Author Topic: Mathematics Question Thread (Read 1626910 times) Tweet Share

Eric11267 · « **Reply #2685 on:** October 05, 2017, 10:13:59 am »

Quote from: davidss on October 05, 2017, 09:48:13 am

hey can someone help with part iii please. thx

(Image removed from quote.)

An easier way of approaching it is first looking at the probability of getting 45 or higher. To get this you would need either (20,25), (25,20) or (25,25).
The probability of (20,25) is 2/36x1/36
The probability of (25,20) is 1/36x2/36
The probability of (25,25) is 1/36x1/36
So the probability of getting 45 or above is 5/1296
Thus the probability of getting less than 45 is 1-5/1296= 1291/1296
Edit: added working

EEEEEEP · « **Reply #2686 on:** October 05, 2017, 11:23:51 am »

Here you go =), the answers for all the q's.

bdobrin · « **Reply #2687 on:** October 05, 2017, 11:54:11 am »

Hi,

Could someone please explain how to do part two of this question for me - i am really confused.

Thanks,
Ben

Shadowxo · « **Reply #2688 on:** October 05, 2017, 01:17:46 pm »

Quote from: bdobrin on October 05, 2017, 11:54:11 am

Hi,

Could someone please explain how to do part two of this question for me - i am really confused.

Thanks,
Ben

In part i) You've found the area of the cross section, which you'll use here.
So you want to find the volume of water that passes through the cross section. It's easiest to imagine this, water is travelling at 0.4 m/s through this cross sectional area (which I'll call A). The volume passing through it in one second is simply area * velocity = A*0.4.
The amount of water passing through it in 10 seconds is volume per second * time = 0.4*A*10 = 4*A

fantasticbeasts3 · « **Reply #2689 on:** October 05, 2017, 01:24:45 pm »

i'm terrible at everything to do with logs - can someone explain why the answer's D?

Shadowxo · « **Reply #2690 on:** October 05, 2017, 01:31:35 pm »

Quote from: fantasticbeasts3 on October 05, 2017, 01:24:45 pm

i'm terrible at everything to do with logs - can someone explain why the answer's D?

All the answers have it as a single term in base 2. So, you want to convert everything to log₂. Remember log_aa=1
$4+\log_2 x=4\log_2 2+\log_2 x\\=\log_2 2^4+\log_2 x\\=\log_2 16+\log_2 x\\=\log_2 16x$
Which is D

inescelic · « **Reply #2691 on:** October 05, 2017, 02:56:34 pm »

Hey can someone please help me with this question part (iii) and (iv)?
I'm really struggling with these derivative graph questions, does anyone have some general tips for them or know where I can practice questions like these (I've really only seen questions like these in the hsc and not in any textbooks)
Much appreciated

JeffChiang · « **Reply #2692 on:** October 05, 2017, 05:47:22 pm »

This seemingly simple question has made quite a kerfuffle for part ii. The answer says top graph minus bottom graph but then that disregards the negative area below the axis. Please explain, thanks!

Thebarman · « **Reply #2693 on:** October 05, 2017, 06:06:19 pm »

I'm watching the max/min HSC revision video right now, and I'm a bit confused regarding the differentiation. Originally y' = 2H(pi)x - (3/R)H(pi)x^2, where H and R are constants. Later on in the video, it was written as y' = 2x - (3x^2)/R. Why so? Shouldn't the pi and H be left in the equation?

RuiAce · « **Reply #2694 on:** October 05, 2017, 07:28:13 pm »

Quote from: inescelic on October 05, 2017, 02:56:34 pm

Hey can someone please help me with this question part (iii) and (iv)?
I'm really struggling with these derivative graph questions, does anyone have some general tips for them or know where I can practice questions like these (I've really only seen questions like these in the hsc and not in any textbooks)
Much appreciated

Please provide the answers to the previous parts. Also, note that there are some exercises in maths in focus in regards to (iv), just not at the same difficulty.

Quote from: JeffChiang on October 05, 2017, 05:47:22 pm

This seemingly simple question has made quite a kerfuffle for part ii. The answer says top graph minus bottom graph but then that disregards the negative area below the axis. Please explain, thanks!

$\text{The formula for the area between }y_2\text{ and }y_1\\ A = \int_a^b (y_2 - y_1) \, dx\\ \text{is made, so that any area below the }x\text{-axis}\\ \text{is }\textbf{automatically compensated for}\text{ in the subtraction}$
$\text{Hence, when dealing with areas below the }x\text{-axis}\\ \text{if the question involves areas between two curves}\\ \textbf{it will not matter.}$
For now, I will not provide an extensive explanation as to why this is the case, but only a rough sketch. That rough sketch is that we know the integral (without absolute values) measures a signed area, and when we subtract something negative we get back to something positive.

RuiAce · « **Reply #2695 on:** October 05, 2017, 07:33:14 pm »

Quote from: Thebarman on October 05, 2017, 06:06:19 pm

I'm watching the max/min HSC revision video right now, and I'm a bit confused regarding the differentiation. Originally y' = 2H(pi)x - (3/R)H(pi)x^2, where H and R are constants. Later on in the video, it was written as y' = 2x - (3x^2)/R. Why so? Shouldn't the pi and H be left in the equation?

Judging by the red lines, Jamon has clearly divided both sides by $\pi$ and $H$[/tex]. Of course, $ \frac{0}{\pi H} = 0$ .

There is no reason to leave $ \pi $ and $ H $ in there if they can be divided out without harm.

Opengangs · « **Reply #2696 on:** October 05, 2017, 08:23:46 pm »

Quote from: inescelic on October 05, 2017, 02:56:34 pm

Hey can someone please help me with this question part (iii) and (iv)?
I'm really struggling with these derivative graph questions, does anyone have some general tips for them or know where I can practice questions like these (I've really only seen questions like these in the hsc and not in any textbooks)
Much appreciated

Assuming that you've done the parts before it, you'll find that the distance travelled for the first 4 seconds is 6 metres.
This is important to note, but we'll save it for a little bit later.

The next thing to note is that the area under the curve between 4 and 5 seconds is exactly the same as the area under the curve between 5 and 6 seconds. This implies it takes as much time going vertically upwards and coming back down to 6 metres; this means we need to find the time it takes to reach 6 metres vertically down.

The final thing we need to note is that the graph given is the velocity graph, or its speed with a vector quantity. It is the velocity at which the particle travels at any point in time. So, once we understand that it's the dx/dt graph, we can use it to find the time. Note that the time is given by the distance over the speed at a particular time.

Thus, the time it takes for the particle to reach 6 metres is given by: distance (6m) / speed (5ms^-1) = 1.2 seconds.
This means, it takes 1.2 seconds to reach a displacement of 6 metres, meaning that for the particle to have no displacement, it will take: 6 + 1.2 = 7.2 seconds.

Graphing the displacement graph shouldn't come at a surprise to you.
You're given that it starts at the origin; it will keep increasing until 5 seconds, where there is a maximum turning point. It will curve until about 6 seconds, where the displacement graph then drops at a constant rate until it reaches the origin again at 7.2 seconds.

inescelic · « **Reply #2697 on:** October 05, 2017, 10:30:08 pm »

Quote from: RuiAce on October 05, 2017, 07:28:13 pm

Please provide the answers to the previous parts. Also, note that there are some exercises in maths in focus in regards to (iv), just not at the same difficulty.

Sure, solution is attached- I think my my main issue is how does a velocity graph tell us when a particle returns to the origin? Is it when the area of the top half of the axis cancels the bottom half of the x axis to = 0? Does that mean if area under the curve of velocity graph is 0, the displacement is also 0?
I have seen a similar question in the 2013 hsc (attached with solutions) which would also be great if you could explain

Thank you

RuiAce · « **Reply #2698 on:** October 05, 2017, 10:43:58 pm »

Quote from: inescelic on October 05, 2017, 10:30:08 pm

Quote from: RuiAce on October 05, 2017, 07:28:13 pm
Please provide the answers to the previous parts. Also, note that there are some exercises in maths in focus in regards to (iv), just not at the same difficulty.

Sure, solution is attached- I think my my main issue is how does a velocity graph tell us when a particle returns to the origin? Is it when the area of the top half of the axis cancels the bottom half of the x axis to = 0? Does that mean if area under the curve of velocity graph is 0, the displacement is also 0?
I have seen a similar question in the 2013 hsc (attached with solutions) which would also be great if you could explain

Thank you

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
________________________________________
$\text{And now to apply this in our example.}\\ \text{We know that the particle starts at the origin.}$
$\text{We have found, via part a) for when }t=4,\\ \int_0^4 \frac{dx}{dt}\, dt \approx 6.\\ \text{Hence, after 4 seconds, the displacement of the particle is roughly }\boxed{x\approx 6}.$
A key point is that this measured the change in displacement, SPECIFICALLY between times t=0 and t=4. This is significant, as otherwise the boundaries of our integral would've been different numbers.
________________________________________
$\text{We then consider the change in displacement between times }t=4\text{ and }t=6\\ \text{So now, instead of }\int_0^4 \frac{dx}{dt}dt\text{, we're interested in }\int_4^6\frac{dx}{dt}\,dt$
$\text{Which is equal to 0. This is what falls from symmetry;}\\ \text{the exact same area being above and below the }t\text{-axis}\\ \text{So }\int_4^6 \frac{dx}{dt}\,dt = 0\text{, and hence overall there's been }\textbf{no}\text{ change in displacement}$
________________________________________
$\text{Then, combining integrals,}\\ \int_0^6 v\,dt = \int_0^4 v\,dt + \int_4^6v\,dt = 6$
$\text{What's left is to find the value of }t_1\text{ such that}\\ \int_0^{t_1} v\, dt = 0$
Reason being, this value $t_1$ will ensure that the change in displacement, from 0, to $t_1$, will be zero. Because the particle starts at the origin, the time when it returns to the origin is the time when the change in displacement is zero.
________________________________________
$\text{Breaking up the integral, we have the equation}\\ \int_0^6 v\,dt + \int_6^{t_1}v\,dt = 0\\ \text{But of course, we know that the first term is approximately 6 from above}\\ \text{So }\int_6^{t_1}v\,dt \approx -6$
$\text{But because }v = -5\textbf{ for all }t \ge 6\text{,}\\ \text{the other integral is just }\int_6^{t_1}-5\,dt\\ \text{which equals }-5(t_1 - 6)$
$\text{So we just solve the equation }-5(t_1-6) \approx-6\\ \text{and then we are done.}$

winstondarmawan · « **Reply #2699 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.

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Eric11267

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