Author Topic: Vcaa 2005 (Read 1031 times) Tweet Share

Sanguinne · « **on:** August 09, 2013, 08:01:34 pm »

Vcaa 2005 Exam 2

Can someone explain how to solve q2cii)

I have no idea

Phy124 · « **Reply #1 on:** August 09, 2013, 08:24:20 pm »

I believe you need to do it via trial and error (There could well be another approach, but I believe this was the recommended one by VCAA anyway)

So you would set it up as $\begin{bmatrix} 0.8 & 0.05\\ 0.2 & 0.95 \end{bmatrix}^n\begin{bmatrix} 0.8 \\ 0.2 \end{bmatrix}$ and plug in integer values for $n$ until the top value in the resultant matrix is less than 0.25. e.g.

$n = 8$

$\begin{bmatrix} 0.8 & 0.05\\ 0.2 & 0.95 \end{bmatrix}^8 \begin{bmatrix} 0.8 \\ 0.2 \end{bmatrix} \approx \begin{bmatrix} 0.26 \\ 0.74 \end{bmatrix}$

Try again:

$n = 9$

$\begin{bmatrix} 0.8 & 0.05\\ 0.2 & 0.95 \end{bmatrix}^9 \begin{bmatrix} 0.8 \\ 0.2 \end{bmatrix} \approx \begin{bmatrix} 0.245 \\ 0.755 \end{bmatrix}$

Therefore it would take 9 months.

SocialRhubarb · « **Reply #2 on:** August 09, 2013, 08:26:49 pm »

You'd probably be able to use the solve function on your CAS calculator.

Sanguinne · « **Reply #3 on:** August 09, 2013, 08:42:11 pm »

my mistake, i meant q3cii) instead of q2. Im sincerely sorry for wasting your time $:-\$

Phy124 · « **Reply #4 on:** August 09, 2013, 08:54:35 pm »

Quote from: SocialRhubarb on August 09, 2013, 08:26:49 pm

You'd probably be able to use the solve function on your CAS calculator.

Yeah you can try solving $\begin{bmatrix} 0.8 & 0.05\\ 0.2 & 0.95 \end{bmatrix}^n \begin{bmatrix} 0.8 \\ 0.2 \end{bmatrix} = \begin{bmatrix} 0.25 \\ 0.75 \end{bmatrix}$ for $n$ then ceiling this value and taking that as your answer, but I'm not sure that CAS calculators can compute that.

Quote from: Sanguinne on August 09, 2013, 08:42:11 pm

my mistake, i meant q3cii) instead of q2. Im sincerely sorry for wasting your time $:-\$

Find the roots of the equation over the given domain, then find the difference of the two non zero values.

solve(100cos(π*(x-400)/600)+50=0,x)|0≤x≤1600

x = 0, 800, 1200

The difference between 1200 and 800 is 400m, which is the length of the bridge.

Sanguinne · « **Reply #5 on:** August 10, 2013, 10:40:30 am »

ah ok thanks

how would i work out q1cii)

SocialRhubarb · « **Reply #6 on:** August 10, 2013, 12:29:30 pm »

$h'(t)=(-t^2+(2-a)t+(a-10))e^{-t}$

We want to find a so that h'(t)<0.

$e^{-t}>0,$ so that shouldn't be a problem.

So what we need to show is that $-t^2+(2-a)t+(a-10)<0$ .

If we do a little bit of rearranging:

$-t^2+(2-a)t+(a-10)=-(t^2+(a-2)t+10-a)$

$=-(t^2+(a-2)t+(\frac{a-2}{2})^2-(\frac{a-2}{2})^2+10-a)$

$=-((t+\frac{a-2}{2})^2-\frac{a^2-4a+4}{4}+10-a)$

$=-((t+\frac{a-2}{2})^2+\frac{-a^2+4a-4}{4}+\frac{40}{4}-\frac{4a}{4})$

$=-((t+\frac{a-2}{2})^2+\frac{-a^2+36}{4})=-(t+\frac{a-2}{2})^2+\frac{a^2-36}{4}$

You'll notice that this is the turning point form of a negative parabola. The maximum of this parabola is $\frac{a^2-36}{4}.$ . We want the maximum of the function to be less than 0.

$\frac{a^2-36}{4}<0$

$-6<a<6$

Phy124 · « **Reply #7 on:** August 11, 2013, 01:38:10 pm »

Quote from: SocialRhubarb on August 10, 2013, 12:29:30 pm

$h'(t)=(-t^2+(2-a)t+(a-10))e^{-t}$

We want to find a so that h'(t)<0.

$e^{-t}>0,$ so that shouldn't be a problem.

So what we need to show is that $-t^2+(2-a)t+(a-10)<0$ .

Although there is nothing wrong with your method from here on it's probably easier to just use the discriminant to find values of $a$ such that the function has no roots (and hence below the axis for all $x \in \mathbb{R}$ due to the parabola's "negativity")

$(2-a)^2-4(-1)(a-10)<0$

$\Rightarrow -6 < a < 6$

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