May 25, 2019, 10:57:35 am

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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4050 on: April 23, 2019, 10:15:31 pm »
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In the future, please add a link to the original paper so we can see the full question in entirety.
$\text{In general, }f(x)\text{ is maximised}\\ \text{when }\frac{1}{f(x)}\text{ is }\textbf{minimised.}$
$\text{Here, }\frac{v\ell}{v^2+x^2}\text{ is maximised}\\ \text{when }\frac{v^2+x^2}{v\ell}\text{ is minimised.}$
$\text{Which is of course when }v^2+x^2\text{ is minimised}\\ \text{which we know is at }x=0\text{, because that new expression is just an easy-to-handle quadratic.}$

Thank you!! Sorry, I'll include a reference next time.

#### georgebanis

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##### Re: 3U Maths Question Thread
« Reply #4051 on: April 24, 2019, 01:25:32 pm »
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Hey guys,

Just need a hand with the working for this 3U applications of calculus question (attached). I have found that 0.5v^2 = k/x +c but am not sure if this is necessary for the rest of the question.

Thanks

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4052 on: April 24, 2019, 01:45:19 pm »
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Hey guys,

Just need a hand with the working for this 3U applications of calculus question (attached). I have found that 0.5v^2 = k/x +c but am not sure if this is necessary for the rest of the question.

Thanks
Your statement $\frac{v^2}{2} = \frac{k}{x}+C$ is correct and definitely necessary. Now the question tells you that when $t=0$, $x=6400$ and $v=0$ which you can use to find your constant of integration...

(Note: I checked the answers. They do a bit of extra simplifying as well that you weren't asked to do.)

#### david.wang28

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##### Re: 3U Maths Question Thread
« Reply #4053 on: April 24, 2019, 09:45:03 pm »
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Hello,
I am stuck on Q10 in the attachments below (my working out is shown, I've gone into a dead end). Can anyone please help me out? Thanks
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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4054 on: April 24, 2019, 10:38:06 pm »
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Hello,
I am stuck on Q10 in the attachments below (my working out is shown, I've gone into a dead end). Can anyone please help me out? Thanks
Here's a hint. Don't worry about the gradient until the very end.
$\text{You're trying to minimise the }\textbf{distance}\\ \text{between a variable point }P\left(t, \frac{t^2}{2}\right)\text{ on the parabola.}\\ \text{and the fixed point }A(4,1)$
$\text{Focus on your use of the distance formula:}\\ AP^2 = (4-t)^2 + \left( 1 - \frac{t^2}{2}\right)^2.\\ \text{You now want to use calculus to }\textbf{minimise}\text{ this expression for the distance-squared.}$
i.e. compute $\frac{d(AP^2)}{dt}$, set it to 0, find the stationary point and then do the usual boring stuff. Alternatively you can just expand the quadratic and read off that if you prefer.

Once you find what that $t$ is, then worry about the gradient stuff.

(Note: Minimising $AP^2$ will minimise $AP$ as well.)

#### annabeljxde

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##### Re: 3U Maths Question Thread
« Reply #4055 on: April 24, 2019, 11:29:12 pm »
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Can someone please explain in detail how you would go about with answering this question...I don't understand the answers in the back of the excel book.

Thank you so much!
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#### david.wang28

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##### Re: 3U Maths Question Thread
« Reply #4056 on: April 24, 2019, 11:43:48 pm »
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Ahhh, now I know; I worked it out. Thanks as always
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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4057 on: April 25, 2019, 12:22:49 am »
+1
Can someone please explain in detail how you would go about with answering this question...I don't understand the answers in the back of the excel book.

Thank you so much!

The hard part in this one is really just that 1-marker.

Note that there is still the boring way of literally listing out every outcome (K,M), where K = what Kim spins and M = what Mel spins. Then just read off your table of 25 elements to see that 10 of them ensure Kim wins (and hence the probability is $\frac25$). In an exam, when in a panic, that would be a go-to option.
$\text{One way of going about it more cleverly is to use a concept}\\ \text{sometimes referred to as }\textbf{symmetric probabilities}.$
\text{There are 3 possible options for Kim, namely win, draw and lose.}\\ \text{Now note that for a draw}\\ \begin{align*}P(\text{Draw}) &= P(11)+ P(22)+P(33)+P(44)+ P(55) \\&= \frac1{25}+\frac1{25}+\frac1{25}+\frac1{25}+\frac1{25}\\ &= \frac15.\end{align*}
Alternatively, we note that the number of ways we can have a draw is $5$ (because it's easy to see that a draw must occur when they both roll 1, 2, 3, 4 or 5). Then the probability of a particular outcome that results in a draw, say, 1 and 1, is $P(11) = \frac15 \times \frac15 = \frac1{25}$. So therefore the probability of a draw is $P(\text{Draw})=5\times \frac1{25} = \frac15$.
$\text{The symmetry now comes into play.}\\ \text{Observe that for the remaining configurations}\\ \text{there is a }\textbf{one-to-one}\text{ correspondence between Kim winning, and Kim losing.}$
$\text{The one-to-one matching occurs because if we start with some outcome}\\ \text{where Kim wins, say Kim rolls 3 and Mel rolls 1,}\\ \text{we can immediately find an option where Kim loses by flipping their roles}\\ \text{namely Kim rolls 1 and Mel rolls 3 instead.}$
$\text{Because we can do this for }\textbf{every}\text{ option Kim wins to one where she loses}\\ \text{we deduce that the number of different ways Kim can win}\\ \text{must }\textbf{equal}\text{ that of her losing.}$
$\text{The consequence: }\boxed{P(\text{Win}) = P(\text{Lose})}.$
Once we have this, we just note that since win, draw and lose are the only 3 outcomes, $P(W) + P(D) + P(L) = 1$.

Therefore $P(W) + \frac15 + P(W) = 1$, so $P(W) = \frac25$.

The second part is now just a straightforward binomial probability question. If you haven't covered that in class you should come back to it later, but otherwise just sub into the formula.

#### jamonwindeyer

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##### Re: 3U Maths Question Thread
« Reply #4058 on: April 28, 2019, 10:22:48 am »
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Help me plz

#### mirakhiralla

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##### Re: 3U Maths Question Thread
« Reply #4059 on: May 03, 2019, 10:42:31 pm »
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Water is being poured into an inverted right conical vessel whose apex angle is 90 degrees at a constant rate of 3cm^3/s. At what rate is the water level rising when the depth is pi cm?

#### jamonwindeyer

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##### Re: 3U Maths Question Thread
« Reply #4060 on: May 03, 2019, 11:28:28 pm »
+1
Water is being poured into an inverted right conical vessel whose apex angle is 90 degrees at a constant rate of 3cm^3/s. At what rate is the water level rising when the depth is pi cm?

Hey! So the chain rule application we need here is:

$\frac{dh}{dt}=\frac{dh}{dV}\times\frac{dV}{dt}$

We're given $\frac{dV}{dt}$ in the question! We just need to find the volume of the cone in terms of its height, and differentiate it to find that first derivative. Now, the volume of a cone is $V=\frac{1}{3}\pi r^2h$. If you do some SOHCAHTOA on the right triangle in the cone, given that the apex angle is 90 degrees, you get:

$\tan{45}=\frac{r}{h}\\\therefore r=h$

So the volume can be found and differentiated in terms of h:

$V=\frac{1}{3}\pi h^3\\\frac{dV}{dh}=\pi h^2\\\frac{dh}{dV}=\frac{1}{\pi h^2}$

So then just sub $h=\pi$ into that and put it all together using the rule above!!

#### saloni.aphale

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##### Re: 3U Maths Question Thread
« Reply #4061 on: May 05, 2019, 09:32:06 am »
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hi,

Does anyone know how to prove:

SIn^-1(x)+tan^-1(x)=pi/2

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4062 on: May 05, 2019, 09:53:46 am »
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hi,

Does anyone know how to prove:

SIn^-1(x)+tan^-1(x)=pi/2
This identity is not true. The correct identity involves cos^-1(x) instead of tan^-1(x). Is there a typo somewhere?

#### georgebanis

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##### Re: 3U Maths Question Thread
« Reply #4063 on: May 05, 2019, 11:53:04 am »
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Hey guys,

I just need a hand with parts C and D for the question attached below (with answers attachment). For part C can you explain to me if there is an equation we should arrive to or just do all by the calculator? (I let 5t = 20ln(t+1)). And for Part D I'm just not sure how to do it.

Thanks