Author Topic: Mathematics Question Thread (Read 1313178 times) Tweet Share

hansolo9 · « **Reply #2805 on:** October 18, 2017, 07:35:41 pm »

Hey

Need help with the last part because I got 11 whereas the 2016 guidelines said 12

Mymy409 · « **Reply #2806 on:** October 18, 2017, 08:46:13 pm »

Can someone help me with part iv)

(c) There’s been a cane beetle infestation at Mortdale! To combat this 5 cane toads
are introduced to start eating the cane beetles. Consequently, the number of
cane beetles decreases according to

B = 5000-10e^(0.01t)

where t is the time in months after the toads are introduced.
The population of cane toads, C, increases according to

dC/dt = 0.01C.

(i) How many beetles were in the town when the toads were introduced. 1
(ii) When will the population of beetles be one tenth of the population at t = 0. 1
(iii) Find an expression for the number of toads C in terms of t. 1
(iv) When is the rate of increase of toads equal to the rate of decrease of beetles? 3

RuiAce · « **Reply #2807 on:** October 18, 2017, 08:58:13 pm »

Quote from: sidzeman on October 18, 2017, 03:33:09 pm

Hey could someone help me with the final question from the 2002 hsc?
I'm not sure how to do part 1 and 3
For part ii, i got the Di/Dx as positive on both sides of x = 0, so I'm not sure what I'm doing wrong.

$\text{We use the chain rule, and then immediately combine the fractions}\\\begin{align*}l&=\frac{1}{b^2+(x+8)^2}+\frac{1}{b^2+(x-8)^2}\\ \frac{dl}{dx}&=-\frac{2(x+8)}{\left(b^2+(x+8)^2\right)}-\frac{2(x+8)}{\left(b^2+(x+8)^2\right)}\\ &= -2\left[\frac{(x+8)\left(b^2+(x-8)^2\right)+(x-8)\left(b^2+(x+8)^2\right)}{\left(b^2+(x+8)^2\right)\left(b^2+(x-8)^2\right)}\right]\\ &= -2\frac{P}{Q}.\end{align*}$
_____________________________
$\text{The }y\text{-axis is representative of the distance from the coast}\\ \text{In Saga's case, this will be when }\boxed{b=15}.$
$\text{So for Saga, using the given factorisation,}\\ \frac{dI}{dx} = -\frac{2x\left(x^2+64+b^2+16\sqrt{64+b^2}\right) \left(x^2+64+b^2-16\sqrt{64+b^2}\right)}{\left(b^2+(x+8)^2\right)\left(b^2+(x-8)^2\right)}$
$\therefore \boxed{\frac{dI}{dx} = -\frac{2x\left(x^2+561\right)\left(x^2+17\right)}{\left(225+(x+8)^2\right)\left(225+(x-8)^2\right)}}$
$\text{Just like with every maxima/minima problem, we set }\frac{dI}{dx} = 0\\ \therefore \frac{2x\left(x^2+561\right)\left(x^2+17\right)}{\left(225+(x+8)^2\right)\left(225+(x-8)^2\right)}=0\\ \text{The only solution to this will be }x=0.$
$\text{Differentiating again is a }\textbf{horrible}\text{ idea, so we will test both sides}\\ \text{When }x=-0.01, \frac{dI}{dx} = +0.00228\dots > 0\\ \text{When }x=+0.01, \frac{dI}{dx} = -0.0022837\dots < 0\\ \text{So indeed we have a maximum}$
$\text{Hence the brightness does indeed increase until 0, and then decreases from there}.$
_____________________________
$\text{In Hero's case, we now have }\boxed{b = 6}.\\ \text{So for Hero, }\boxed{\frac{dI}{dx} = -\frac{2x\left(x^2+260\right)\left(x^2-60\right)}{\left(225+(x+8)^2\right)\left(225+(x-8)^2\right)}}$
$\text{Once again, we set }\frac{dI}{dx} = 0\\ \text{This time, we will have }\textbf{three}\text{ solutions, namely }\boxed{x=-\sqrt{60}, 0, +\sqrt{60}}.$
$\text{To identify the nature of these stationary points, we test a bit to each side as well.}$
Note from the calculator, $\sqrt{60}\approx 7.74 $.
$\text{When }x = -9, \frac{dI}{dx} = 0.487\dots > 0\\ \text{When }x=-4, \frac{dI}{dx} = -0.713\dots < 0\\ \text{So the brightness increases until Hero is }\sqrt{60}\text{ kilometres away, and then decreases.}$
$\text{When }x=-4, \frac{dI}{dx} = -0.713\dots < 0\\ \text{When }x=4, \frac{dI}{dx} = 1.672\dots > 0\\ \text{So the brightness keeps decreasing until Hero is at }x=0\text{, at which point it starts increasing again}$
$\text{When }x=4, \frac{dI}{dx} = 1.672\dots > 0\\ \text{When }x=9, \frac{dI}{dx} = -2.523\dots < 0\\ \text{So the brightness continues to increase until }x=\sqrt{60}\text{, and from then on it keeps decreasing.}$
tagged: 2002
(Will most likely be added to the compilation at some point.)

RuiAce · « **Reply #2808 on:** October 18, 2017, 09:04:40 pm »

Quote from: Mary_a on October 18, 2017, 05:47:28 pm

Okay, so I thought averaging 62 in all the past papers I did was bad, but I just sat the 2016 paper and I got 42.

No matter how much work I'm doing, it's not paying off!

Where do I go from here?

Following from your previous post, what have you done? What are the causes for the lost marks and how have you been trying to improve

Quote from: jaskirat on October 18, 2017, 07:34:10 pm

need help with this question

$\begin{align*}y&=ax^3+bx^2+cx+7\\ \frac{dy}{dx}&= 3ax^2+2bx+c\\ \frac{d^2y}{dx^2}&=6ax+2b\end{align*}$
$\text{Since S.P.s occur when }\frac{dy}{dx} = 0\text{ and P.O.I.s occur when }\frac{d^2y}{dx^2}=0\\ \text{we have the simultaneous equations}\\ \begin{align*}0&=48a+8b+c\\ 0&=12a-4b+c\\ 0&=6a+2b\end{align*}$
Remark: We also have a fourth equation $ -19 = a + b + c + 7 $ from just substituting the point in. This becomes $ -26 = a + b + c$, and can be convenient to use.
$\text{These now form a system of simultaneous equations which you should be able to solve.}\\ \text{If you do not arrive at the given answer, please post your working.}$

allmyfranzaredead · « **Reply #2809 on:** October 18, 2017, 09:21:20 pm »

Hi, this is probs easy as,but I am awful at trig. In the 2014 multiple choice there was this, and it stumped me.

RuiAce · « **Reply #2810 on:** October 18, 2017, 09:25:37 pm »

Quote from: allmyfranzaredead on October 18, 2017, 09:21:20 pm

Hi, this is probs easy as,but I am awful at trig. In the 2014 multiple choice there was this, and it stumped me.

The compilation briefly mentions why the answer is not 3, but rather 2. If you need any help beyond that, please provide more details.

RuiAce · « **Reply #2811 on:** October 18, 2017, 09:28:34 pm »

Quote from: hansolo9 on October 18, 2017, 07:35:41 pm

Hey

Need help with the last part because I got 11 whereas the 2016 guidelines said 12

$\text{Be careful to make sure you did not misinterpret the question.}\\ \text{Recall from the previous part that the probability}\\ \text{it ends before the }\textbf{4th}\text{ roll was}\\ \frac18 + \frac18\left(\frac78\right) + \frac18\left(\frac78\right)^{\textbf{2}}\\ \text{Emphasis on how the powers only go up to }\textbf{2},\text{ not }\textbf{3}$
$\text{This means, in general, if the game were to end before the }n\text{-th roll}\\ \text{we would have}\\ \frac18 + \frac18\left(\frac78\right) + \frac18\left(\frac78\right)^{2}+\dots + \frac18 \left(\frac{7}{8}\right)^{\bf{n-2}}$
$\text{Since it only goes up to }n-2\text{, the sum of the series is actually}\\ \frac{1}{8}\left(\frac{1-\left(\frac78\right)^{n-1}}{1-\frac78}\right)$
If you solve this the same way you tried solving it originally, you will now get n=12. That is to say, solve the equation
$\frac{1}{8}\left(\frac{1-\left(\frac78\right)^{n-1}}{1-\frac78}\right) < \frac34$
with that n-1, instead of n.

RuiAce · « **Reply #2812 on:** October 18, 2017, 09:41:13 pm »

Quote from: Mymy409 on October 18, 2017, 08:46:13 pm

Can someone help me with part iv)

(c) There’s been a cane beetle infestation at Mortdale! To combat this 5 cane toads
are introduced to start eating the cane beetles. Consequently, the number of
cane beetles decreases according to

B = 5000-10e^(0.01t)

where t is the time in months after the toads are introduced.
The population of cane toads, C, increases according to

dC/dt = 0.01C.

(i) How many beetles were in the town when the toads were introduced. 1
(ii) When will the population of beetles be one tenth of the population at t = 0. 1
(iii) Find an expression for the number of toads C in terms of t. 1
(iv) When is the rate of increase of toads equal to the rate of decrease of beetles? 3

$\text{Literally sub in }t=0\text{ to get the initial quantity }B=4990$
____________________________________
$\text{So we require when }B=499\\ \begin{align*}499&=5000-10e^{0.01t}\\ 10e^{0.01t}&=4501\\ e^{0.01t}&=450.1\\ 0.01t&=\ln 450.1\\ t&=\frac{\ln 450.1}{0.01}\end{align*}$
____________________________________
$\text{If }\frac{dC}{dt}=0.01C\text{ then quoting the exponential growth formula}\\ \text{we have }C=C_0e^{0.01t}$
$\text{Initially there are 5 cane toads, i.e. when }t=0, C=5\\ \text{So }C_0 = 5\text{ and thus }C=5e^{0.01t}$
All of these three parts are fairly standard and you need to get used to doing them.
____________________________________
$\frac{dC}{dt} = 0.05e^{0.01t}\text{ and }\frac{dB}{dt} = -0.1e^{0.01t}$
$\text{So since we need rate of decrease of beetles to equal}\\ \text{the rate of increase of toads, we must have }\frac{dC}{dt} = -\frac{dB}{dt}\\ \text{i.e. }0.05e^{0.01t} = 0.1e^{0.01t}$
$\text{But this has no solution, so they are never equal}$
Which makes me suspect that you made a typo somewhere.

Mymy409 · « **Reply #2813 on:** October 18, 2017, 09:53:11 pm »

Quote from: RuiAce on October 18, 2017, 09:41:13 pm

$\text{Literally sub in }t=0\text{ to get the initial quantity }B=4990$
____________________________________
$\text{So we require when }B=499\\ \begin{align*}499&=5000-10e^{0.01t}\\ 10e^{0.01t}&=4501\\ e^{0.01t}&=450.1\\ 0.01t&=\ln 450.1\\ t&=\frac{\ln 450.1}{0.01}\end{align*}$
____________________________________
$\text{If }\frac{dC}{dt}=0.01C\text{ then quoting the exponential growth formula}\\ \text{we have }C=C_0e^{0.01t}$
$\text{Initially there are 5 cane toads, i.e. when }t=0, C=5\\ \text{So }C_0 = 5\text{ and thus }C=5e^{0.01t}$
All of these three parts are fairly standard and you need to get used to doing them.
____________________________________
$\frac{dC}{dt} = 0.05e^{0.01t}\text{ and }\frac{dB}{dt} = -0.1e^{0.01t}$
$\text{So since we need rate of decrease of beetles to equal}\\ \text{the rate of increase of toads, we must have }\frac{dC}{dt} = -\frac{dB}{dt}\\ \text{i.e. }0.05e^{0.01t} = 0.1e^{0.01t}$
$\text{But this has no solution, so they are never equal}$
Which makes me believe that you made a typo somewhere.

I only needed help with part iv)

Nope, that's how the question is exactly. Error in the question maybe?

RuiAce · « **Reply #2814 on:** October 18, 2017, 10:03:31 pm »

Quote from: Mymy409 on October 18, 2017, 09:53:11 pm

I only needed help with part iv)
Nope, that's how the question is exactly. Error in the question maybe?

Lol, sorry. Probably went blind after already typing a lot of others before I got around to your's.

But in that case, the correct answer is definitely "never".

caitlinlddouglas · « **Reply #2815 on:** October 19, 2017, 07:06:35 am »

Hey i a was wondering why for this question you couldn't just do the top curve minus the bottom curve? Was there a reason they had to split it at y=4 and then add them? Thanks!

RuiAce · « **Reply #2816 on:** October 19, 2017, 07:14:08 am »

Quote from: caitlinlddouglas on October 19, 2017, 07:06:35 am

Hey i a was wondering why for this question you couldn't just do the top curve minus the bottom curve? Was there a reason they had to split it at y=4 and then add them? Thanks!

Because you are rotating it about the y-axis.

If you were rotating about the x-axis, then you would have an area between two curves, or rather volume between two curves problem. Since you're rotating about the y-axis, the analogy is different.

gilliesb18 · « **Reply #2817 on:** October 19, 2017, 09:17:59 am »

Quote from: Jess1113 on October 18, 2017, 02:55:18 pm

Hi!

- Draw a diagram to help visualise, using the condition given in the question (If it is always 4 units away from the line, that must mean that the locus is parallel to the equation: the perpendicular distance between the equation and the locus is 4)
- Substitute the values of the equation and an arbitrary point P(x,y) into the perpendicular distance formula (there should be two equations for the locus: one above and one below 3x+4y+5=0)

I have attached my working below, hope this helps

Wow! Thanks so much... Very helpful.

gilliesb18 · « **Reply #2818 on:** October 19, 2017, 09:41:03 am »

Hello again,
I just need help on the following question...
Find the equation of the locus of a point that moves so that it is equidistant from the line 4x - 3y + 2 =0 and the line 3x + 4y - 7 =0
Thanks once again!!

RuiAce · « **Reply #2819 on:** October 19, 2017, 09:48:10 am »

Quote from: gilliesb18 on October 19, 2017, 09:41:03 am

Hello again,
I just need help on the following question...
Find the equation of the locus of a point that moves so that it is equidistant from the line 4x - 3y + 2 =0 and the line 3x + 4y - 7 =0
Thanks once again!!

$\text{From the perpendicular distance formula, we require}\\ \text{the point }P(x,y)\text{ to satisfy}\\ \begin{align*}\frac{|4x-3y+2|}{\sqrt{4^2+3^2}} &= \frac{|3x+4y-7|}{\sqrt{3^2+4^2}}\\ |4x-3y+2|&=|3x+4y-7|\end{align*}$
$\text{This gives the two cases}\\ 4x-3y+2 = 3x+4y-7 \implies \boxed{x-7y+9=0}\\ 4x-3y+2 = -(3x+4y-7)\implies \boxed{7x+y-5=0}$
______________________________
$\text{Alternatively, rearrange the equations of the lines to be}\\ y = \frac{1}{3}(2+4x)\text{ and }y=\frac{1}{4} (7-3x)$
$\text{Using the }PA=PB\text{ method}\\ \text{we still have }P\text{ is the point }(x,y)\\ \text{but }A\text{ and }B\text{ are now }\left(x, \frac13(2+4x)\right)\text{ and }\left(x, \frac14 (7-3x)\right)$
\begin{align*}PA^2&=PB^2\\ (x-x)^2+\left(y - \frac{1}{3}(2+4x)\right)^2&= (x-x)^2 + \left(y-\frac14(7-3x)\right)^2\end{align*}
\[ \text{These will give the same equations as above.} \]

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