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June 18, 2019, 12:38:06 pm

Author Topic: 4U Maths Question Thread  (Read 247714 times)  Share 

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pmcnair

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Re: 4U Maths Question Thread
« Reply #2265 on: April 29, 2019, 05:11:22 pm »
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Hello everyone,
Just stuck on this question, which has a weird sequence I can't really figure out. Any help with part (iii) would be very much appreciated.
It's from the 2003 HSC, question 6

RuiAce

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Re: 4U Maths Question Thread
« Reply #2266 on: April 29, 2019, 08:09:02 pm »
+3
Hello everyone,
Just stuck on this question, which has a weird sequence I can't really figure out. Any help with part (iii) would be very much appreciated.
It's from the 2003 HSC, question 6
\[ \text{Since this is a second order recurrence relation}\\ \text{we require a stronger version of induction that requires two base cases.} \]
\[ \text{When }n=1, \, s_1 =1 \geq \sqrt{1!}\text{ as required.}\\ \text{When }n=2, \, s_2 = 2 \geq \sqrt{2!}\text{ as required.} \]
\[ \text{With two base cases, now assume that the statement holds}\\ \text{when }n=k\text{ and when }n=k-1.\\ \text{i.e. }\boxed{s_k \geq \sqrt{k!}}\text{ and }\boxed{s_{k-1}\geq \sqrt{(k-1)!}} \]
\[\text{Then when }n=k+1,\\ \begin{align*} s_{k+1} &= s_k + k s_{k-1} \tag{definition}\\ &\geq \sqrt{k!} + k \sqrt{(k-1)!} \tag{assumption} \\ &= \sqrt{k (k-1)!} + k \sqrt{(k-1)!} \tag{classic factorial trick}\\ &= \sqrt{k}\sqrt{(k-1)!} + k \sqrt{(k-1)!} \tag{split product inside sqrt}\\ &= \left( \sqrt{k} + k\right) \sqrt{(k-1)!}\tag{factorise}\\ &\geq \sqrt{k(k+1)}\sqrt{(k-1)!}\tag{part ii)}\\ &= \sqrt{(k+1)k(k-1)!}\tag{recombine product into sqrt}\\ &= \sqrt{(k+1)!}\tag{classic factorial trick} \end{align*}\\ \text{as required.} \]
Note: The factorial trick used was that \(n! = n(n-1)!\).
« Last Edit: April 29, 2019, 08:11:42 pm by RuiAce »

pmcnair

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Re: 4U Maths Question Thread
« Reply #2267 on: April 29, 2019, 08:43:20 pm »
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Thanks very much Rui! I haven't seen a second order recurrence relation before but it does make sense why you need it.
Also gotta love the classic factorials

david.wang28

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Re: 4U Maths Question Thread
« Reply #2268 on: May 07, 2019, 05:57:57 pm »
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Hello,
I am stuck on this mechanics question in the link below. Can anyone please help me out? Thanks :)
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Re: 4U Maths Question Thread
« Reply #2269 on: May 07, 2019, 08:38:38 pm »
+1
Hello,
I am stuck on this mechanics question in the link below. Can anyone please help me out? Thanks :)
\[ \text{Your displacement bit looks fudgy.}\\ \text{I'm with you up to and including this step:}\\ \frac{1}{2k} \left[ \ln (c^2+v^2)\right]_c^0 = -x .\]
I can't see why \(c^2\) should equal 1 here - not sure where your explanation comes from. The next step here is to do what you would normally do in 2U, i.e. sub in the boundaries.
\begin{align*} \frac1{2k} \left[ \ln (c^2) - \ln (c^2+c^2) \right] &= -x\\ \frac{1}{2k} \ln \left(\frac{c^2}{c^2+c^2}\right)&=-x \tag{log laws}\\ \frac{1}{2k}\ln \frac12 &=-x\\ \therefore x &= -\frac1{2k}\ln \frac12\\ &= \frac1{2k}\ln 2. \end{align*}
\[ \text{Also I think there's a mistake with the velocity.}\\ \text{From }\int -dt = \frac1k\int \frac{dv}{c^2+v^2}\text{ you should've obtained }\boxed{-t = \frac{1}{k} \left( \frac{1}{c}\tan^{-1} \frac{v}{c}\right)+C} .\]
Note that whilst it is true that \( \int \frac{x}{a^2+x^2}dx = \frac12 \ln (a^2+x^2)+C \), when the numerator is instead a constant you have \( \int \frac{1}{a^2+x^2}\,dx = \frac1a \tan^{-1} \frac{x}a \). The first is a \( \int \frac{f^\prime(x)}{f(x)}\,dx \) pattern, but the second is a 3U standard integral from the inverse trig topic.
« Last Edit: May 07, 2019, 09:35:44 pm by RuiAce »

david.wang28

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Re: 4U Maths Question Thread
« Reply #2270 on: May 07, 2019, 08:50:10 pm »
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\[ \text{Your displacement bit looks fudgy.}\\ \text{I'm with you up to and including this step:}\\ \frac{1}{2k} \left[ \ln (c^2+v^2)\right]_c^0 = -x .\]
I can't see why \(c^2\) should equal 1 here - not sure where your explanation comes from. The next step here is to do what you would normally do in 2U, i.e. sub in the boundaries.
\begin{align*} \frac1{2k} \left[ \ln (c^2) - \ln (c^2+c^2) \right] &= -x\\ \frac{1}{2k} \ln \left(\frac{c^2}{c^2+c^2}\right)&=-x \tag{log laws}\\ \frac{1}{2k}\ln \frac12 &=-x\\ \therefore x &= -\frac1{2k}\ln \frac12\\ &= \frac1{2k}\ln 2. \end{align*}
\[ \text{Also I think there's a mistake with the velocity.}\\ \text{From }\int -dt = \frac1k\int \frac{dv}{c^2+v^2}\text{ you should've obtained }\boxed{-x = \frac{1}{k} \left( \frac{1}{c}\tan^{-1} \frac{v}{c}\right)+C} .\]
Note that whilst it is true that \( \int \frac{x}{a^2+x^2}dx = \frac12 \ln (a^2+x^2)+C \), when the numerator is instead a constant you have \( \int \frac{1}{a^2+x^2}\,dx = \frac1a \tan^{-1} \frac{x}a \). The first is a \( \int \frac{f^\prime(x)}{f(x)}\,dx \) pattern, but the second is a 3U standard integral from the inverse trig topic.
After a moment of refining my working out, I got the answer for the first part. I have trouble understanding and doing the second part, can you help me out please? Thanks :)
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Re: 4U Maths Question Thread
« Reply #2271 on: May 07, 2019, 09:39:40 pm »
+1
After a moment of refining my working out, I got the answer for the first part. I have trouble understanding and doing the second part, can you help me out please? Thanks :)
(There's a slight typo in that I wrote \(x\) instead of \(t\) for that last boxed expression. Fixed that now.)

When \(t=0\), \(v=c\), so \( C = -\frac{1}{ck}\tan^{-1}\frac{c}{c} = -\frac{1}{ck}\times\frac\pi4\). Then just sub \(v=0\). I'm not sure what else you want elaboration with?

david.wang28

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Re: 4U Maths Question Thread
« Reply #2272 on: May 07, 2019, 11:18:25 pm »
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(There's a slight typo in that I wrote \(x\) instead of \(t\) for that last boxed expression. Fixed that now.)

When \(t=0\), \(v=c\), so \( C = -\frac{1}{ck}\tan^{-1}\frac{c}{c} = -\frac{1}{ck}\times\frac\pi4\). Then just sub \(v=0\). I'm not sure what else you want elaboration with?
The part which says 'Find its velocity when it reaches a point C, having travelled for a time pi/12ck from A. Find the distance AC'. The very bottom part.
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Re: 4U Maths Question Thread
« Reply #2273 on: May 08, 2019, 08:22:30 am »
+1
The part which says 'Find its velocity when it reaches a point C, having travelled for a time pi/12ck from A. Find the distance AC'. The very bottom part.
Completely missed that bit oops.
\[ \text{After showing that}\\ \boxed{t = \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)}\\ \text{if we sub }t = \frac\pi{12ck}\text{ we have:} \]
\begin{align*}\frac\pi{12ck} &= \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)\\ \frac\pi{12} &= \frac\pi4 - \tan^{-1} \frac{v}{c}\\ \tan^{-1} \frac{v}{c} &= \frac\pi4 - \frac\pi{12}\\ &= \frac\pi6\\  \therefore \frac{v}{c} &= \frac1{\sqrt3}\\ v &= \frac{c}{\sqrt3} \end{align*}
From here, we should be able to just sub straight into \( x= -\frac1{2k} \ln (c^2+v^2) + \frac1{2k} \ln (2c^2) \). The computations look a bit nasty but I don't see any curveball in this part either.

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Re: 4U Maths Question Thread
« Reply #2274 on: May 08, 2019, 05:57:04 pm »
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Completely missed that bit oops.
\[ \text{After showing that}\\ \boxed{t = \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)}\\ \text{if we sub }t = \frac\pi{12ck}\text{ we have:} \]
\begin{align*}\frac\pi{12ck} &= \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)\\ \frac\pi{12} &= \frac\pi4 - \tan^{-1} \frac{v}{c}\\ \tan^{-1} \frac{v}{c} &= \frac\pi4 - \frac\pi{12}\\ &= \frac\pi6\\  \therefore \frac{v}{c} &= \frac1{\sqrt3}\\ v &= \frac{c}{\sqrt3} \end{align*}
From here, we should be able to just sub straight into \( x= -\frac1{2k} \ln (c^2+v^2) + \frac1{2k} \ln (2c^2) \). The computations look a bit nasty but I don't see any curveball in this part either.
Ok, thanks for the help :)
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Re: 4U Maths Question Thread
« Reply #2275 on: May 11, 2019, 10:38:37 am »
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Hello,
I am stuck on both parts of this question in the attachment below. I have done part of the question but I can't seem to integrate the function. Can anyone please help me out? Thanks :)
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Re: 4U Maths Question Thread
« Reply #2276 on: May 12, 2019, 10:01:20 am »
+1
Hello,
I am stuck on both parts of this question in the attachment below. I have done part of the question but I can't seem to integrate the function. Can anyone please help me out? Thanks :)
Sorry must've missed this. Here's the integration bit done for you:
\begin{align*} \int_U^0 \frac{v}{V+v}\,dv &= \int_U^0 \frac{V+v}{V+v} - \frac{V}{V+v}\,dv\\ &= \int_U^0 1 - \frac{V}{V+v}\,dv\\ &= \left[v - V \ln(V+v) \right]_U^0 \end{align*}
(Basically the classic 'add something and subtract the same thing' trick)

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Re: 4U Maths Question Thread
« Reply #2277 on: May 12, 2019, 09:53:37 pm »
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Sorry must've missed this. Here's the integration bit done for you:
\begin{align*} \int_U^0 \frac{v}{V+v}\,dv &= \int_U^0 \frac{V+v}{V+v} - \frac{V}{V+v}\,dv\\ &= \int_U^0 1 - \frac{V}{V+v}\,dv\\ &= \left[v - V \ln(V+v) \right]_U^0 \end{align*}
(Basically the classic 'add something and subtract the same thing' trick)
Ok, thanks :)
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Re: 4U Maths Question Thread
« Reply #2278 on: May 14, 2019, 10:51:35 am »
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Hello,
I am confused with this question in the attachment below. I know how to find the area of the triangle (stated in my working out), but I don't know why Area (OLM) is independent of the point P. Can anyone please explain this thoroughly? Thanks :)
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Re: 4U Maths Question Thread
« Reply #2279 on: May 14, 2019, 11:04:31 am »
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Hello,
I am confused with this question in the attachment below. I know how to find the area of the triangle (stated in my working out), but I don't know why Area (OLM) is independent of the point P. Can anyone please explain this thoroughly? Thanks :)
The point \(P\) depends entirely on its parameter \(t\). Since your expression for the area does not have this parameter appearing, i.e. it is independent of whatever \(t\) is, equivalently speaking it is independent of the point \(P\).

(Note that \(c\) is assumed to be a constant, not some parameter that is allowed to vary.)