HSC 2015 MX1 Marathon (archive) (6 Viewers)

photastic · Oct 25, 2015

Re: HSC 2015 3U Marathon

malcolm21 said:
how do we do this?

thomasdo1 · Oct 25, 2015

Re: HSC 2015 3U Marathon

anyone know how to do this? http://puu.sh/kWQRX/39aa4f9a6b.jpg

Since the question tells you the particles collide, could I make x[a] = x (particles a and b) and eliminate t? so Vsin(alpha) = Ucos(beta) --> only when they collide

And prove LHS = RHS when subbing in the value for T into t (in displacement equations) and using Vsin(alpha) = Ucos(beta)?

photastic · Oct 25, 2015

Re: HSC 2015 3U Marathon

thomasdo1 said:
anyone know how to do this? http://puu.sh/kWQRX/39aa4f9a6b.jpg

Since the question tells you the particles collide, could I make x[a] = x (particles a and b) and eliminate t? so Vsin(alpha) = Ucos(beta) --> only when they collide

And prove LHS = RHS when subbing in the value for T into t (in displacement equations) and using Vsin(alpha) = Ucos(beta)?

Correct.

Well what's special about collisions, this will occur when the two particles have the same time of flight and some point in time the same vertical and horizontal displacement so to do the q, you must incorporate both vertical and horizontal components. Still stuck, i'll (someone may) post a solution.

rand_althor · Oct 25, 2015

Re: HSC 2015 3U Marathon

$$I'm assuming you've got the first part done. For the projectile released at $O$: $x=Vt\cos{\alpha}$, $y=\frac{-gt^2}{2}+Vt\sin{\alpha}$. For the projectile released at $A$: $x=Ut\cos{\beta}$, $y=\frac{-gt^2}{2}+Ut\sin{\beta}+20$. When the particles collide at time $T$ their vertical and horizontal displacements are equal. $VT\cos{\alpha}=UT\cos{\beta}$ and $\frac{-gT^2}{2}+VT\sin{\alpha}=\frac{-gT^2}{2}+UT\sin{\beta}+20$. Now we can simplify the equation involving vertical displacements to $T(V\sin{\alpha}-U\sin{\beta})=20$.$

$$Notice that the equation you're required to show does not involve $V$. To eliminate $V$ from our equation we can use the horizontal displacement equations: $VT\cos{\alpha}=UT\cos{\beta} \to V=\frac{U\cos{\beta}}{\cos{\alpha}}$. Now subbing this into our equation: $T(V\sin{\alpha}-U\sin{\beta})=20 \to T\left (\frac{U\sin{\alpha}\cos{\beta}}{\cos{\alpha}}-U\sin{\beta} \right )=20$. Rearranging this equation and using the expansion of $\sin{(\alpha-\beta)}$ you get the required equation.$

thomasdo1 · Oct 25, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$I'm assuming you've got the first part done. For the projectile released at $O$: $x=Vt\cos{\alpha}$, $y=\frac{-gt^2}{2}+Vt\sin{\alpha}$. For the projectile released at $O$: $x=Ut\cos{\beta}$, $y=\frac{-gt^2}{2}+Ut\sin{\beta}+20$. When the particles collide at time $T$ their vertical and horizontal displacements are equal. $VT\cos{\alpha}=UT\cos{\beta}$ and $\frac{-gT^2}{2}+VT\sin{\alpha}=\frac{-gT^2}{2}+UT\sin{\beta}+20$. Now we can simplify the equation involving vertical displacements to $T(V\sin{\alpha}-U\sin{\beta})=20$.$

$$Notice that the equation you're required to show does not involve $V$. To eliminate $V$ from our equation we can use the horizontal displacement equations: $VT\cos{\alpha}=UT\cos{\beta} \to V=\frac{U\cos{\beta}}{\cos{\alpha}}$. Now subbing this into our equation: $T(V\sin{\alpha}-U\sin{\beta})=20 \to T\left (\frac{U\sin{\alpha}\cos{\beta}}{\cos{\alpha}}-U\sin{\beta} \right )=20$. Rearranging this equation and using the expansion of $\sin{(\alpha-\beta)}$ you get the required equation.$

facepalm, I'm not used to no diagrams, I interpreted the launches to be both from A, so when I equated their y-values, It all cancelled haha

next question

$simplify $S=e^{-2t}+2e^{-4t}+3e^{-6t}+4e^{-8t}+..........$

t>0

InteGrand · Oct 25, 2015

Re: HSC 2015 3U Marathon

thomasdo1 said:
facepalm, I'm not used to no diagrams, I interpreted the launches to be both from A, so when I equated their y-values, It all cancelled haha

next question

$simplify $S=e^{-2t}+2e^{-4t}+3e^{-6t}+4e^{-8t}+..........$

$Note that $S=\mathrm{e}^{-2t}\left(1 + 2 \left(\mathrm{e}^{-2t} \right) + 3 \left(\mathrm{e}^{-2t} \right)^2 + 4 \left(\mathrm{e}^{-2t} \right)^3 + \cdots \right)$.$

$To evaluate the infinite sum, use this series with $x=\mathrm{e}^{-2t}$: $1+2x+3x^2 + 4x^3 + \cdots = \frac{1}{(1-x)^2}$, valid for $|x|<1$ (the sum is divergent otherwise).$

$This can be derived by considering $1+x+x^2 + \cdots = \frac{1}{1-x},|x|<1$, and differentiating both sides w.r.t. $x$, which we can do term-by-term on the infinite sequence if restricted to $|x|<1$.$

$So your sum $S$ will be convergent if and only if $t$ satisfies $\Big{|}\mathrm{e}^{-2t} \Big{|}<1\Longleftrightarrow t>0$.$

thomasdo1 · Oct 25, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$Note that $S=\mathrm{e}^{-2t}\left(1 + 2 \left(\mathrm{e}^{-2t} \right) + 3 \left(\mathrm{e}^{-2t} \right)^2 + 4 \left(\mathrm{e}^{-2t} \right)^3 + \cdots \right)$.$

$To evaluate the infinite sum, use this series with $x=\mathrm{e}^{-2t}$: $1+2x+3x^2 + 4x^3 + \cdots = \frac{1}{(1-x)^2}$, valid for $|x|<1$ (the sum is divergent otherwise).$

$This can be derived by considering $1+x+x^2 + \cdots = \frac{1}{1-x},|x|<1$, and differentiating both sides w.r.t. $x$, which we can do term-by-term on the infinite sequence if restricted to $|x|<1$.$

$So your sum $S$ will be convergent if and only if $t$ satisfies $\Big{|}\mathrm{e}^{-2t} \Big{|}<1\Longleftrightarrow t>0$.$

Yeahh, I realised that 5 seconds after I posted it, I added t>0 just about after you got to it

rand_althor · Oct 25, 2015

Re: HSC 2015 3U Marathon

thomasdo1 said:
facepalm, I'm not used to no diagrams, I interpreted the launches to be both from A, so when I equated their y-values, It all cancelled haha

next question

$simplify $S=e^{-2t}+2e^{-4t}+3e^{-6t}+4e^{-8t}+..........$

t>0

$\begin{align*}\textup{If we let }x&=e^{-2t} \textup{ we have:} \\S&=x+2x^2+3x^3+4x^4+... \\Sx&=x^2+2x^3+3x^4+4x^5+... \\S-Sx&=x+2x^2+3x^3+4x^4+...-x^2-2x^3-3x^4-4x^5-... \\S(1-x)&=x+x^2+x^3+x^4+... \\&=\frac{x}{1-x} \textup{ using }S_\infty=\frac{a}{1-r} \\S&=\frac{x}{(1-x)^2} \\&=\frac{e^{-2t}}{(1-e^{-2t})^2} \\&=\frac{\dfrac{1}{e^{2t}}}{\left (1-\dfrac{1}{e^{2t}} \right )^2} \\&=\frac{\dfrac{1}{e^{2t}}}{\left (\dfrac{e^{2t}-1}{e^{2t}} \right )^2} \\&=\frac{\dfrac{1}{e^{2t}}}{\left (\dfrac{\left (e^{2t}-1 \right )^2}{\left (e^{2t} \right )^2} \right )} \\&=\frac{e^{2t}}{\left (e^{2t}-1 \right )^2}\end{align*}$

Limiting sum exists as since t>0, $0<\frac{1}{e^{2t}}<1$ .

kawaiipotato · Oct 25, 2015

Re: HSC 2015 3U Marathon

Another alternate method
S = x (1 + 2x + 3x^2 + 4x^3 + ...)

S/x = 1 + x + x^2 + x^3 + x^4 + ...
.............+ x + x^2 + x^3 + x^4 + ...
...................+ x^2 + x^3 + x^4 + ...
.............................+ x^3 + x^4 +
.......................................+x^4 + ...
= 1/(1-x) + x/(1-x) + x^2 /(1-x) + ...
= 1/(1-x)/(1-x) = 1/(1-x)^2
S = x/(1-x)^2
Where x = e^(-2t)

davidgoes4wce · Oct 26, 2015

Re: HSC 2015 3U Marathon

for the explanation for Angle QCD= Angle QAD step

Could I explain it by saying "Angles at the circumference are equal when subtended by the same arc QD") ?

In the solution it says "since angle standing on the same arc QD are equal".

Zlatman · Oct 26, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
for the explanation for Angle QCD= Angle QAD step

Could I explain it by saying "Angles at the circumference are equal when subtended by the same arc QD") ?

In the solution it says "since angle standing on the same arc QD are equal".

Yep, they're just different ways of saying the same thing.

thomasdo1 · Oct 26, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
for the explanation for Angle QCD= Angle QAD step

Could I explain it by saying "Angles at the circumference are equal when subtended by the same arc QD") ?

In the solution it says "since angle standing on the same arc QD are equal".

http://i4.minus.com/jfpQrNM8MT9cy.jpg

davidgoes4wce · Oct 26, 2015

Re: HSC 2015 3U Marathon

thomasdo1 said:
http://i4.minus.com/jfpQrNM8MT9cy.jpg

That's pretty handy to have. Think I will print out to kepnancopynfor myself.

Crisium · Oct 26, 2015

Re: HSC 2015 3U Marathon

lol @ leehuan the snake he managed to get a post into the MX1 general thoughts thread before Rafy locked it

On another note, I've seen two triangles that have been used for the t-formula

Triangle 1:

The sides have 2t , 1 + t^2 and 1 - t^2

Triangle 2:

The sides have t , 1 , (1 + t^2 )^0.5

Could somebody post up an example of when triangle 2 would be used?

rand_althor · Oct 26, 2015

Re: HSC 2015 3U Marathon

Crisium said:
lol @ leehuan the snake he managed to get a post into the MX1 general thoughts thread before Rafy locked it [emoji38]

On another note, I've seen two triangles that have been used for the t-formula

Triangle 1:

The sides have 2t , 1 + t^2 and 1 - t^2

Triangle 2:

The sides have t , 1 , (1 + t^2 )^0.5

Could somebody post up an example of when triangle 2 would be used?

In the second triangle tan(x/2)=t. Using this triangle, you can find sinx, cosx and tanx in terms of t. The first triangle is the sides of a triangle with sinx, cosx and tanx in terms of t.

InteGrand · Oct 26, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
In the second triangle tan(x/2)=t. Using this triangle, you can find sinx, cosx and tanx in terms of t. The first triangle is the sides of a triangle with sinx, cosx and tanx in terms of t.

For the second triangle, it's t = tan(x), and you can find sin(x), cos(x) (and trivially tan(x)) in terms of tan(x) = t.

InteGrand · Oct 26, 2015

Re: HSC 2015 3U Marathon

$E.g. $\sin x = \frac{t}{\sqrt{1 + t^2}}$, where $t\equiv \tan x$ and $x$ is such that $\sin x$ and $\tan x$ have the same sign (i.e. in quadrant 1 or 4).$

leehuan · Oct 31, 2015

Re: HSC 2015 3U Marathon

Crisium said:
lol @ leehuan the snake he managed to get a post into the MX1 general thoughts thread before Rafy locked it

On another note, I've seen two triangles that have been used for the t-formula

Triangle 1:

The sides have 2t , 1 + t^2 and 1 - t^2

Triangle 2:

The sides have t , 1 , (1 + t^2 )^0.5

Could somebody post up an example of when triangle 2 would be used?

It got deleted.

HSC 2015 MX1 Marathon (archive) (6 Viewers)

photastic

Well-Known Member

thomasdo1

Member

photastic

Well-Known Member

rand_althor

Active Member

thomasdo1

Member

InteGrand

Well-Known Member

thomasdo1

Member

rand_althor

Active Member

kawaiipotato

Well-Known Member

davidgoes4wce

Well-Known Member

Zlatman

Member

thomasdo1

Member

davidgoes4wce

Well-Known Member

Crisium

Pew Pew

rand_althor

Active Member

InteGrand

Well-Known Member

InteGrand

Well-Known Member

leehuan

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 6)