Author Topic: Mathematics Question Thread (Read 1626461 times) Tweet Share

sidzeman · « **Reply #2865 on:** October 20, 2017, 02:27:35 pm »

For part 1 of this, my solution for k was - ln(1/2) / 1600. Would that still be correct?

Fahim486 · « **Reply #2866 on:** October 20, 2017, 03:05:11 pm »

Hey so I thought the answer to this was A but it is in fact B. I have a feeling it is something to do with the All Stations To Central rule but I can't seem to understand it so could someone pls explain why it is B. Thanks!

julia_warren13 · « **Reply #2867 on:** October 20, 2017, 03:15:05 pm »

Quote from: Fahim486 on October 20, 2017, 03:05:11 pm

Hey so I thought the answer to this was A but it is in fact B. I have a feeling it is something to do with the All Stations To Central rule but I can't seem to understand it so could someone pls explain why it is B. Thanks!

When using the m = tan x rule, x is the angle the line makes with the POSITIVE X AXIS

m = tan (180 - 60)
m = tan 120
m = - srt 3 = B

Mymy409 · « **Reply #2868 on:** October 20, 2017, 04:01:22 pm »

Little help with this qn, please. I don't understand the solution. TIA.

kemi · « **Reply #2869 on:** October 20, 2017, 04:15:55 pm »

Quote from: sidzeman on October 20, 2017, 02:27:35 pm

For part 1 of this, my solution for k was - ln(1/2) / 1600. Would that still be correct?

Yes

Remember the log rule, log (M^k) = klog (M)

So if you raise (1/2) to the power (-1), you will get k = (ln2)/1600

Shadowxo · « **Reply #2870 on:** October 20, 2017, 04:22:11 pm »

Quote from: Mymy409 on October 20, 2017, 04:01:22 pm

Little help with this qn, please. I don't understand the solution. TIA.

It's rotated about the y axis so the volume is
$\pi \int_{y=0}^{y=2}x^2dy\\y=\log_3 x, \text{ so }x=3^y\\x^2=(3^y)^2=(3^2)^y=9^y\\e^{\log_e (anything)}=anything\text{ so} \\x^2=9^y=e^{log_e 9^y}=e^{y*log_e 9}\\\text{Sub this in and you'll get the answer for i). ii) is just finding the integral of }e^{y*constant} $
If you're still confused, let me know where

Also the reason why they did the whole "e to a power" thing was so it's easier to find the derivative.
Hope this helps

Quote from: Fahim486 on October 20, 2017, 03:05:11 pm

Hey so I thought the answer to this was A but it is in fact B. I have a feeling it is something to do with the All Stations To Central rule but I can't seem to understand it so could someone pls explain why it is B. Thanks!

You should also be able to tell since the line is sloped down, the gradient will be negative. That's one of the reasons you can eliminate A

sophiegmaher · « **Reply #2871 on:** October 20, 2017, 04:37:27 pm »

I don't know how to do question (iii) from the attached question, it's from the CSSA trial

sophiegmaher · « **Reply #2872 on:** October 20, 2017, 04:56:58 pm »

Also not sure how to answer part (ii) of this attached question from the CSSA trial

Mymy409 · « **Reply #2873 on:** October 20, 2017, 05:17:37 pm »

Quote from: Shadowxo on October 20, 2017, 04:22:11 pm

It's rotated about the y axis so the volume is
$\pi \int_{y=0}^{y=2}x^2dy\\y=\log_3 x, \text{ so }x=3^y\\x^2=(3^y)^2=(3^2)^y=9^y\\e^{\log_e (anything)}=anything\text{ so} \\x^2=9^y=e^{log_e 9^y}=e^{y*log_e 9}\\\text{Sub this in and you'll get the answer for i). ii) is just finding the integral of }e^{y*constant} $
If you're still confused, let me know where Also the reason why they did the whole "e to a power" thing was so it's easier to find the derivative.
Hope this helps

Thanks, I'm just confused about how we get x=3^y

RuiAce · « **Reply #2874 on:** October 20, 2017, 05:19:29 pm »

Deleted previous bit - misread.

Quote from: Mymy409 on October 20, 2017, 05:17:37 pm

Thanks, I'm just confused about how we get x=3^y

$\text{This is the }\textbf{definition}\text{ of the logarithm.}$
$y = \log_e x\textbf{ means the same thing as }x = e^y\textbf{ by definition}.$
$\text{Similarly, }\textbf{by definition}\\ y=\log_3 x \text{ and }x = 3^y\text{ mean the }\textbf{same thing}$

Mymy409 · « **Reply #2875 on:** October 20, 2017, 05:19:50 pm »

Also, how do we solve sin2x=cosx?

RuiAce · « **Reply #2876 on:** October 20, 2017, 05:21:14 pm »

Quote from: Mymy409 on October 20, 2017, 05:19:50 pm

Also, how do we solve sin2x=cosx?

You cannot do this via 2U methods. The best you can do is sketch both of them and approximate a solution.

RuiAce · « **Reply #2877 on:** October 20, 2017, 05:23:16 pm »

Quote from: caitlinlddouglas on October 20, 2017, 02:09:31 pm

Hi thanks for that I can't see the discussion for iii) though?

When I click the hyperlink, it links to a huge amount of text related to part iii).

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
________________________________________
$\text{And now to apply this in our example.}\\ \text{We know that the particle starts at the origin.}$
$\text{We have found, via part a) for when }t=4,\\ \int_0^4 \frac{dx}{dt}\, dt \approx 6.\\ \text{Hence, after 4 seconds, the displacement of the particle is roughly }\boxed{x\approx 6}.$
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.
________________________________________
$\text{Breaking up the integral, we have the equation}\\ \int_0^6 v\,dt + \int_6^{t_1}v\,dt = 0\\ \text{But of course, we know that the first term is approximately 6 from above}\\ \text{So }\int_6^{t_1}v\,dt \approx -6$
$\text{But because }v = -5\textbf{ for all }t \ge 6\text{,}\\ \text{the other integral is just }\int_6^{t_1}-5\,dt\\ \text{which equals }-5(t_1 - 6)$
$\text{So we just solve the equation }-5(t_1-6) \approx-6\\ \text{and then we are done.}$

RuiAce · « **Reply #2878 on:** October 20, 2017, 05:32:59 pm »

Quote from: sophiegmaher on October 20, 2017, 04:37:27 pm

I don't know how to do question (iii) from the attached question, it's from the CSSA trial

$\text{Seeing as though }v=6 + \frac{18t}{3t^2+1}\\ \text{The limiting velocity will be }\lim_{t \to \infty}\left(6+ \frac{18t}{3t^2+1}\right) = 6 + 0 = 6$
I strongly believe that this question was unfair and I'm actually appalled that they put that in a 2U paper. Please be advised the limit they made you evaluate was actually a 3U limit.

RuiAce · « **Reply #2879 on:** October 20, 2017, 05:42:35 pm »

Quote from: sophiegmaher on October 20, 2017, 04:56:58 pm

Also not sure how to answer part (ii) of this attached question from the CSSA trial

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