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September 20, 2019, 04:44:58 am

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#### pmcnair

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##### Re: 4U Maths Question Thread
« Reply #2265 on: April 29, 2019, 05:11:22 pm »
0
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 »
0
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 »
0
Hello,
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#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2269 on: May 07, 2019, 08:38:38 pm »
+1
Hello,
$\text{Your displacement bit looks fudgy.}\\ \text{I'm with you up to and including this step:}\\ \frac{1}{2k} \left[ \ln (c^2+v^2)\right]_c^0 = -x .$
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 »
0
$\text{Your displacement bit looks fudgy.}\\ \text{I'm with you up to and including this step:}\\ \frac{1}{2k} \left[ \ln (c^2+v^2)\right]_c^0 = -x .$
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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#### RuiAce

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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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#### RuiAce

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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.

#### david.wang28

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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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#### david.wang28

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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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#### RuiAce

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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)

#### david.wang28

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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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#### david.wang28

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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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#### RuiAce

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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.)