HSC 2015 MX2 Marathon (archive) (1 Viewer)

Kaido · May 16, 2015

Re: HSC 2015 4U Marathon

integral of sqrt(5-x^2+4x)

InteGrand · May 16, 2015

Re: HSC 2015 4U Marathon

Kaido said:
integral of sqrt(5-x^2+4x)

$5-x^2 + 4x = -(x^2 - 4x - 5)=-(x^2 -4x + 4)+9 = 9-(x-2)^2$

$\Rightarrow \int \sqrt{5-x^2 + 4x}\text{ d}x=\int \sqrt{9-(x-2)^2}\text{ d}x, \text{ which is easy (albeit somewhat tedious) to find.}$

Kaido · May 16, 2015

Re: HSC 2015 4U Marathon

Yeah i got to there, but im not sure where to go lol

should i let u=x-2
then do another trig sub, then reverse everything back

InteGrand · May 16, 2015

Re: HSC 2015 4U Marathon

Kaido said:
Yeah i got to there, but im not sure where to go lol

should i let u=x-2
then do another trig sub, then reverse everything back

Yeah, that's one way, or you can use integration by parts after letting $u=x-2$ .

InteGrand · May 16, 2015

Re: HSC 2015 4U Marathon

Actually, trig. sub. should be easier in this case.

InteGrand · May 16, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$5-x^2 + 4x = -(x^2 - 4x - 5)=-(x^2 -4x + 4)+9 = 9-(x-2)^2$

$\Rightarrow \int \sqrt{5-x^2 + 4x}\text{ d}x=\int \sqrt{9-(x-2)^2}\text{ d}x, \text{ which is easy (albeit somewhat tedious) to find.}$

So the trig. sub. method is as follows.

$Let $I=\int \sqrt{a^2 - u^2}\text{ d}u$ (in our case, $u=x-2,\mathrm{d}u=\mathrm{d}x,a=3$. This is just for the general case (also, $a>0$)).$

$Let $u=a\sin \theta, \theta \in \left[-\frac{\pi}{2},\frac{\pi}{2} \right]$ (so $\theta=\sin ^{-1}\frac{u}{a},\cos \theta \geq 0$). Then $\mathrm{d}u=a\cos \theta$ $\mathrm{d}\theta$. Also, $\sqrt{a^2 - u^2}=\sqrt{a^2(1-\sin ^2 \theta)}=a\sqrt{\cos^2 \theta}=a\cos \theta$, as $\cos \theta \geq 0$.$

$Hence $I=\int a\cos \theta \cdot a \cos \theta \text{ d}\theta$$

$\Rightarrow I=a^2 \int \cos^2 \theta \text{ d}\theta$

$=a^2 \int \frac{1}{2}(1+\cos 2\theta) \text{ d}\theta$

$=\frac{a^2}{2}\left(\theta + \frac{\sin 2 \theta}{2} \right) + c$

$=\frac{a^2}{2}\left(\theta + \sin \theta \cos \theta \right) + c$

$=\frac{a^2}{2}\left(\sin ^{-1}\frac{u}{a}+ \frac{u}{a} \cdot \sqrt{1-\left(\frac{u}{a} \right)^2}\right) + c$, using $\cos^2 \theta = \sqrt{1-\sin^2 \theta}$ for $\cos \theta \geq 0$

$$=\frac{a^2}{2}\left(\sin ^{-1}\frac{u}{a}+ \frac{u}{a} \cdot \sqrt{\frac{a^2 -u^2}{a^2}}\right)+ c$$

$$=\frac{a^2}{2}\left(\sin ^{-1}\frac{u}{a}+ \frac{u\sqrt{a^2 - u^2}}{a^2}\right)+ c$.$

$$=\frac{a^2}{2}\sin ^{-1}\frac{u}{a}+ \frac{u\sqrt{a^2 - u^2}}{2}+ c$.$

Drsoccerball · May 16, 2015

Re: HSC 2015 4U Marathon

Integrand where the mechanics questions at doe?

InteGrand · May 16, 2015

Re: HSC 2015 4U Marathon

Here was the integration by parts method:

$I=\int \sqrt{a^2 \pm u^2}\text{ d}u$

$=u\sqrt{a^2 \pm u^2}-\int u\times \frac{\pm 2u}{2\sqrt{a^2 \pm u^2}}\text{ d}u$

$=u\sqrt{a^2 \pm u^2}-\int \frac{\pm u^2}{\sqrt{a^2 \pm u^2}}\text{ d}u$

$=u\sqrt{a^2 \pm u^2}-\int \frac{a^2\pm u^2}{\sqrt{a^2 \pm u^2}}\text{ d}u + \int \frac{a^2}{\sqrt{a^2 \pm u^2}}\text{ d}u$

$\Rightarrow I =u\sqrt{a^2 \pm u^2}-\underbrace{\int\sqrt{a^2 \pm u^2}\text{ d}u}_{=I} + a^2 \cdot(\text{Standard integral})$

$\Rightarrow 2I = u\sqrt{a^2 \pm u^2}+ a^2 \cdot(\text{Standard integral})$

$\Rightarrow I = \frac{1}{2}[u\sqrt{a^2 \pm u^2}+ a^2 \cdot(\text{Standard integral})]+c$

$(The standard integrals are $\int \frac{1}{\sqrt{a^2 + u^2}}\text{ d}u= \ln (u+\sqrt{a^2 + u^2})+c \text{ }(=\sinh ^{-1}\frac{u}{a}+C)$ and $\int \frac{1}{\sqrt{a^2 - u^2}}\text{ d}u= \sin^{-1}\frac{u}{a}+\mathcal{C}$.)$

InteGrand · May 16, 2015

Re: HSC 2015 4U Marathon

$\underline{MECHANICS QUESTION}$

$Consider a projectile thrown vertically upwards with initial speed $u$ that is under the influence of gravity and a drag (resistive) force $F_\text{drag} (v) = -mkv^2$, where $v$ is the velocity of the projectile at time $t>0$, $k>0$ is a constant, and $m>0$ is the mass of the projectile. Does it take longer for the projectile to reach its maximum height, or to come back down to its original height from the maximum height?$

Drsoccerball · May 16, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$\underline{MECHANICS QUESTION}$

$Consider a projectile thrown vertically upwards with initial speed $u$ that is under the influence of gravity and a drag (resistive) force $F_\text{drag} (v) = -mkv^2$, where $v$ is the velocity of the projectile at time $t>0$, $k>0$ is a constant, and $m>0$ is the mass of the projectile. Does it take longer for the projectile to reach its maximum height, or to come back down to its original height from the maximum height?$

Didnt braintic answer this

InteGrand · May 16, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Didnt braintic answer this

He answered the extra challenge Q that I'd given before (which wouldn't be asked in HSC).

Drsoccerball · May 16, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
He answered the extra challenge Q that I'd given before (which wouldn't be asked in HSC).

Is the answr it takes longer on its way up since its being opposed by two vectors: weight and resistance while on the way down the weight vector is positive therefore quicker on the way down ?

InteGrand · May 16, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Is the answr it takes longer on its way up since its being opposed by two vectors: weight and resistance while on the way down the weight vector is positive therefore quicker on the way down ?

It takes longer on the way down.

And in the HSC, I'm guessing they'd expect you to prove this by finding the time of ascent and time of descent.

InteGrand · May 16, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Is the answr it takes longer on its way up since its being opposed by two vectors: weight and resistance while on the way down the weight vector is positive therefore quicker on the way down ?

Here's braintic's answer:

braintic said:
Extra challenge: As resistance is a non-conservative force (since it is dependent on velocity and not position), only reducing the total energy of the body, at any point in its downward path it must have less kinetic energy and hence less speed than at that point in its upward path (because potential energy is the same at both times). Since it is slower all the way down, it takes longer on the way down.

But you wouldn't need to know this for HSC.

Kaido · May 18, 2015

Re: HSC 2015 4U Marathon

need help with this:
volume obtained by rotating (x-1)^2+y^2/4=1 about y-axis using slices

Drsoccerball · May 18, 2015

Re: HSC 2015 4U Marathon

Kaido said:
need help with this:
volume obtained by rotating (x-1)^2+y^2/4=1 about y-axis using slices

Since i had troubles with both methods at the beginning ill help yah
So first sketch the circle and draw a line perpendicular to the axis of rotation and it it intercepts twice
In order to find where they intercept you have to make x the subject
$\therefore x=1 +- \sqrt{1-\frac{y^2}{4}}$

$\pi ( $ outer radius )^2 $ - \pi ( $ inner radius $ )^2$

$\pi \int_{-2}^2{(1+\sqrt{1-\frac{y^2}{4}})^2 - (1-\sqrt{1-\frac{y^2}{4}})^2 }dy$

Kaido · May 19, 2015

Re: HSC 2015 4U Marathon

Ahh i see, i was just confused due to the +/- sign, so the outer radius assumes the larger x value - therefore a + sign?

Drsoccerball · May 19, 2015

Re: HSC 2015 4U Marathon

Kaido said:
Ahh i see, i was just confused due to the +/- sign, so the outer radius assumes the larger x value - therefore a + sign?

yeah

kawaiipotato · May 19, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
i) $F=mg-kv^2$

$\therefore a= g- \frac{kv^2}{m}$

$\frac{dv}{dt}= g-\frac{kv^2}{m}$

$\frac{dv}{dt} = \frac{1}{m}(mg-kv^2)$

$\frac{dv}{dt} = \frac{1}{m}(F-kv^2)$

ii) $\frac{v \cdot dv}{dx} = \frac{gm-kv^2}{m}$

$\frac{dx}{dv}= \frac{mv}{gm-kv^2}$

$x= \frac{m}{2k}[ln(gm-kv^2)]_{v_f}^{v_i}$

$D= \frac{m}{2k}ln(\frac{F-kv^2_i}{F-kv^2_f})$

Shouldn't the first step be something like
$F_{net} = ma$

$F_{net} = F - kv^{2}$

$ma = F - kv^{2}$

$\frac{dv}{dt} = \frac{1}{m} [F - kv^{2}]$

Do we need to assume mg acts on it?

InteGrand · May 19, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
Shouldn't the first step be something like
$F_{net} = ma$

$F_{net} = F - kv^{2}$

$ma = F - kv^{2}$

$\frac{dv}{dt} = \frac{1}{m} [F - kv^{2}]$

You're right, that's what the working should be (and there should be no mg).

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