Author Topic: VCE Specialist 3/4 Question Thread! (Read 2804678 times) Tweet Share

SocialRhubarb · « **Reply #1785 on:** June 10, 2013, 04:32:58 pm »

Quote from: Kanye East on June 10, 2013, 12:25:10 pm

Still a bit confused though, in the second step x^2*y'=1-ln|x|, how did you remove the square from the x and cancel it down to x*y=ln|x|?

Your original equation was $y=\frac{ln(|x|)}{x}$ , which he transposed to form $xy=ln(|x|)$ .

He then added this to his equation for y' and this gave the end answer.

götze · « **Reply #1786 on:** June 11, 2013, 04:39:15 pm »

Hi guys does anybody anyhow to express a equation in terms of x on the ti inspire cas calculator ?

b^3 · « **Reply #1787 on:** June 11, 2013, 04:41:26 pm »

If the other variable is $y$ then solve for $y$ using solve(......., $y$ ). (Normally for questions like these there is a nice neat way to do them by hand though).

EDIT: On second thought if it's to do with differential equations, then you want to find say $y$ inside your derivative, and then substitute the expression back in. Really depends on your context here.

götze · « **Reply #1788 on:** June 11, 2013, 05:18:33 pm »

yes for diff questions

b^3 · « **Reply #1789 on:** June 11, 2013, 05:31:10 pm »

You're normally meant to manipulate the expression to find your original expression inside it, or something along those lines in that case. Do you have an example..? (It's not something that you would normally rely on the calc to do).

götze · « **Reply #1790 on:** June 11, 2013, 05:41:07 pm »

a cubiod has a total surface area of 150cm^2with a square base side of x cm
a)show that the height h cm of the cubiod is given by 75-x^2/2x
b) express the volume in of the cubiod in terms of x
this question has me really bugged so thanks for the help appreciate it

b^3 · « **Reply #1791 on:** June 11, 2013, 05:49:11 pm »

a) Surface area of the cuboid, we have 4 faces that have length $x$ and height $h$ while we have 2 faces which have width $x$ and lenght $x$ , so our total surface area will be what?

Spoiler

$\begin{alignedat}{1}4\left(x\right)\left(h\right)+2\left(x\right)\left(x\right) & =150 \\ 4xh+2x^{2} & =150 \\ 4xh & =150-2x^{2} \\ h & =\frac{150-2x^{2}}{4x} \\ \implies h & =\frac{75-x^{2}}{2x} \end{alignedat} $

b) The volume of the cuboid will be given by $V=lwh$ , so substituting $h$ in terms of $x$ in gives what?

Spoiler

$\begin{alignedat}{1}V & =lwh \\ & =x\times x\times\left(\frac{75-x^{2}}{2x}\right) \\ & =\frac{75x-x^{3}}{2} \\ \frac{75-x^{2}}{2x} & >0 \\ 75-x^{2} & >0 \\ 0<x<\sqrt{75} \\ 0<x<5\sqrt{3} \end{alignedat} $

götze · « **Reply #1792 on:** June 11, 2013, 07:36:55 pm »

how do you find the maximum i differentiate it and then solved for 0 but the x values are 0 lol

b^3 · « **Reply #1793 on:** June 11, 2013, 07:46:08 pm »

Check your derivative.

Spoiler

$\begin{alignedat}{1}\frac{dV}{dx} & =\frac{1}{2}\left(75-3x^{2}\right)\end{alignedat} $

götze · « **Reply #1794 on:** June 11, 2013, 08:29:04 pm »

but which equation do we diff im so confused

b^3 · « **Reply #1795 on:** June 11, 2013, 08:34:52 pm »

If you want to maximise VOLUME then you differentiate the equation for VOLUME in terms of another variable, in this case we have volume in terms of $x$ , so we would differentiate the equation for the volume with respect to $x$ to find a value of $x$ for which we have maximum volume. That is we differentiate $V=\frac{1}{2}(75x-x^{3})$

Holmes · « **Reply #1796 on:** June 13, 2013, 05:05:55 pm »

Complex numbers:

If $P(z) = x^3 - 5\sqrt{3} x^2 + 13x + 31\sqrt{3}$ and $P(-\sqrt{3})=0$ , is it possible to find the other two roots over the complex field?
Thanks a lot for helping!

Daenerys Targaryen · « **Reply #1797 on:** June 13, 2013, 05:09:55 pm »

Quote from: Holmes on June 13, 2013, 05:05:55 pm

Complex numbers:

If $P(z) = x^3 - 5\sqrt{3} x^2 + 13x + 31\sqrt{3}$ and $P(-\sqrt{3})=0$ , is it possible to find the other two roots over the complex field?
Thanks a lot for helping!

$\text{If } P(-\sqrt{3})=0 \text{ then } z+\sqrt{3} \text{ is a factor too}$
$\implies z-{3} \text{ as all coefficients of z are R no.s}$
Expand these two roots, then divide into the P(z) to find the other roots

b^3 · « **Reply #1798 on:** June 13, 2013, 05:17:11 pm »

As $P(-\sqrt{3})=0$ , we know that $x+\sqrt{3}$ is a factor. So you can long divide your polynomial by this factor to give a quadratic expression.

Spoiler

$\begin{alignedat}{1} & \: x^{2}-6\sqrt{3}x+31 \\ x+\sqrt{3} & |\overline{x^{3}-5\sqrt{3}x^{2}+13x+31\sqrt{3}} \\ (-) & \:\:\underline{x^{3}+\sqrt{3}x^{2}} \\ & \:\:\:\:\:\:\:\:\:\:\:-6\sqrt{3}x^{2}+13x \\ (-) & \:\:\:\:\:\:\:\:\,\,\:\:\:\underline{-6\sqrt{3}x^{2}-18x} \\ & \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\qquad 31x \\ (-) & \:\:\:\:\:\qquad\:\:\:\:\:\:\:\:\:\;\:\:\:\:\:\qquad \underline{31x\:\:} \\ & \qquad\qquad\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:0 \end{alignedat}$

Now to factorise the resulting quadratic, complete the square, you will get a term where you don't have a minus between the bracket and the other term, but what we can do is make this term, in this case $4$ into a minus by multiplying it by $i^{2}$ (as $i^{2}=1$ , then you can factorise it nicely, and solve for $x$ .

Spoiler

$\begin{alignedat}{1}x^{2}-6\sqrt{3}x+31 & =0 \\ x^{2}-6\sqrt{3}x+\left(3\sqrt{3}\right)^{2}-\left(3\sqrt{3}\right)^{2}+31 & =0 \\ \left(x-3\sqrt{3}\right)^{2}+31-27 & =0 \\ \left(x-3\sqrt{3}\right)^{2}+4 & =0 \\ \left(x-3\sqrt{3}\right)^{2}-4i^{2} & =0 \\ \left(x-3\sqrt{3}-2i\right)\left(x-3\sqrt{3}+2i\right) & =0 \\ x=3\sqrt{3}\pm2i \end{alignedat}$

EDIT: Also you have $P(z)=x.....$ so not sure if it's meant to be $P(z)$ or $P(x)$ .

Holmes · « **Reply #1799 on:** June 13, 2013, 05:52:41 pm »

THANKYOU so much! You're superb, this was brilliantly worked out.

Quote from: b^3 on June 13, 2013, 05:17:11 pm

Also you have $P(z)=x.....$ so not sure if it's meant to be $P(z)$ or $P(x)$ .

Sorry about that, I should have kept it to one of the two variables.

Thanks a lot!

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