Author Topic: Mathematics Question Thread (Read 1625940 times) Tweet Share

RuiAce · « **Reply #2700 on:** October 05, 2017, 10:57:26 pm »

Quote from: winstondarmawan on October 05, 2017, 10:53:01 pm

Hello!
I was just wondering for Series Application questions, particularly time repayments and the like, which steps of working to we ALWAYS have to include? I have lost marks for omitting steps in the past and I want to make sure it doesn't happen again.
Thanks in advance.

1. Establishing the very first case $A_1$
2. Explicitly showing a pattern beginning to form in $A_2$. Some computations should be kept there - do not assume the pattern, prove it.
3. Show how a rearrangement of the terms demonstrates the same pattern in $A_3$. (Recommended: Keep the explicit computations, but I'm not sure how markers care about this)
4. Only after you've made the pattern explicit in the first three cases, can you jump straight to the n-th case.
5. Mention the first term and common ratio of your geometric series
6. Fair game

Also step 0: Make clear what your appropriate interest rate is, based off the compounding frequency.

inescelic · « **Reply #2701 on:** October 05, 2017, 11:05:36 pm »

Quote from: RuiAce on October 05, 2017, 10:43:58 pm

Some information on tackling that 2013 question was provided in the compilation thread already. But if it's not enough, you may come back and further your question.

tagged: 2007
$\text{Remember that since the velocity is the derivative of the displacement, i.e. }v=\frac{dx}{dt}\\ \text{the velocity measures the (instantaneous) }\textit{change in displacement}\text{ with respect to time.}$
$\text{If we integrate the velocity function, we obtain information about the displacement function}\\ \text{Of course, as we have no rule here, we need to rely on an approximation}$
$\text{Hence the use of Simpson's rule from earlier.}$
________________________________________
$\text{We start by commencing an arbitrary example}\\ \text{just to illustrate the point.}\\ \text{Suppose the velocity of a function satisfies }v=f^\prime(t)$
where f(t) is any random function that we plucked out of thin air because maths is magic.
$\text{Upon integrating, }x = f(t) + C\\ \text{When }t=0,\, x = 0\text{, so upon substitution } C = -f(0)\\ \therefore \boxed{x= f(t) - f(0)}$
$\text{Now that we have an equation for }x\text{, we may find the displacement at any time}\\ \text{e.g. at time }t=2, \, x = f(2) - f(0)$
$\text{And now for why we do this}$
________________________________________
$\text{The }\textit{definite integral}\text{ is more powerful than the }\textit{indefinite integral}\\ \text{in that it helps us }\textit{kill two birds with one stone.}$
$\text{Consider my example from earlier. When }t=0\,, x = 0.\\ \text{Suppose that we want to find the displacement at any arbitrary time, say }t=T$
$\text{Our earlier calculations would suggest that the displacement is }x=f(T) - f(0)\\ \text{But if we just use a }\textbf{definite integral},\\ \text{we see that at time }t=T\\ x = \int_0^T f^\prime(t)\, dt = [f(t)]_0^T = f(T) - f(0)$
The point: We got back to the same answer, and we didn't have to do that useless finding +C stuff
________________________________________
$\text{And now to apply this in our example.}\\ \text{We know that the particle starts at the origin.}$
$\text{We have found, via part a) for when }t=4,\\ \int_0^4 \frac{dx}{dt}\, dt \approx 6.\\ \text{Hence, after 4 seconds, the displacement of the particle is roughly }\boxed{x\approx 6}.$

Ok, I'm a little confused by all that, I understand i) and ii) but how does that help us with part 3?
Also how can I find that 2013 q when this thread has 180 pages haha
Appreciate the help

RuiAce · « **Reply #2702 on:** October 05, 2017, 11:06:45 pm »

Quote from: inescelic on October 05, 2017, 11:05:36 pm

Ok, I'm a little confused by all that, I understand i) and ii) but how does that help us with part 3?
Also how can I find that 2013 q when this thread has 180 pages haha
Appreciate the help

I had added in more information.

The compilation thread is a separate thread.

pokemonlv10 · « **Reply #2703 on:** October 05, 2017, 11:42:37 pm »

Unsure how to do the 2nd part of this question? I got 3 points of intercepts for part a - (-2,0) (-1,0) and (3,0)

RuiAce · « **Reply #2704 on:** October 06, 2017, 12:01:18 am »

Quote from: pokemonlv10 on October 05, 2017, 11:42:37 pm

Unsure how to do the 2nd part of this question? I got 3 points of intercepts for part a - (-2,0) (-1,0) and (3,0)

$\text{Reading off your sketch in the previous part}\\ \text{we are interested in }A = \left|\int_{-2}^{-1}f^\prime(x)\,dx\right| + \int_{-1}^3 f^\prime(x)\,dx$
$\text{Note that we }\textbf{know}\text{ information about the primitive of }f^\prime(x),\\ \text{that is, }f(x)\text{, at these points}\\ \int_{-2}^{-1}f^\prime(x)\,dx = [f(x)]_{-2}^{-1} = f(-1) - f(-2) =-3 - (-1) = -2\\ \int_{-1}^3f^\prime(x)\,dx = [f(x)]_{-1}^{3} = f(3) - f(-1) = 5 - (-3) = 8$
Remember, you can read off the graph they actually gave you to deduce that f(-2) = -3, f(-1) = -1, f(3) = 5.

davidss · « **Reply #2705 on:** October 06, 2017, 12:07:22 pm »

Hello can someone help me with last part for this

)

Tempestuous · « **Reply #2706 on:** October 06, 2017, 12:56:35 pm »

Hey davidss,
From part ii we know that angle BDC = angle BCD (assuming you've done the first 2 parts)
Thus BC = BD (equal sides opposite equal angles in triangle BCD)
Then, AE:ED = AB:BC (a line parallel to one side of a triangle divides the other two sides in the same ratio)
Therefore AE:ED = AB:BD

jaskirat · « **Reply #2707 on:** October 06, 2017, 01:23:55 pm »

Need help with this question

RuiAce · « **Reply #2708 on:** October 06, 2017, 02:17:10 pm »

Quote from: jaskirat on October 06, 2017, 01:23:55 pm

Need help with this question

$\text{Because the area of }\triangle ACB\text{ is double that of }\triangle AST\\ \text{writing }\angle A = \theta\text{, from the area formula we have}\\ \frac{1}{2} b c \cos\theta = 2 \times \frac12 xy\cos\theta \implies \boxed{xy=\frac12bc}$
___________________________
$\text{The cosine rule gives }z^2 = x^2 + y^2 - 2xy \cos \theta\\ \text{so we just sub in the result from above}$
$z^2 = x^2 + \left(\frac{bc}{2x}\right)^2 - bc\cos\theta\\ \text{which clearly rearranges to the desired result.}$
___________________________
$\text{This is the usual max-min process that you need to get used to doing}\\ \text{Albeit, it will be quite messy.}$
$\text{We wish to minimise }z^2\text{, so we just differentiate straight away}\\ \begin{align*}\frac{d(z^2)}{dx^2} &= 2x - \frac{b^2c^2}{2x^3}\end{align*}$
$\text{Set equal to 0:}\\ \begin{align*}2x-\frac{b^2c^2}{2x^3}&=0\\ 2x^4 &= \frac{b^2c^2}{2}\\ x^4&=\frac{b^2c^2}{4}\\ x&=\pm \sqrt[4]{\frac{b^2c^2}{4}}\\ &= \pm\sqrt{\frac{bc}{2}}\end{align*}\\ \text{Discarding the nonsensical negative solution gives }\boxed{x=\sqrt{\frac{bc}2}}$
___________________________
$\text{The minimum value of }z^2\text{ is then}\\ \begin{align*}z^2_{min}&= \frac{bc}{2} + \frac{b^2c^2}{4} \times \frac{2}{bc} - bc\cos A\\ &= \frac{bc}{2}+\frac{bc}{2} - bc\cos A\\ &= bc(1-\cos A)\end{align*}$
$\text{But the length of the fence is }z\\ \text{and note that }z\text{ is minimised when }z^2\text{ is minimised}\\ \text{so }z_{min} = \sqrt{bc (1- \cos A)}$
The last part is the challenging part as it involves doing battle with algebra and manipulation
$\text{As per the cosine rule, }\cos A = \frac{b^2+c^2-a^2}{2bc}\\ \text{but the final result does not involve }a\text{, so we must use }P\text{ to eliminate it}$
$\text{Since }P=a+b+c\text{ we have }a=P-(b+c)\text{ and thus}\\ \begin{align*}\cos A &= \frac{b^2 + c^2 - (P - (b+c))^2}{2bc}\\ &= \frac{b^2+c^2 - P^2 + 2P(b+c) - (b+c)^2}{2bc}\\ &= \frac{b^2+c^2-P^2+2Pb+2Pc-b^2-2bc-c^2}{2bc}\\ &= \frac{-P^2 + 2Pb + 2Pc - 2bc}{2bc}\end{align*}$
$\text{Hence}\\ \begin{align*}1-\cos A&= \frac{2bc}{2bc} - \frac{-P^2+2Pb+2Pc-2bc}{2bc}\\ &= \frac{P^2-2Pb-2Pc+4bc}{2bc}\\ &= \frac{(P-2b)(P-2c)}{2bc}\tag{upon factorising}\end{align*}$
Substitute back in and you are done.

sidzeman · « **Reply #2709 on:** October 06, 2017, 03:18:06 pm »

Quote from: RuiAce on October 06, 2017, 02:17:10 pm

$\text{Because the area of }\triangle ACB\text{ is double that of }\triangle AST\\ \text{writing }\angle A = \theta\text{, from the area formula we have}\\ \frac{1}{2} b c \cos\theta = 2 \times \frac12 xy\cos\theta \implies \boxed{xy=\frac12bc}$
___________________________
$\text{The cosine rule gives }z^2 = x^2 + y^2 - 2xy \cos \theta\\ \text{so we just sub in the result from above}$
$z^2 = x^2 + \left(\frac{bc}{2x}\right)^2 - bc\cos\theta\\ \text{which clearly rearranges to the desired result.}$
___________________________
$\text{This is the usual max-min process that you need to get used to doing}\\ \text{Albeit, it will be quite messy.}$
$\text{We wish to minimise }z^2\text{, so we just differentiate straight away}\\ \begin{align*}\frac{d(z^2)}{dx^2} &= 2x - \frac{b^2c^2}{2x^3}\end{align*}$
$\text{Set equal to 0:}\\ \begin{align*}2x-\frac{b^2c^2}{2x^3}&=0\\ 2x^4 &= \frac{b^2c^2}{2}\\ x^4&=\frac{b^2c^2}{4}\\ x&=\pm \sqrt[4]{\frac{b^2c^2}{4}}\\ &= \pm\sqrt{\frac{bc}{2}}\end{align*}\\ \text{Discarding the nonsensical negative solution gives }\boxed{x=\sqrt{\frac{bc}2}}$
___________________________
$\text{The minimum value of }z^2\text{ is then}\\ \begin{align*}z^2_{min}&= \frac{bc}{2} + \frac{b^2c^2}{4} \times \frac{2}{bc} - bc\cos A\\ &= \frac{bc}{2}+\frac{bc}{2} - bc\cos A\\ &= bc(1-\cos A)\end{align*}$
$\text{But the length of the fence is }z\\ \text{and note that }z\text{ is minimised when }z^2\text{ is minimised}\\ \text{so }z_{min} = \sqrt{bc (1- \cos A)}$
The last part is the challenging part as it involves doing battle with algebra and manipulation
$\text{As per the cosine rule, }\cos A = \frac{b^2+c^2-a^2}{2bc}\\ \text{but the final result does not involve }a\text{, so we must use }P\text{ to eliminate it}$
$\text{Since }P=a+b+c\text{ we have }a=P-(b+c)\text{ and thus}\\ \begin{align*}\cos A &= \frac{b^2 + c^2 - (P - (b+c))^2}{2bc}\\ &= \frac{b^2+c^2 - P^2 + 2P(b+c) - (b+c)^2}{2bc}\\ &= \frac{b^2+c^2-P^2+2Pb+2Pc-b^2-2bc-c^2}{2bc}\\ &= \frac{-P^2 + 2Pb + 2Pc - 2bc}{2bc}\end{align*}$
$\text{Hence}\\ \begin{align*}1-\cos A&= \frac{2bc}{2bc} - \frac{-P^2+2Pb+2Pc-2bc}{2bc}\\ &= \frac{P^2-2Pb-2Pc+4bc}{2bc}\\ &= \frac{(P-2b)(P-2c)}{2bc}\tag{upon factorising}\end{align*}$
Substitute back in and you are done.

Sorry I think I'm being stupid but I don't understand the last part. It says to find the min length of the fence, which is ST or "z" right?. So why did you use the cosine rule to find the angle A?

davidss · « **Reply #2710 on:** October 06, 2017, 03:31:29 pm »

Quote from: Tempestuous on October 06, 2017, 12:56:35 pm

Hey davidss,
From part ii we know that angle BDC = angle BCD (assuming you've done the first 2 parts)
Thus BC = BD (equal sides opposite equal angles in triangle BCD)
Then, AE:ED = AB:BC (a line parallel to one side of a triangle divides the other two sides in the same ratio)
Therefore AE:ED = AB:BD

Thanksss!!! Didn't know about that rule :ooo

_____ · « **Reply #2711 on:** October 06, 2017, 03:52:38 pm »

What's the technique to solve this?

I got y' as a concave up parabola with vertex at x = (the point where the y'' graph touches the x axis), and y as a cubic with a point of inflexion at this point, which is correct. But how do I determine elevation (y value) at this point for y' and y?

---------

When I'm asked to graph say "y = SQRT(16-x^2)" do I assume this just means the positive square root and not the negative square root because it is a function? Do I only sketch the full circle if the question specifically states that it's a relation or a full circle?

Thanks

RuiAce · « **Reply #2712 on:** October 06, 2017, 04:19:28 pm »

Quote from: sidzeman on October 06, 2017, 03:18:06 pm

Sorry I think I'm being stupid but I don't understand the last part. It says to find the min length of the fence, which is ST or "z" right?. So why did you use the cosine rule to find the angle A?

I always was solving for z.

But z was written in terms of cos A. So I needed to get rid of it somehow

Quote from: davidss on October 06, 2017, 03:31:29 pm

Thanksss!!! Didn't know about that rule :ooo

It relies on similar triangles. Quite risky to use, even though I would happily give marks for it.

bdobrin · « **Reply #2713 on:** October 06, 2017, 04:45:58 pm »

Hi

Can someone please explain to me how the answer to this question is D.

I thought it would be 'b' because the velocity is negative meaning it moves in a negative direction and the acceleration is positive so it is picking up speed.

Thanks,
Ben

Eric11267 · « **Reply #2714 on:** October 06, 2017, 04:50:41 pm »

Quote from: bdobrin on October 06, 2017, 04:45:58 pm

Hi

Can someone please explain to me how the answer to this question is D.

I thought it would be 'b' because the velocity is negative meaning it moves in a negative direction and the acceleration is positive so it is picking up speed.

Thanks,
Ben

In this case the positive and negative signs refer to a direction, with negative being left and positive being right. So in this case it is moving to the left but is accelerating to the right. So it is moving to the left with decreasing speed.

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