Author Topic: methods help (Read 1026 times) Tweet Share

lalor · « **on:** November 03, 2012, 04:40:31 pm »

Could someone show me a worked solution to vcaa methods exam 1 2011 Q9 i'm really stuck and the vcaa solutions don't explain how it's done

b^3 · « **Reply #1 on:** November 03, 2012, 04:55:41 pm »

We have two bits of information we can pull from this quesiton, to obtain 2 equations to solve for m and a.
- The shaded region, so we want the integral of the top function minus the bottom function to give 64.
- The lines intersect at $x=m$

So for the intersecting lines part.
$\begin{alignedat}{1}m^{3}-ma & =ma \\ m^{3}-2ma & =0 \\ m\left(m^{2}-2a\right) & =0 \\ m=0 & \:or\: m^{2}=2a \end{alignedat}$
But $m>0$ (look at the diagram)
So
$\therefore m^{2}=2a...[1]$

Now for the area part.
$\begin{alignedat}{1}\int_{0}^{m}ax-\left(x^{3}-ax\right)dx & =64 \\ \int_{0}^{m}2ax-x^{3}dx & =64 \\ \left[ax^{2}-\frac{1}{4}x^{4}\right]_{0}^{m} & =64 \\ am^{2}-\frac{1}{4}m^{4} & =64 \\ 4am^{2}-m^{4} & =256 \\ m^{4}-4am^{2}+256 & =0....[2] \end{alignedat}$

Substitute equ [1] into equ [2].
$\begin{alignedat}{1}\left(2a\right)^{2}-4a\left(2a\right)+256 & =0 \\ 4a^{2}-8a^{2}+256 & =0 \\ 4a^{2} & =256 \\ a^{2} & =64 \\ a & =\pm8 \\ \mathrm{but\:}a>0 \\ \therefore a=8 \\ m^{2} & =2a \\ m & =\pm\sqrt{16} \\ m & =\pm4 \\ \mathrm{but\:},m>0 \\ \therefore m=4 \end{alignedat}$

lalor · « **Reply #2 on:** November 04, 2012, 10:47:14 am »

Thanks a lot i was on the right track just got a bit confused but it's all good now

pyro-53 · « **Reply #3 on:** November 04, 2012, 10:57:32 am »

sorry to barge in, but how do you type up the math formulas?

Laecs · « **Reply #4 on:** November 04, 2012, 01:57:35 pm »

Quote from: pyro-53 on November 04, 2012, 10:57:32 am

sorry to barge in, but how do you type up the math formulas?

LaTeX - https://sites.google.com/site/scienceandmathguide/
Good reference to learn

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Author Topic: methods help (Read 1026 times) Tweet Share

lalor

methods help

b^3

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lalor

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pyro-53

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Laecs

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