Author Topic: VCE Methods Question Thread! (Read 5724553 times) Tweet Share

brightsky · « **Reply #2070 on:** June 10, 2013, 03:12:35 pm »

y=(2x-1)/(x+1) = ((2x+2)-3)/(x+1) = 2 - 3/(x+1)
so you have a rectangular hyperbola with asymptotes x = -1, and y = 2. the curves are in the 2nd and 4th 'quadrants'. now on draw, on the same set of axes, the lines y = 0 and y = 2. we want to find the x-values for which y = (2x-1)/(x+1) is between these two horizontal lines. so...

intersection: (2x-1)/(x+1) = 0, x = 1/2. so from the graph we conclude that x > 1/2.

2^(3y) -2^(2y+3) - 2^(y+2) + 2^5 = 0
(2^y)^3 - 8(2^y)^2 - 4(2^y) +32 = 0
let x = 2^y
x^3 - 8x^2 - 4x + 32 = 0
etc

i'm assuming that last term of the polynomial is +2^5 as opposed to -2^5...which makes more sense given the way in which the question is structured. let me know if it wasn't a typo..

Zealous · « **Reply #2071 on:** June 10, 2013, 03:31:49 pm »

Quote from: shadows on June 10, 2013, 01:31:53 pm

V= 6pi(r)^2 + (2/3)pi(r)^3 r=4

Show that a increase of q% in the radius will result in 2.3% of the volume.

Thank you.

$V=6\pi{r}^2 +\frac{2}{3}\pi{r}^3$

Original volume when r = 4:
$V=6\pi({4}^2) +\frac{2}{3}\pi({4}^3) = 435.634$

An increase of 1% of the radius will be:

$4 \times(1+\frac{1}{100}) = 4.04$

Sub in r = 4.04:
$V=6\pi({4.04}^2) +\frac{2}{3}\pi({4.04}^3) = 445.758$

$\frac{new\hspace{2 mm}volume}{original\hspace{2 mm}volume}=\frac{445.758}{435.634}=1.02324$

1.023 is equivalent to a 2.3% increase.

fleet street · « **Reply #2072 on:** June 10, 2013, 04:59:15 pm »

Hi, I have a calculus question:

A right circular cylinder is placed inside a sphere of radius 5cm. Find the largest possible surface area of the cylinder.

If anyone is able to help with this, then thank you very much!

EspoirTron · « **Reply #2073 on:** June 10, 2013, 06:56:33 pm »

Quote from: brightsky on June 10, 2013, 03:12:35 pm

y=(2x-1)/(x+1) = ((2x+2)-3)/(x+1) = 2 - 3/(x+1)
so you have a rectangular hyperbola with asymptotes x = -1, and y = 2. the curves are in the 2nd and 4th 'quadrants'. now on draw, on the same set of axes, the lines y = 0 and y = 2. we want to find the x-values for which y = (2x-1)/(x+1) is between these two horizontal lines. so...

intersection: (2x-1)/(x+1) = 0, x = 1/2. so from the graph we conclude that x > 1/2.

2^(3y) -2^(2y+3) - 2^(y+2) + 2^5 = 0
(2^y)^3 - 8(2^y)^2 - 4(2^y) +32 = 0
let x = 2^y
x^3 - 8x^2 - 4x + 32 = 0
etc

i'm assuming that last term of the polynomial is +2^5 as opposed to -2^5...which makes more sense given the way in which the question is structured. let me know if it wasn't a typo..

Hey thanks for that, but it was -2^5, so how would I go about it using my previous answer?

fleet street · « **Reply #2074 on:** June 10, 2013, 07:05:41 pm »

Quote from: kezzab on June 10, 2013, 06:56:33 pm

Hey thanks for that, but it was -2^5, so how would I go about it using my previous answer?

I'm pretty sure that it was a typo. The exact answer to your question (see: https://www.wolframalpha.com/input/?i=2^%283y%29%20-2^%282y%2B3%29%20-%202^%28y%2B2%29%20-%202^5%20%3D%200)) is extremely long, and I have a feeling it involves using the cubic equation for 2^y.

Alwin · « **Reply #2075 on:** June 10, 2013, 08:40:31 pm »

Quote from: fleet street on June 10, 2013, 04:59:15 pm

Hi, I have a calculus question:

A right circular cylinder is placed inside a sphere of radius 5cm. Find the largest possible surface area of the cylinder.

If anyone is able to help with this, then thank you very much!

Firstly, draw a 2D cross section diagram:

Now, consider the red triangle. From pythagoras,
$r^2+(\frac{h}{2})^2 = 5^2$
$r^2=25-(\frac{h}{2})^2$
$\therefore r=\sqrt{25-(\frac{h}{2})^2}$
Note that only positive square root is chosen since r>0

The surface area of the cylinder is given by:
$SA=2\pi r^2 + 2 \pi r h$

From the previous equation linking r and h,
$SA=2\pi (25-(\frac{h}{2})^2) + 2\pi \sqrt{25-(\frac{h}{2})^2} h$
$SA = \frac{1}{2} \pi (-h^2+2 \sqrt{100-h^2} h+100)$ by collecting like terms.

To find the maximum value of SA, differentiate. Note I have replace SA with A for simplicity.
$\frac{dA}{dh} = -\frac{2 \pi h^2}{\sqrt{100-h^2}}+\frac{100 \pi}{\sqrt{100-h^2}}-\pi h$

Solving for stationary points,
$-\frac{2 \pi h^2}{\sqrt{100-h^2}}+\frac{100 \pi}{\sqrt{100-h^2}}-\pi h=0$

$h = -\sqrt{10 (5+\sqrt{5})}$ Reject obviously, since h>0
$\therefore h = \sqrt{50-10 \sqrt{5}}$

Substituting this value of h back into SA,
$2\pi (25-(\frac{h}{2})^2) + 2\pi \sqrt{25-(\frac{h}{2})^2} h = 25 (1+\sqrt{5}) \pi \text{ at } h = \sqrt{50-10 \sqrt{5}}$

You can prove this is a minimum if you want, but there is a graph just in case you don't believe me

Hence, the largest possible value surface area of the cylinder is $25 (1+\sqrt{5}) \pi cm^2$
provided I didn't make a mistake somewhere of course

fleet street · « **Reply #2076 on:** June 11, 2013, 07:47:44 am »

I understand most of this, but how did you solve
$-\frac{2 \pi h^2}{\sqrt{100-h^2}}+\frac{100 \pi}{sqrt{100-h^2}}-\pi h=0$

Alwin · « **Reply #2077 on:** June 11, 2013, 08:38:22 am »

Quote from: fleet street on June 11, 2013, 07:47:44 am

I understand most of this, but how did you solve
$-\frac{2 \pi h^2}{\sqrt{100-h^2}}+\frac{100 \pi}{sqrt{100-h^2}}-\pi h=0$

Apologies, I made a bit of a typo and missed the "\" (went back and changed it). It's meant to be:
$-\frac{2 \pi h^2}{\sqrt{100-h^2}}+\frac{100 \pi}{\sqrt{100-h^2}}-\pi h=0$

I just CASed it when I solved it the first time. However, it is possible by hand:

(From Wolfram Alpha) -Sorry, I have school in a couple of minutes, no time to LaTex. It just gives you an idea of how to do it

EDIT: Reuploaded the image using a third party website. Sorry b^3, the image just disappeared.. randomly. Good suggestion though, here is the original link: http://www.wolframalpha.com/input/?i=-%5Cfrac%7B2+%5Cpi+h%5E2%7D%7B%5Csqrt%7B100-h%5E2%7D%7D%2B%5Cfrac%7B100+%5Cpi%7D%7B%5Csqrt%7B100-h%5E2%7D%7D-%5Cpi+h%3D0&h=1 You have to sign up to the site to get the working tho..

b^3 · « **Reply #2078 on:** June 11, 2013, 01:26:25 pm »

Since his image has disappeared, (and well he could have just linked the wolfram link where he got it from anyway -.-), I'll add the working by hand here, but this is a question that you should be using CAS for.
$\begin{alignedat}{1}-\frac{2\pi h^{2}}{\sqrt{100-h^{2}}}+\frac{100\pi}{\sqrt{100-h^{2}}}-\pi h & =0 \\ -2\pi h^{2}+100\pi-\pi h\sqrt{100-h^{2}} & =0 \\ -2\pi h^{2}+100\pi & =\pi h\sqrt{100-h^{2}} \\ \left(-2\pi h^{2}+100\pi\right)^{2} & =\pi^{2}h^{2}\left(100-h^{2}\right) \\ \text{We need to be careful about adding} \\ \text{new solutions when we do this, so} \\ \text{as the RHS is always positive} \\ -2\pi h^{2}+100\pi>0 \\ 2\pi\left(50-h^{2}\right)>0 \\ 0<h<5\sqrt{2} \\ 4\pi^{2}h^{4}-400\pi^{2}h^{2}+10000\pi^{2} & =100\pi^{2}h^{2}-\pi^{2}h^{4} \\ 5\pi^{2}h^{4}-500\pi^{2}h^{2}+10000\pi^{2} & =0 \\ 5\pi^{2}\left(h^{4}-100h^{2}+2000\right) & =0 \\ \text{Let }u=h^{2} \\ u^{2}-100u+2000 & =0 \\ u^{2}-100u+\left(50\right)^{2}-\left(50\right)^{2}+2000 & =0 \\ \left(u-50\right)^{2}-500 & =0 \\ \left(u-50-\sqrt{5\times100}\right)\left(u-50+\sqrt{5\times100}\right) & =0 \\ \left(u-50-10\sqrt{5}\right)\left(u-50+10\sqrt{5}\right) & =0 \\ u & =50\pm10\sqrt{5} \\ h^{2} & =50\pm10\sqrt{5} \\ h & =\pm\sqrt{50\pm10\sqrt{5}} \\ \text{but }0<h<5\sqrt{2} \\ \implies h & =\sqrt{40-10\sqrt{5}} \end{alignedat} $

fleet street · « **Reply #2079 on:** June 11, 2013, 02:46:40 pm »

Quote from: b^3 on June 11, 2013, 01:26:25 pm

Since his image has disappeared, (and well he could have just linked the wolfram link where he got it from anyway -.-), I'll add the working by hand here, but this is a question that you should be using CAS for.

Thanks! The latex must have taken you forever! That's an interesting way of solving!

Quote from: Alwin on June 11, 2013, 08:38:22 am

Apologies, I made a bit of a typo and missed the "\" (went back and changed it). It's meant to be:
$-\frac{2 \pi h^2}{\sqrt{100-h^2}}+\frac{100 \pi}{\sqrt{100-h^2}}-\pi h=0$

I just CASed it when I solved it the first time. However, it is possible by hand:

(From Wolfram Alpha) -Sorry, I have school in a couple of minutes, no time to LaTex. It just gives you an idea of how to do it

Thanks! Too bad the image didn't work out though.

b^3 · « **Reply #2080 on:** June 11, 2013, 04:31:42 pm »

Quote from: Alwin on June 11, 2013, 08:38:22 am

EDIT: Reuploaded the image using a third party website. Also, b^3 this is what happened to the original image: http://www4c.wolframalpha.com/Calculate/MSP/MSP17691fdbfbh7ah3d17f600005d8ea9g787081446?MSPStoreType=image/gif&s=58&w=312.&h=131.&cdf=RangeControl... You have to sign up to the site to get the working too..

I know, I meant the wolfram link for the solving, it's worth it to sign up for wolfram anyways, then they could get the working from there then anyways.

ashoni · « **Reply #2081 on:** June 11, 2013, 04:56:20 pm »

was just wondering how to do this question..
solve for x and y.

log base 3 of x + 2log base 9 of y = 3

x + y = 12

thanks in advance!!

Alwin · « **Reply #2082 on:** June 11, 2013, 05:08:30 pm »

Quote from: ashoni on June 11, 2013, 04:56:20 pm

was just wondering how to do this question..
solve for x and y.

log base 3 of x + 2log base 9 of y = 3

x + y = 12

thanks in advance!!

3 Important log laws for this question:
$\log_{a}b = \frac{\log_{c} b}{\log_{c} a}$ (change of base)
$\log_{a}n + \log_{a}m=\log_{a}mn$ (addition of two logs)
$\log_{a} a =1$

Now to your question,
$\log_{3} x + 2 \log_{9} y = 3$

$\log_{3} x + 2\cdot \frac{\log_{3} y}{\log_{3} 9} = 3\cdot\log_{3}3$

$\log_{3} x + 2\cdot \frac{\log_{3} y}{2} = \log_{3}3^{3}$

$\log_{3} x + \log_{3}y = \log_{3}27$

$\log_{3} xy = \log_{3}27$

$\therefore xy=27$

You now have two simultaneous equations,

$x+y=12 ...[1]$
$xy = 27 ...[2]$

I think you can do it from here

Otherwise, see the spoiler for the answer:

Spoiler

$x = 3, y = 9 \text{ or } x = 9, y = 3$

Holmes · « **Reply #2083 on:** June 13, 2013, 05:16:17 pm »

If $f(x)=(1/x)+2$ and $g(x) = 1/(x-2)$ , how do we find the range of $gof(x)$ ? And do restrictions come into play in the range?
Thanks a lot!

b^3 · « **Reply #2084 on:** June 13, 2013, 05:25:29 pm »

For the composite function to be defined we need 'the range of the second to be equal to or a subset of the domain of the first', way to remember it : R2D1.
So here what are our ranges and domains for $f$ and $g$ ?

Spoiler

$\text{Dom \:} f=R\setminus \{0\}$
$\text{Ran \:}f=R \setminus\{2\}$
$\text{Dom \:} g=R\setminus\{2\}$
$\text{Ran \:} g=R\setminus\{0\}$

So we need $\text{Ran \:} f\subseteq \text{dom \:} g$
In our case they are both equal, so we don't need to restrict the function, but the domain will be the domain of the second, that is the domain of $f$ , so from there you should be able to do a little sketch and work out the range of the resulting function with that domain.

Hope that helps.

EDIT: My bad, read fog not gof, fixed.

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Author Topic: VCE Methods Question Thread! (Read 5724553 times) Tweet Share

brightsky

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