2 unit maths revision (1 Viewer)

hscishard · Feb 21, 2011

jm01 said:
You need to split the area into two integrals. Take one of them with abs value maybe.

Oh yea. Didn't think that 1.5 is less than e. Still, you won't have equal strip widths for this question. This questions pretty hard

Anyway your fav question:
Evaluate: |x^2-4|= -1

Bored Of Fail1 · Feb 22, 2011

New Question

Find the period for y=tan(-3x)

SpiralFlex · Feb 22, 2011

Squishxmishyx said:
Thank you Matthew Goodwin, you're my hero!
And SpiralFlex, are you doing maths accelerated o.o.
How come you know how to do this...

I'll continue the questions so I don't seem like a whore.
The distance between the sun and the earth is approximately $1.49 \times 10^{8} km$ . The sun subtends an angle of approximately $31^{o}$ at a point on the earth. Calculate the approximate diameter of the sun.

No, I am not doing acceleration. I did three unit in Year 9-10 (not the course), I was bored.

SpiralFlex · Feb 22, 2011

Period = $60$ degrees.

Existential · Feb 23, 2011

on a slightly related topic, what 2U topic are we all studying and how many HSC 2U topics have we all done so far?

Riproot · Feb 23, 2011

Existential said:
on a slightly related topic, what 2U topic are we all studying and how many HSC 2U topics have we all done so far?

In 2 unit, I have done 2 many 2 count.
I think we'll be finished with 2 unit soon.

slyhunter · Feb 23, 2011

New question.

Show that the volume of the cone generated when the line y=mx is rotated about the x-axis over the interval from x=0 to x=h, where h is the height of the cone, is $\frac{1}{3}\pi r^{2}h$

Riproot · Feb 23, 2011

slyhunter said:
New question.

Show that the volume of the cone generated when the line y=mx is rotated about the x-axis over the interval from x=0 to x=h, where h is the height of the cone, is $\frac{1}{3}\pi r^{2}h$

$A=\pi\int_{0}^{h}m^2x^2dx$
$A=\pi[\frac{m^2x^3}{3}]\frac{h}{0}]$
$A=\pi[\frac{m^2h^3}{3}-\frac{0}{3}]$
$m=r$
$\therefore A=\pi\frac{r^2h}{3}$
$\therefore A=\frac{1}{3}\pi r^2h$

Squishxmishyx · Feb 23, 2011

Evaluate the following limits:

A) $\lim_{x \to 0 } \! \frac{ 3x}{sin2x}$

B) $\lim_{x \to 0 } \! \frac{ x^{2}}{2-2 cos 2x}$

Riproot · Feb 23, 2011

Squishxmishyx said:
Evaluate the following limits:

A) $\lim_{x \to 0 } \! \frac{ 3x}{sin2x}$

B) $\lim_{x \to 0 } \! \frac{ x^{2}}{2-2 cos 2x}$

No.

boredoffail6 · Feb 23, 2011

Riproot said:
No.

you are shit

ohexploitable · Feb 23, 2011

Squishxmishyx said:
Evaluate the following limits:

A) $\lim_{x \to 0 } \! \frac{ 3x}{sin2x}$

$\\=\frac32\times\lim_{x\to0}\frac{2}{\sin2x}=\frac32$

B) $\lim_{x \to 0 } \! \frac{ x^{2}}{2-2 cos 2x}$

$\\=\lim_{x\to0}\frac{x^2}{4\sin^2x}=\frac14$

I don't think this is 2U...

slyhunter · Feb 23, 2011

Yes, it's 3U.

Riproot · Feb 23, 2011

boredoffail6 said:
you are shit

You're fugly.
Also, no one picked up on my earlier mistake.

jessica2011hsc · Feb 23, 2011

<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{-1}^{0}x(x- 1)(x @plus; 1)dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{-1}^{0}x(x- 1)(x + 1)dx" title="\int_{-1}^{0}x(x- 1)(x + 1)dx" /></a>

by expanding the brackets where necessary, evaluate the following definite integral

slyhunter · Feb 23, 2011

$\int_{-1}^{0}x(x- 1)(x + 1)dx$

$=\int_{-1}^{0}(x^3-x)dx$

$=\left [ \frac{x^{4}}{4}-\frac{x^2}{2} \right ]^{0}_{-1}$

$=\frac{1}{4}$

slyhunter · Feb 23, 2011

jessica2011hsc · Feb 23, 2011

okay maybe that was too easy haha
what about:
a) show that <a href="http://www.codecogs.com/eqnedit.php?latex=\int_{2}^{k}3dx=3k -6" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{2}^{k}3dx=3k -6" title="\int_{2}^{k}3dx=3k -6" /></a>

b) hence find the value of k if<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{2}^{k}3dx=18" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{2}^{k}3dx=18" title="\int_{2}^{k}3dx=18" /></a>

SpiralFlex · Feb 23, 2011

a) = $3x$

= $3k - 3(2)$
= $3k - 6$

b) $3k - 6 = 18$

$k = 8$

boredoffail6 · Feb 23, 2011

jessica2011hsc said:
okay maybe that was too easy haha

ohh the irony

you say that then you post an even easier question XD

2 unit maths revision (1 Viewer)

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