2}+2ab\left({a}^{2}+{b}^{2}\right)-2ab\left(2ab\right)-4{a}^{2}{b}^{2}&\ge0\\ \iff{\left({a}^{2}+{b}^{2}\right)}^{2}+2ab\left({a}^{2}-2ab+{b}^{2}\right)-4{a}^{2}{b}^{2}&\ge0\\ \iff{a}^{4}+2{a}^{2}{b}^{2}+{b}^{4}+2ab{\left(a-b\right)}^{2}-4{a}^{2}{b}^{2}&\ge0\\ \iff{\left({a}^{2}-{b}^{2}\right)}^{2}+2ab{\left(a-b\right)}^{2}&\ge0\end{align*}$
$\text{By construction it should be possible to see why the last line is true.}$

RuiAce · « **Reply #286 on:** July 03, 2016, 09:37:04 am »

Quote from: amandali on July 03, 2016, 02:18:32 am

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need help with these inequality ques thanks

$\text{The last one depends on how well you saw the AM-GM inequality }\frac{a+b}{2} \ge \sqrt{ab}$
$\text{By multiplying three of them together:}\\ \left(\frac{a+b}{2}\right)\left(\frac{a+c}{2}\right)\left(\frac{b+c}{2}\right)\ge \sqrt{ab}\sqrt{ac}\sqrt{bc}\\ \text{And the result falls out immediately.}$
$\text{Proof of what I started with is left as your exercise.}$

amandali · « **Reply #287 on:** July 03, 2016, 09:25:26 pm »

RuiAce · « **Reply #288 on:** July 04, 2016, 12:02:42 am »

Quote from: amandali on July 03, 2016, 09:25:26 pm

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I dare say that heaps of information is missing somewhere to make the 3 variable AM-GM inequality easy to prove.

(Once again I'm not taking credit for anything above the long line. I asked for help on this one so please don't mind how he used a, b, c instead of x, y, z)
$\text{Let }a,b,c \geq 0\text{. Then we want to show that }\left(a+b+c\right)^3 \geq 27 {abc}\\ \text{since if we can show this, we'll be able to conclude that }\frac{a+b+c}{3}\geq \sqrt[3]{abc}\text{. We have}$
$\begin{align*}\left(a+b+c\right)^3 &= \left(a+b+c\right)^2 \left(a+b+c\right) \\&= \left(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\right)\left(a+b+c\right) \\ &\geq \left(3ab + 3bc + 3ca\right)\left(a+b+c\right) \quad (1) \\ &= 3\left(ab+bc+ca\right)\left(a+b+c\right) \\&= 3 \left(a^2b + ab^2+ abc + abc + b^2 c + bc^2 + ca^2 + abc + c^2 a\right)\\&= 3\left(a^2b+ab^2b^2 c+ bc^2+ca^2+ c^2a+3abc\right)\\ &\geq 3 \left(6abc + 3abc\right)\quad (2)\\&=27abc\end{align*}$
$\text{ as required.}$
$\textbf{Note.}\text{ Here are the proofs of results (1) and (2).}$
$\textbf{Result (1).} a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \geq 3ab + 3bc + 3ca$
$\textbf{Proof.}\text{ This follows from the well-known fact that }a^2 + b^2 + c^2 \geq ab + bc + ca\\ \text{This in turn follows from the fact that }a^2 + b^2 \geq 2ab\\ \text{, and similarly }b^2 + c^2 \geq 2bc\text{ and }c^2 + a^2 \geq 2ca.\\ \text{Then we sum these three inequalities and divide through by 2.}$
$\textbf{Result (2).}a^2b+ab^2 b^2c+bc^2+ca^2+c^2a \geq 6abc$
$\textbf{Proof.}\text{ Here, we have}$
$\begin{align*}LHS&=\left(a^2b+c^2b\right)+\left(b^2a+c^2a\right)+\left(a^2c+b^2c\right)\\&=\left(a^2+c^2\right)b+\left(b^2+c^2\right)a+\left(a^2+b^2\right)c\\&\geq2acb+2bca+2abc\quad(\text{usingtwo-variableAM-GM})\\&=6abc,\end{align*}$
$\text{ as desired.}$
It was considerably harder than expected to convert LaTeX (different packages) between two different sources.
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The purpose of introducing the three-variable AM-GM inequality is to also help do the first question.
$x+y+z\geq 3\sqrt[3]{xyz}$
Let:
For eqn 1: x=a, y=b, z=c
For eqn 2: x=a, y=b, z=d
For eqn 3: x=a, y=c, z=d
For eqn 4: x=b, y=c, z=d
$\begin{align}a+b+c&\geq 3\sqrt[3]{abc}\\ a+b+d&\geq3\sqrt[3]{abd} \\a+c+d&\geq3\sqrt[3]{acd} \\b+c+d&\geq3\sqrt[3]{bcd}\end{align}$
$\text{Multiplying all these equations together}\\ (a+b+c)(a+b+d)(a+c+d)(b+c+d)\ge 81\sqrt[3]{a^3b^3c^3d^3}=81abcd$

amandali · « **Reply #289 on:** July 04, 2016, 02:36:41 pm »

is my starting working correct and how do i progress

for resistance force, do i normally include k as a constant

jakesilove · « **Reply #290 on:** July 04, 2016, 05:11:33 pm »

Quote from: amandali on July 04, 2016, 02:36:41 pm

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is my starting working correct and how do i progress

for resistance force, do i normally include k as a constant

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Hey! I don't have time for a full solution (I'm sure Rui will jump in), so I'll just give some quick pointers. The start of the first question looks fine, great job! What you want to do next is integrate between definite limits. So, the right hand side (dv) can be integrated between 30 and v, whilst the left hand side can be integrated between time 0 and T. Once you've done that, you know that the highest point will be when v is zero (ie. the object is stationary). Set v=0 and an answer should pop out neatly! Great job setting up the question.

As for the next question, your problem lies in your integration. I don't know how your school does it, but using indefinite integrals, and finding the +C term each time, is generally more difficult than just using definite integrals, like we did above. Can you see which limits to use?

Hope this helps!

Jake

RuiAce · « **Reply #291 on:** July 04, 2016, 05:17:50 pm »

Quote from: jakesilove on July 04, 2016, 05:11:33 pm

Hey! I don't have time for a full solution (I'm sure Rui will jump in), so I'll just give some quick pointers. The start of the first question looks fine, great job! What you want to do next is integrate between definite limits. So, the right hand side (dv) can be integrated between 30 and v, whilst the left hand side can be integrated between time 0 and T. Once you've done that, you know that the highest point will be when v is zero (ie. the object is stationary). Set v=0 and an answer should pop out neatly! Great job setting up the question.

As for the next question, your problem lies in your integration. I don't know how your school does it, but using indefinite integrals, and finding the +C term each time, is generally more difficult than just using definite integrals, like we did above. Can you see which limits to use?

Hope this helps!

Jake

Oops. Didn't see. Thanks for the bump haha
_____________________________
In regards to Q14
$\text{Observe that the value of }k=\frac{1}{5} \text{ has already been given to you.}\\\text{We do not consider }F_R=\frac{kv^2}{5}\textbf{ for this question}\text{ as this is actually a redundancy. }k\text{ is just a constant anyhow.}\\ \text{Additionally, keep in mind that }g=10\text{ and the mass }m=20\text{ is given.}\\\text{Commence our resolution of forces:}$
$\text{The upwards direction is the positive direction. The resistive force and gravity BOTH work AGAINST the particle.}$
$\text{Aside: You are correct in choosing }\ddot{x}=\frac{dv}{dt}\text{ as we are interested in time here.}$
$\begin{align*}\sum F &= -F_R -F_G\\ m\ddot{x} &= -kv^2-mg\\ \frac{dv}{dt}&= -\frac{v^2}{5m}-10\\ &= -\frac{v^2}{100}-10\\ &= -\frac{v^2+1000}{100}\\ \therefore \frac{dv}{v^2+1000}&=-\frac{dt}{100}\end{align*}$
$\text{When }t=0,v=30\\ \text{When }t=T, v=0\\ \text{Hence via seperation of variables}$
$\begin{align*}\int_{30}^0\frac{dv}{v^2+1000}&= -\int_0^T \frac{dt}{100}\\ \frac{1}{100}\left[t\right]_0^T&= \left[\frac{1}{10\sqrt{10}}\tan^{-1}\left(\frac{v}{10\sqrt{10}}\right)\right]_0^{30}\\ T&= \sqrt{10}\tan^{-1}\frac{3}{\sqrt{10}}+0\end{align*}$

Emphasis: F_R=kv² generally speaking, but you were given the value for k here.

RuiAce · « **Reply #292 on:** July 04, 2016, 05:46:23 pm »

Quote from: amandali on July 04, 2016, 02:36:41 pm

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is my starting working correct and how do i progress

for resistance force, do i normally include k as a constant

(Image removed from quote.)

In regards to Q21
$\text{Your starting point is most certainly correct upon correctly resolving the forces, so I will jump a few steps.}\\ \ddot{x}=-g\left(1+\frac{v}{V}\right)\\ \text{It helps to do some tidying up now rather than later when the integral symbol is brought in.}\\ v\frac{dv}{dx}=-\frac{g}{V}(V+v)$
$\text{Let the greatest height be }H.\\ \text{When }x=H, v=0\\ \text{When }x=0, v=u$
$\begin{align*}\frac{v}{V+v}dv&=-\frac{g}{V}dx\\ \therefore \int_0^u \frac{v}{V+v}dv &= \int_H^0 -\frac{g}{V}dx\end{align*}$
$\text{Recall our strategy from integration to cater for these bad forms.}$
$\begin{align*}\left[\frac{g}{V}x\right]_0^H &= \int_0^u \left(1-\frac{V}{V+v}\right)dv\\\frac{gH}{V}-0&= \left[v - V\ln (V+v)\right]_0^u\\ H&=\frac{V}{g}\left(u-V\ln{\frac{V+u}{V+0}}\right)\end{align*}$

amandali · « **Reply #293 on:** July 07, 2016, 12:28:48 pm »

how to do part (ii) for ques 4

RuiAce · « **Reply #294 on:** July 07, 2016, 01:24:54 pm »

Quote from: amandali on July 07, 2016, 12:28:48 pm

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how to do part (ii) for ques 4

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$\text{So your integrand looks good for Q5:}\\ \begin{align*} \int_U^0 \frac{v}{v^2+c^2}\, dv&=-k \int_0^s dx \\ k[x]_0^s &= \left[ \frac{1}{2}\ln \left(v^2+c^2\right)\right]_0^U \\ s&=\frac{1}{2k}\ln \left(\frac{c^2+U^2}{c^2}\right) \tag{1} \end{align*}$
$\text{Your second integration boundaries may have been what created something unpleasant. I chose}\\\int_{\frac{U}{3}}^0 \frac{v}{v^2+c^2}\, dv=-k \int_0^{\frac{s}{2}} dx$ $\text{Which lead to }s=\frac{1}{k}\ln \left(\frac{c^2+\frac{U^2}{9}}{c^2}\right)=\frac{1}{k}\ln \left(\frac{9c^2+U^2}{9c^2}\right) \tag{2}$
$\text{Equating (1) with (2):}\\ \begin{align*}\frac{1}{k}\ln \left(\frac{9c^2+U^2}{9c^2}\right) &= \frac{1}{2k}\ln \left(\frac{c^2+U^2}{c^2}\right) \\ \ln\left(\frac{c^2+U^2}{c^2}\right)&= \ln \left(\frac{9c^2+U^2}{9c^2}\right)^2 \\ \frac{c^2+U^2}{c^2} &= \frac{\left(9c^2+U^2\right)^2}{81c^4} \\ 81c^4+81c^2U^2&=81c^4+18c^2U^2+U^4\\ &=63c^2=U^2 \end{align*}$

RuiAce · « **Reply #295 on:** July 07, 2016, 01:33:50 pm »

Quote from: amandali on July 07, 2016, 12:28:48 pm

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how to do part (ii) for ques 4

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$\text{Sorry but a bit of nitpicking on part (i) first, your integral should be something like }\int_0^D -k \, dx \\ \text{, as pairing an indefinite integral with a definite one is not mathematically correct.}\\ \text{Leaving it as indefinite meant you would've had }-kx+C$
$\text{I think for part (ii) you just missed the tangent inverse.}\\ \text{Keep in mind the particle arrives at }B\text{ the same time it comes to rest.}$
$\begin{align*}\int_c^0 \frac{dv}{c^2+v^2}&= \int_0^T -k \, dt\\ k[t]_0^T&=\left[\frac{1}{c}\tan^{-1}\left(\frac{v}{c}\right)\right]_0^c \\ T&=\frac{1}{kc}\left(\tan^{-1}1-\tan^{-1}0\right)\\ &=\frac{\pi}{4kc} \end{align*}$

katherine123 · « **Reply #296 on:** July 07, 2016, 04:17:01 pm »

Quote from: RuiAce on July 07, 2016, 01:24:54 pm

$\text{So your integrand looks good for Q5:}\\ \begin{align*} \int_U^0 \frac{v}{v^2+c^2}\, dv&=-k \int_0^s dx \\ k[x]_0^s &= \left[ \frac{1}{2}\ln \left(v^2+c^2\right)\right]_0^U \\ s&=\frac{1}{2k}\ln \left(\frac{c^2+U^2}{c^2}\right) \tag{1} \end{align*}$
$\text{Your second integration boundaries may have been what created something unpleasant. I chose}\\\int_{\frac{U}{3}}^0 \frac{v}{v^2+c^2}\, dv=-k \int_0^{\frac{s}{2}} dx$ $\text{Which lead to }s=\frac{1}{k}\ln \left(\frac{c^2+\frac{U^2}{9}}{c^2}\right)=\frac{1}{k}\ln \left(\frac{9c^2+U^2}{9c^2}\right) \tag{2}$
$\text{Equating (1) with (2):}\\ \begin{align*}\frac{1}{k}\ln \left(\frac{9c^2+U^2}{9c^2}\right) &= \frac{1}{2k}\ln \left(\frac{c^2+U^2}{c^2}\right) \\ \ln\left(\frac{c^2+U^2}{c^2}\right)&= \ln \left(\frac{9c^2+U^2}{9c^2}\right)^2 \\ \frac{c^2+U^2}{c^2} &= \frac{\left(9c^2+U^2\right)^2}{81c^4} \\ 81c^4+81c^2U^2&=81c^4+18c^2U^2+U^4\\ &=63c^2=U^2 \end{align*}$

why is the boundary between 0 and U/3 for the second integration not between U and U/3 since the initial speed is U

RuiAce · « **Reply #297 on:** July 07, 2016, 05:42:54 pm »

Oh shoot

Quote from: katherine123 on July 07, 2016, 04:17:01 pm

why is the boundary between 0 and U/3 for the second integration not between U and U/3 since the initial speed is U

$\text{Using the correct boundaries and a similar process we arrive at this instead}$
$\begin{align*}\frac{1}{k}\ln \left(\frac{9c^2+9U^2}{9c^2+U^2}\right) &= \frac{1}{2k}\ln \left(\frac{c^2+U^2}{c^2}\right)\\ \frac{c^2+U^2}{c^2}&= \frac{81(c^2+U^2)^2}{(9c^2+U^2)^2}\\ (9c^2+U^2)^2&=81c^2(c^2+U^2)\\ 81c^4+18c^2U^2+U^4&=81c^4+81c^2U^2\\ U^4&=63c^2U^2\\ U^2&=63c^2 \end{align*}$

amandali · « **Reply #298 on:** July 08, 2016, 07:24:12 pm »

ive done part i but not sure how to do part ii (2nd picture)

RuiAce · « **Reply #299 on:** July 08, 2016, 08:45:04 pm »

Quote from: amandali on July 08, 2016, 07:24:12 pm

ive done part i but not sure how to do part ii (2nd picture)

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$\text{We keep in mind }V_T^2 = \frac{g}{k}\text{ for later.}$
$\text{Assume that the }h\text{ in part (i) was the maximum height of the particle.}\\ \text{Then, }h=-\frac{1}{2k}\ln \left( \frac{g-kV_R^2}{g} \right)$
$\text{Otherwise, continuing your resolution of forces (watch out - you dropped a negative):} \\ x=0 \implies v=V_I \qquad x=h \implies v=0 \\ \begin{align*}v\frac{dv}{dx}&=-(g+kv^2)\\ \int_h^0 -dx &= \int_0^{V_I} \frac{v \, dv}{g+kv^2} \\ h &= \frac{1}{2k}\ln \left(\frac{g+kV_I^2}{g}\right) \end{align*}\\ \text{skipping some lines}$
$\text{So upon equating:}\\ \begin{align*}\frac{1}{2k}\ln \left(\frac{g+kV_I^2}{g}\right)&=-\frac{1}{2k}\ln \left( \frac{g-kV_R^2}{g} \right) \\ \frac{g+kV_I^2}{g}&= \frac{g}{g-kV_R^2}\\ g^2 + g(kV_I^2 - kV_R^2) - k^2V_I^2V_R^2&=g^2\\ g(V_I^2-V_R^2)&=kV_I^2V_R^2\\ \frac{k}{g} &= \frac{V_I^2-V_R^2}{V_I^2V_R^2}\\ \frac{1}{V_T^2}&=\frac{1}{V_R^2}-\frac{1}{V_I^2}\\ \frac{1}{V_R^2}&=\frac{1}{V_I^2}+\frac{1}{V_T^2}\end{align*}$
$\text{Obviously }g\neq 0, k \neq 0$

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