Author Topic: 3U Maths Question Thread (Read 1502196 times) Tweet Share

RuiAce · « **Reply #2460 on:** July 18, 2017, 06:19:50 pm »

Quote from: abba9554 on July 18, 2017, 06:05:22 pm

Hi!
I need help with this question. I am having trouble understanding the notation. Thanks guys

This is obviously stretching the boundaries way too far for a 3U question. But it should be doable. What is the source of this question?
$\text{In }f(x+y)=f(x)f(y)\\ \text{Let }x=0\text{ and }y=0.\text{ Then}\\ \begin{align*}f(0+0)&=f(0)f(0)\\ [f(0)]^2-f(0)=0\\ f(0)\left[f(0)-1\right]&=0\end{align*}$
$\text{So the quadratic yields }f(0)=0\text{ or }f(0)=1\\ \text{But }f(0)\neq 0\text{ as for all real }x,\text{ we were told }f(x)\neq 0\\ \text{Hence }f(0)=1$
___________________________
$\text{This part was made needlessly confusing by the choice}\\ \text{of differentiating w.r.t. }x\text{, when differentiating}\\ \text{w.r.t. }y\text{ is superior.}$
$\text{Fix }y\in \mathbb{R}\\ \text{Taking }\frac{d}{dx}:\\ \begin{align*}\frac{d}{dx} f(x+y)&=\frac{d}{dx}f(x)f(y)\\ f^\prime(x+y)&=f^\prime(x)f(y)\end{align*}$
$\text{Put }x=0:\\ \begin{align*}f^\prime(y)&=f^\prime(0)f(y)\\ &= af(y)\end{align*}$
$\text{Of course, as }f^\prime(y)=af(y)\\ \text{We really are saying }f^\prime(x)=af(x)\text{ as well.}$
___________________________
$\text{Relating back to the exponential growth topic}\\ \text{We know as a fact that the solution to }\frac{dN}{dt}=kN\text{ is }N=Ae^{kt}$
$\text{In a similar way, the solution to }f^\prime(x) = af(x)\\ \text{is }f(x) = Ae^{ax}$
$\text{Using our initial condition, i.e. }f(0)=1\text{ we have }A=1\\ \text{Hence }f(x) = e^{ax}$

Note that the dot simply means multiply. The dot is another commonly used symbol instead of the cross for multiplication.

abba9554 · « **Reply #2461 on:** July 18, 2017, 08:55:38 pm »

thank u

My friend showed it to me

Sukakadonkadonk · « **Reply #2462 on:** July 19, 2017, 01:09:58 pm »

Hey guys,

Could anyone help me with this?
Totally forgot how to do these types.

Thanks

RuiAce · « **Reply #2463 on:** July 19, 2017, 02:00:32 pm »

Quote from: Sukakadonkadonk on July 19, 2017, 01:09:58 pm

Hey guys,

Could anyone help me with this?
Totally forgot how to do these types.

Thanks

$\frac{d}{dx}\left(\frac{1}{2}v^2\right)=-n^2 x\implies v^2 = -n^2x^2 + C$
$\text{When }x=0, v=10\sqrt{3}\text{ so }300=C\\ \text{Hence }\boxed{v^2=300-n^2x^2}$
$\text{At a point of 'instantaneous rest' we have }v=0\\ \text{Setting }v=0\text{ gives }0=300-n^2x^2 \implies x = \pm \frac{10\sqrt3}{n}$
$\text{Hence a point of instantaneous rest is }x=\frac{10\sqrt3}{n}\\ \text{Halfway between this and the equilibrium is }x=\frac{5\sqrt3}{n}\\ \text{which is what you should plug into the boxed formula}$

aaronster · « **Reply #2464 on:** July 20, 2017, 02:38:54 pm »

Do this pls

jamonwindeyer · « **Reply #2465 on:** July 20, 2017, 02:47:20 pm »

Quote from: aaronster on July 20, 2017, 02:38:54 pm

Do this pls

Welcome to the forums! Have you made any progress with it?

RuiAce · « **Reply #2466 on:** July 20, 2017, 02:54:48 pm »

Quote from: aaronster on July 20, 2017, 02:38:54 pm

Do this pls

Excuse me, but please be mindful of your tone when you say "Do this".

Lemma: On top of $OP = PA$, noting that $\angle OTA = 90^\circ $ (radius is perpendicular to tangent at point of contact) we have $OP = PA = PT$
Proof: YouTube link attached
$\text{Now, besides }OP=PT\text{ from above}\\ \text{we also have }OP=OT\text{ from equal radii}\\ \text{Hence }\triangle OPT\text{ is an equilateral triangle}\\ \text{So }\boxed{\alpha = 60^\circ}$
$OP=OS\text{ (equal radii) so }\triangle OSP\text{ is isosceles}\\ \text{and thus }\angle OSP=\beta \text{ (base angles)}\\ \text{Using the exterior angle of an isosceles triangle equals the sum of opp. interior angles}\\ \text{we have }2\beta = \alpha \implies \boxed{\beta=30^\circ}$
$PT=PA\text{ (from above) so similarly }\triangle PTA\text{ is isosceles}\\ \text{Using a similar argument to the above}\\ \text{we have }\boxed{\gamma = 30^\circ}$

aaronster · « **Reply #2467 on:** July 20, 2017, 04:43:25 pm »

sorry for the tone, didn't mean to be rude

aaronster · « **Reply #2468 on:** July 20, 2017, 04:55:55 pm »

Your answers are correct but I would like to know why OP=OA and what is the prove behind PT being equal to the radius of the circle. I got the answer if I assumed that triangle OPT is equilateral or PTA is isosceles but I was told you can't do that. Thank you for the help.

RuiAce · « **Reply #2469 on:** July 20, 2017, 05:00:32 pm »

Quote from: aaronster on July 20, 2017, 04:55:55 pm

Your answers are correct but I would like to know why OP=OA and what is the prove behind PT being equal to the radius of the circle. I got the answer if I assumed that triangle OPT is equilateral or PTA is isosceles but I was told you can't do that. Thank you for the help.

OP=OA was a typo for OP=PA that has been fixed.
$\text{The YouTube video demonstrates how to prove that } OP=PT\\ OP=PT\text{ is the key ingredient to proving that }\triangle OPT\text{ is isosceles}$
$\text{The YouTube video is relevant as it proves it for a right-angled triangle}\\ \text{In our case, }\triangle OTA\text{ is a right-angled triangle at }T.$

beau77bro · « **Reply #2470 on:** July 20, 2017, 06:38:08 pm »

help ii. i think it makes the assumption (or proves idk how) that tps is collinear? i really cant see how to do it.

RuiAce · « **Reply #2471 on:** July 20, 2017, 07:22:49 pm »

Quote from: beau77bro on July 20, 2017, 06:38:08 pm

help ii. i think it makes the assumption (or proves idk how) that tps is collinear? i really cant see how to do it.

Join PQ. You'll see two angles in a semi-circle, and they add to pi. Hence $\angle TPS$ is a straight angle and there's your collinearity.

Ali_Abbas · « **Reply #2472 on:** July 20, 2017, 07:41:31 pm »

Quote from: RuiAce on July 20, 2017, 07:22:49 pm

Join PQ. You'll see two angles in a semi-circle, and they add to pi. Hence $\angle TPS$ is a straight angle and there's your collinearity.

Did you assume that TQ is a diameter of circle TPQ?

junzhang · « **Reply #2473 on:** July 20, 2017, 08:18:20 pm »

Hello

Could you help me with this question, particularly the last part?
Ty!

RuiAce · « **Reply #2474 on:** July 20, 2017, 08:35:56 pm »

I swear last time I checked it was given. That seriously makes me wonder if this question lacks information for a 2 marker though, because it's damn obvious that we do have a diameter but how are they expecting us to prove it for so little reward?

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