Author Topic: VCE Methods Question Thread! (Read 6018360 times) Tweet Share

Stevensmay · « **Reply #9120 on:** March 08, 2015, 02:53:30 pm »

Quote from: Apink! on March 08, 2015, 02:27:45 pm

Hello!
Could someone please help me with this question?
if x+ y =m and x-y =n, then (x²-y²)-2x is equal to (find in terms of m and n)

Thank you!

Spoilers!

Spoiler

$x^2 - y^2 - 2x = (x+y)(x-y)-2x$

$n*m-2x = n*m - x - x$

$n*m - (m-y) - (y-n) = n*m -m+y-y+n$

$n*m -m +n$

ally12579 · « **Reply #9121 on:** March 08, 2015, 03:04:05 pm »

Hi
Could someone help me with this question
8e^-x −e^x =2

Thanks

lzxnl · « **Reply #9122 on:** March 08, 2015, 03:08:34 pm »

Let e^x = u, solve the resulting quadratic for u, solve for x again

Stevensmay · « **Reply #9123 on:** March 08, 2015, 03:12:13 pm »

Quote from: ally12579 on March 08, 2015, 03:04:05 pm

Hi
Could someone help me with this question
8e^-x −e^x =2

Thanks

Spoiler

$8e^{-x}-e^{x}=2$
Multiply through by e^x
$8-e^{2x}=2e^{x}$
Rearrange and let a=e^x
$0=a^{2} + 2a - 8$
Factorize
$0=(a+4)(a-2)$
So now we have a=e^x= -4 or 2 and we know e^x > 0 so the only solution is a = 2

$a = 2 = e^{x}$
$x = \log{2}$

knightrider · « **Reply #9124 on:** March 08, 2015, 03:14:02 pm »

How would you do this question?

Eiffel · « **Reply #9125 on:** March 08, 2015, 03:57:34 pm »

with coordinates is it ever alright to use square brackets ? or must it always be round, with endpoints with square brackets?

what about a square root endpoint of 5,0 with no reflections is the domain [5,inf) or (5,0) given f: r - > r

kinslayer · « **Reply #9126 on:** March 08, 2015, 04:15:38 pm »

Quote from: knightrider on March 08, 2015, 03:14:02 pm

How would you do this question?

The volume of the cylinder is $\pi r^2 h$ where r is the radius of the cylinder and h is its height.

The relationship between the height and radius of the cylinder can be found by starting at the centre of the cylinder/circle, drawing a right-angled triangle with sidelengths $r$ (from the centre out horizontally), height $\frac h2$ (from the edge of the cylinder up to the top), and hypotenuse 8 cm (from the centre radially out towards the edge of the sphere).

Then we have that $r^2 + \frac{h^2}{4} = 8^2 = 64$ , or $r^2 = 64 - \frac{h^2}{4}$ . Now substitute into the volume function to get $V(h) = \pi\left(64 - \frac{h^2}{4} \right) h = 64h \pi - \frac{\pi h^3}{4}.$

Now that we have a function in one variable we can differentiate to find its maximum: $V^\prime (h) &= 64\pi - \frac{3\pi h^2}{4}$ . Setting this equal to zero we have

$\begin{alignedat}1 \frac{3}{4} \pi h^2 - 64\pi &= 0 \\ h^2 - \frac{256}{3} &= 0 \\ \left(h - \frac{16}{\sqrt 3}\right)\left(h + \frac{16}{\sqrt 3}\right) &= 0\\ h = \frac{16\sqrt 3}{3} \end{alignedat}$

(note h > 0).

Finally,the volume is
$\begin{alignedat}1V &= \frac{64\pi \times 16\sqrt 3}{3} - \frac{\pi (16\sqrt 3)^3}{4\times 27} \\ &= \frac{\pi\sqrt 3(3\times 64 \times 16 - 16^3/4)}{9} \\ &= \frac{2048 \pi \sqrt 3}{9}\;\mathrm{cm}^{3} \\ &\approx 1238.22\; \mathrm{cm}^{3} \end{alignedat}$

Eiffel · « **Reply #9127 on:** March 08, 2015, 04:43:19 pm »

if a parabola has turning point (z,1), y int of -15 and one x int of 5 what is the value of z.

got up to 0 = -2az+5a-3

Stevensmay · « **Reply #9128 on:** March 08, 2015, 04:58:09 pm »

Quote from: Eiffel on March 08, 2015, 04:43:19 pm

if a parabola has turning point (z,1), y int of -15 and one x int of 5 what is the value of z.

got up to 0 = -2az+5a-3

$y=ax^2+bx+c$

Substitute in (0,-15), (z,1) and (5,0) to give
$-15=c$ and $0 = 25a + 5b -15$ and $1 = az^2 + bz -15$

We also have the turning point so derivative is
$y' = 2ax + b$
$0 = 2az + b$

Rearrange z in terms of a and b from the last equation
$z = -b/(2a)$

Rearrange and you should be able to solve for z.

knightrider · « **Reply #9129 on:** March 08, 2015, 05:01:09 pm »

Quote from: kinslayer on March 08, 2015, 04:15:38 pm

The volume of the cylinder is $\pi r^2 h$ where r is the radius of the cylinder and h is its height.

The relationship between the height and radius of the cylinder can be found by starting at the centre of the cylinder/circle, drawing a right-angled triangle with sidelengths $r$ (from the centre out horizontally), height $\frac h2$ (from the edge of the cylinder up to the top), and hypotenuse 8 cm (from the centre radially out towards the edge of the sphere).

Then we have that $r^2 + \frac{h^2}{4} = 8^2 = 64$ , or $r^2 = 64 - \frac{h^2}{4}$ . Now substitute into the volume function to get $V(h) = \pi\left(64 - \frac{h^2}{4} \right) h = 64h \pi - \frac{\pi h^3}{4}.$

Now that we have a function in one variable we can differentiate to find its maximum: $V^\prime (h) &= 64\pi - \frac{3\pi h^2}{4}$ . Setting this equal to zero we have

$\begin{alignedat}1 \frac{3}{4} \pi h^2 - 64\pi &= 0 \\ h^2 - \frac{256}{3} &= 0 \\ \left(h - \frac{16}{\sqrt 3}\right)\left(h + \frac{16}{\sqrt 3}\right) &= 0\\ h = \frac{16\sqrt 3}{3} \end{alignedat}$

(note h > 0).

Finally,the volume is
$\begin{alignedat}1V &= \frac{64\pi \times 16\sqrt 3}{3} - \frac{\pi (16\sqrt 3)^3}{4\times 27} \\ &= \frac{\pi\sqrt 3(3\times 64 \times 16 - 16^3/4)}{9} \\ &= \frac{2048 \pi \sqrt 3}{9}\;\mathrm{cm}^{3} \\ &\approx 1238.22\; \mathrm{cm}^{3} \end{alignedat}$

Thank you so much kinslayer

really awesome explanation

Eiffel · « **Reply #9130 on:** March 08, 2015, 05:10:33 pm »

Quote from: Stevensmay on March 08, 2015, 04:58:09 pm

$y=ax^2+bx+c$

Substitute in (0,-15), (z,1) and (5,0) to give
$-15=c$ and $0 = 25a + 5b -15$ and $1 = az^2 + bz -15$

We also have the turning point so derivative is
$y' = 2ax + b$
$0 = 2az + b$

Rearrange z in terms of a and b from the last equation
$z = -b/(2a)$

Rearrange and you should be able to solve for z.

could you show me a non calculus way please?

Stevensmay · « **Reply #9131 on:** March 08, 2015, 05:17:05 pm »

Quote from: Eiffel on March 08, 2015, 05:10:33 pm

could you show me a non calculus way please?

Instead of the derivative part you can use the idea that the turning point is given by (-b/2a, f(-b/2a)) from year 9/10 theory.

This gives you the same end result as the derivative step though, except you just recall memorized theory.

Floatzel98 · « **Reply #9132 on:** March 08, 2015, 05:25:59 pm »

I need help with d) ii). Are we supposed to look at the graphs or equate and solve the two equations? I'm kind of lost with it.

Eiffel · « **Reply #9133 on:** March 08, 2015, 06:30:58 pm »

Quote from: Stevensmay on March 08, 2015, 05:17:05 pm

Instead of the derivative part you can use the idea that the turning point is given by (-b/2a, f(-b/2a)) from year 9/10 theory.

This gives you the same end result as the derivative step though, except you just recall memorized theory.

could someone please show me full working out, i still dont get it :/

wunderkind52 · « **Reply #9134 on:** March 08, 2015, 06:44:06 pm »

Quote from: Eiffel on March 08, 2015, 06:30:58 pm

could someone please show me full working out, i still dont get it :/

Consider a parabola $y=ax^2+bx+c$
This can be rewritten as $y=a(x^2+\frac{b}{a}x+\frac{c}{a}$
Completing the square, you get $a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a}$
Rewritten as $y=a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a})$
Or $y=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c$
Now it is in turning point form, with a TP at $(\frac{-b}{2a},\frac{-b^2}{4a}+c)$

Just a fun fact (or not so fun) - Quadratic formula can be deduced from here -
$a((x+\frac{b}{2a})^2-\frac{b^2}{4a}+c)=0$
Rearranging
$a(x+\frac{b}{2a})^2=\frac{b^2}{4a}-c$
Simplifying the fraction -
$a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}$
Dividing both sides by a
$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$
Taking the square root of both sides
$x+\frac{b}{2a}=+-\sqrt{\frac{b^2-4ac}{4a^2}}$
Isolating x
$x=-\frac{b}{2a}+-\sqrt{\frac{b^2-4ac}{4a^2}}$
Just simplifying further -
$x=\frac{-b+-\sqrt{b^2-4ac}}{2a}$

Search the forums now!

We have moved!

Author Topic: VCE Methods Question Thread! (Read 6018360 times) Tweet Share

Stevensmay

Re: VCE Methods Question Thread!

ally12579

Re: VCE Methods Question Thread!

lzxnl

Re: VCE Methods Question Thread!

Stevensmay

Re: VCE Methods Question Thread!

knightrider

Re: VCE Methods Question Thread!

Eiffel

Re: VCE Methods Question Thread!

kinslayer

Re: VCE Methods Question Thread!

Eiffel

Re: VCE Methods Question Thread!

Stevensmay

Re: VCE Methods Question Thread!

knightrider

Re: VCE Methods Question Thread!

Eiffel

Re: VCE Methods Question Thread!

Stevensmay

Re: VCE Methods Question Thread!

Floatzel98

Re: VCE Methods Question Thread!

Eiffel

Re: VCE Methods Question Thread!

wunderkind52

Re: VCE Methods Question Thread!

Recent Posts

User:
Password: