Author Topic: Mathematics Question Thread (Read 1303367 times) Tweet Share

RuiAce · « **Reply #240 on:** July 24, 2016, 03:43:21 pm »

Quote from: Jakeybaby on July 24, 2016, 03:20:30 pm

$\text{Whilst having }v(t)\text{ instead of just }v\text{ being a feasible notation, for displacement}\\ \text{it is generally tagged as }x(t)$
That aside, here's the add-on.

Quote from: dtinaa on July 24, 2016, 02:50:53 pm

Is anyone else struggling with Applications of Calculus?? Does anyone have any tips on how to do those describe the motion of a particle questions??

$\text{It is easy to determine whether we have a maximum velocity or minimum velocity by testing the nature of the stationary points.}\\ \text{I think this is what the allusion was with 'sign tests'}$
$\text{Also keep in mind that we are talking about the displacement, velocity and acceleration here.}\\ \text{The 'speed' the particle travels at is the magnitude of the velocity.}\\ \text{Keep in mind that speed is therefore independent of direction.}$
$\text{If a particle turns around, its displacement will go the other way but it's }\textit{distance}\text{ keeps increasing.}\\ \text{If you have to calculate the distance, you need to be careful of where the particle turns around.}\\ \text{If the particle is travelling to the left, you may need absolute values.}\\ \text{(An alternate way to find distance travelled is to just find the area under the velocity curve}\\ \text{taking absolute values for negative areas.)}$
$\textit{Often, there will be initial conditions when you are required to integrate.}\\ \textit{Take the time to actually understand what is going on. It helps you find the constant of integration.}$
$\textbf{If you're still stuck, please send example questions.}$

Jakeybaby · « **Reply #241 on:** July 24, 2016, 04:13:56 pm »

Quote from: RuiAce on July 24, 2016, 03:43:21 pm

$\text{Whilst having }v(t)\text{ instead of just }v\text{ being a feasible notation, for displacement}\\ \text{it is generally tagged as }x(t)$

Oh alright, just different notation between the states

RuiAce · « **Reply #242 on:** July 24, 2016, 06:27:07 pm »

Quote from: Jakeybaby on July 24, 2016, 04:13:56 pm

Oh alright, just different notation between the states

Yep. All good here; I'm not entirely why sure why x, r and s get juggled to represent displacement. (Incidentally the latter two are found in the HSC physics courses.)

But yeah, at the 2U level the information you provided formed the bulk of what is necessary.

The one other thing to keep in mind (regarding notation) though is how v as a function of t really gets defined. In Extension 1, because they also do velocity as a function of displacement (so v(x)), we might have problems if we just say v(t)=2t so v(x)=2x which, depending on the situation can easily be mistaken.
It's probably why Leibniz's notation (or just Newton's notation - the dot) is more favourable. (Of course, in 2U this is hardly problematic in my opinion)

isaacdelatorre · « **Reply #243 on:** July 24, 2016, 07:18:55 pm »

Hey,
This question might be super simple, but I've been having some trouble with this question, mainly part ii.
Could someone please explain how to do it?

RuiAce · « **Reply #244 on:** July 24, 2016, 07:49:18 pm »

Quote from: isaacdelatorre on July 24, 2016, 07:18:55 pm

Hey,
This question might be super simple, but I've been having some trouble with this question, mainly part ii.
Could someone please explain how to do it?

$\text{The point }A\text{ lies on the tangent for }x=0\\ y=(-t-b)(t-b)=b^2-t^2\\ \text{The point }B\text{ lies on the tangent for }y=0\\ 0=(2x-t-b)(t-b)\iff 2x-t-b=0 \iff x=\frac{t+b}{2}\\ \text{Keeping in mind that }t\neq b\text{ so we can safely cancel out the }t-b$
$\text{Now, the area of }\triangle ABO\text{ using }A=\frac{1}{2}bh\text{ is:}\\ \begin{align*}A&=\frac{1}{2}\times OA \times OB\\ &= \frac{1}{2}\left((b^2-t^2)-0\right)\left(\frac{t+b}{2}-0\right)\\ &= \frac{(b^2-t^2)(b+t)}{4}\\ &=\frac{1}{4}\left(b^3+b^2t-bt^2-t^3\right)\\ \therefore \frac{dA}{dt}&=\frac{1}{4}\left(b^2-2bt-3t^2\right)\\ &= \frac{1}{4}(b-3t)(b+t)\end{align*}$
$\text{To find where }A\text{ is maximised, set }\frac{dA}{dt}=0\\ \frac{1}{4}(b-3t)(b+t)=0\\ \therefore t=\frac{b}{3}\text{ or }t=-b\\ \text{But since both }t\text{ and }b\text{ are positive and non-zero, reject the latter.}\\ \therefore\text{ the area is possibly maximised } t=\frac{b}{3}$
$\frac{d^2A}{dt^2}=\frac{1}{4}(-2b-6t)< 0 \text{ for all positive }t\\ \text{Hence given the concavity of }A\text{ as a function of }t\text{, the area is indeed maximised when }t=\frac{b}{3}$
$\text{Substitute this in back for }T\text{ to get }\left(\frac{b}{3}, \frac{4b^2}{9}\right)$
For part (i) just gotta be careful when factorising by completing the square. In point gradient form the tangent has equation

y-(t-b)^2 = 2(t-b)(x-t)

So y = (t-b)(t-b)+(t-b)(2x-2t)
So y=(t-b)(t-b+2x-2t)
=(t-b)(2x-t-b)

isaacdelatorre · « **Reply #245 on:** July 25, 2016, 06:42:13 pm »

Quote from: RuiAce on July 24, 2016, 07:49:18 pm

$\text{The point }A\text{ lies on the tangent for }x=0\\ y=(-t-b)(t-b)=b^2-t^2\\ \text{The point }B\text{ lies on the tangent for }y=0\\ 0=(2x-t-b)(t-b)\iff 2x-t-b=0 \iff x=\frac{t+b}{2}\\ \text{Keeping in mind that }t\neq b\text{ so we can safely cancel out the }t-b$
$\text{Now, the area of }\triangle ABO\text{ using }A=\frac{1}{2}bh\text{ is:}\\ \begin{align*}A&=\frac{1}{2}\times OA \times OB\\ &= \frac{1}{2}\left((b^2-t^2)-0\right)\left(\frac{t+b}{2}-0\right)\\ &= \frac{(b^2-t^2)(b+t)}{4}\\ &=\frac{1}{4}\left(b^3+b^2t-bt^2-t^3\right)\\ \therefore \frac{dA}{dt}&=\frac{1}{4}\left(b^2-2bt-3t^2\right)\\ &= \frac{1}{4}(b-3t)(b+t)\end{align*}$
$\text{To find where }A\text{ is maximised, set }\frac{dA}{dt}=0\\ \frac{1}{4}(b-3t)(b+t)=0\\ \therefore t=\frac{b}{3}\text{ or }t=-b\\ \text{But since both }t\text{ and }b\text{ are positive and non-zero, reject the latter.}\\ \therefore\text{ the area is possibly maximised } t=\frac{b}{3}$
$\frac{d^2A}{dt^2}=\frac{1}{4}(-2b-6t)< 0 \text{ for all positive }t\\ \text{Hence given the concavity of }A\text{ as a function of }t\text{, the area is indeed maximised when }t=\frac{b}{3}$
$\text{Substitute this in back for }T\text{ to get }\left(\frac{b}{3}, \frac{4b^2}{9}\right)$
For part (i) just gotta be careful when factorising by completing the square. In point gradient form the tangent has equation

y-(t-b)^2 = 2(t-b)(x-t)

So y = (t-b)(t-b)+(t-b)(2x-2t)
So y=(t-b)(t-b+2x-2t)
=(t-b)(2x-t-b)

Wow, thanks RuiAce.
I don't know why I was struggling with it.
Thanks again

Deng · « **Reply #246 on:** July 28, 2016, 11:23:43 pm »

Hey was looking for help with a couple questions ive been stuck on
Part ii for the area under the curve questions

Thanks!

jamonwindeyer · « **Reply #247 on:** July 29, 2016, 01:32:15 am »

Quote from: Deng on July 28, 2016, 11:23:43 pm

Hey was looking for help with a couple questions ive been stuck on
Part ii for the area under the curve questions

Thanks!

Hey Deng!!

Interestingly, I'd actually approach that first question not as an integration area question, but with some geometry! Let's have a think. The area we want is actually just the area of the triangle enclosed by the axes and the tangent, then subtract the area of the quarter circle! Let me know if that's unclear what I mean there!

So, we need the intercepts of the tangent so we can find the area of that triangle. Use whatever method you like, I'll use substitution:

$x=0 \implies -4y=15 \quad \therefore y=\frac{-15}{4} \\ y=0 \implies 3x=15 \quad \therefore x=5$

So the triangle has a width of 5 and a height of 15/4, let's use that to find the area we need:

$A = A_\text{triangle}-A_{quarter-circle} \\ =\frac{1}{2}bh-\frac{1}{4}\pi r^2 \\=\left(\frac{1}{2}\times\frac{15}{4}\times5\right)-\frac{9\pi}{4}$

I'll let you take it from there

For Question 2, we can massively simplify that expression by recognising that sine squared plus cosine squared of the same value is just 1. So, using that Pythagorean Identity, the expression becomes:

$\frac{30}{3\sin^2{\theta}+2\cos^2{\theta}}=\frac{30}{2+\sin^2{\theta}}$

Now we just need to think about this intuitively (easier than rigorous mathematics). We want the smallest value of this expression over the given domain, which means we want the denominator as large as possible. Well, the maximum value of sin^2 is 1, so that means the minimum value of the expression is:

$\frac{30}{2+1} = 10$

Part (ii) of that last question is a little tricky! Basically, we can get it by understanding that the integral of the first derivative will get us back to the function itself!

I'll leave you to have the full attempt. But let me get you started. So, from your sketch in Part (i), you know that the graph of f'(x) has intercepts at x=-2, x=-1, and x=3. First, let's find the area enclosed between x=3 and x=-1. Use our regular method, and remember, we are integrating a derivative! The result will be our original function:

$A_1=\int^3_{-1}f'(x)dx = \left[f(x)\right]^3_{-1}$

We now just need to evaluate this at 3 and -1, which we can do, because we are given these values in the graph:

$\therefore A_1 = f(3)-f(-1) = 5+3 = 8u^2$

The same principle would apply to the other part of the area between x=-1 and x=-2, but with an absolute value sign as is normal for these sorts of questions

then just add the two areas together to get your answer!

That last one was a bit of a doozy, does that make sense Deng?

Deng · « **Reply #248 on:** July 29, 2016, 01:14:39 pm »

Yeah it does
Thanks!

conic curve · « **Reply #249 on:** July 29, 2016, 02:59:52 pm »

What does it mean by "magnitude" in maths?

I searched it up and it said that "magnitude is the size of a mathematical object" but then, I'm thinking, what? This is so confusing

Thanks

jakesilove · « **Reply #250 on:** July 29, 2016, 03:04:37 pm »

Quote from: conic curve on July 29, 2016, 02:59:52 pm

What does it mean by "magnitude" in maths?

I searched it up and it said that "magnitude is the size of a mathematical object" but then, I'm thinking, what? This is so confusing

Thanks

Magnitude is literally the 'size' of something. The magnitude of 2 is 2. It just strips a vector value of it's direction; so, say you had a value which was 2km East. The MAGNITUDE of that value is 2km. The DIRECTION is East.

anotherworld2b · « **Reply #251 on:** July 30, 2016, 02:12:10 am »

Hi I have a few questions to ask if it is okay

i wasnt sure how to do q1 and q3. I tried to prove q2 but im not sure whether it is right. For q11 I was able to get the answer by using a tree diagram but im am unsure how to use a venn diagram (how to put the info in the venn diagram) and how to use The rules approach?

RuiAce · « **Reply #252 on:** July 30, 2016, 08:17:21 am »

Once again, I need to emphasise the fact that study of set theory is NOT required in the HSC mathematics courses.

These diagrams are pretty small and I can't see everything typed on Q11. Don't shrink diagrams to upload them; just upload them to imgur or something and post the links to the images next time please. Also, try not to ask too many questions on the exact same topic; it's burdening

$\text{Q1 is mostly just formula work.}\\ \text{By the addition rule of probability we start on}\\ Pr(A \cup B)=Pr(A)+Pr(B)-Pr(A \cap B)\\ \text{Using the definition of complementary probabilities we have }Pr(\overline{A \cap B}) = 1-Pr(A \cap B)\\ \text{The definition of conditional probability states that }Pr(A|B)=\frac{Pr(A \cap B)}{Pr(B)}$
$\text{For part d), you start with }Pr(\overline{A}|B)=\frac{Pr(\overline{A}\cap B)}{Pr(B)}\\ \text{However by the law of total probability }Pr(B)=Pr(A \cap B)+Pr(\overline{A} \cap B)\\ \text{so you need to find the probability of not A and B first.}\\ \text{The last one can be done similarly.}$
$\text{For Q2, recall that statistical independence occurs when }Pr(A|B)=Pr(A)\\ Pr(A|B)=\frac{Pr(A \cap B)}{Pr(B)}=\frac{Pr(A)+Pr(B)-Pr(A \cup B)}{0.2} = \frac{0.45+0.2+0.56}{0.2}=0.45=Pr(A)$
$\text{The fact that they are all mutually exclusive in Q3 means that probability of any two equals }0\\ \text{So if you add the probabilities they immediately equal 1 as we have no overlap.}\\ 2p+3p+5p=1 \iff p=\frac{1}{10}$

dreamdog10 · « **Reply #253 on:** July 30, 2016, 02:09:24 pm »

Hi! just wondering how you would go about graphing

dv/dt=1−(√2)sin(πt/60)
between 0<=t<=120

RuiAce · « **Reply #254 on:** July 30, 2016, 02:21:24 pm »

Quote from: dreamdog10 on July 30, 2016, 02:09:24 pm

Hi! just wondering how you would go about graphing

dv/dt=1−(√2)sin(πt/60)
between 0<=t<=120

$\text{Without information to find the constant of integration, the graph of acceleration v.s. time is assumed.}\\ \text{The amplitude of the graph is clearly }\sqrt{2}\\ \text{The period of the graph will be }T=\frac{2\pi}{\pi/60}=120\\ \text{So we know the graph completes one full cycle for every }120\text{ seconds}$
$\text{There is no shift in phase, so the graph definitely starts at }1\\ \text{Clearly, the extreme cases will therefore be }\ddot{x}=1+\sqrt{2},\, \ddot{x}=1-\sqrt{2}$
$\text{Hence, the graph will look like this.}$

$\text{Note that because we have negative sine, the graph was flipped about the line }\ddot{x}=1$
$\text{The dotted lines }\ddot{x}=1+\sqrt{2}\text{ and }\ddot{x}=1-\sqrt{2}\text{ are present for reference only.}$

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