how big a change is yr 11 maths to yr 10 maths (1 Viewer)

ZombieApocalypse · Jul 16, 2011

Drongoski said:
see guys! Why waste money on tutors? Use Maths On-Line (courtesy Maccas).

My school made an account for everyone who studies maths at our school. I don't think they have 4u content though...

benji_10 · Jul 16, 2011

AAEldar said:
I'm not exactly sure which way you did it, but the answer is right (you have a typo though, should be $\frac{x^3}{3}$ , which is obviously what you meant from the line before).

$\int \frac{x^6-1}{x^4+x^3-x-1}dx$
$=\int \frac{(x+1)(x^2-x+1)(x^3-1)}{x(x^3-1)+1(x^3-1)}dx$
$=\int \frac{(x+1)(x^2-x+1)(x^3-1)}{(x+1)(x^3-1)}dx$
$=\int (x^2-x+1)dx$
$\therefore\ =\frac{x^3}{3} - \frac{x^2}{2} +x+C$

And a fundamental mechanics question: Prove that the angle of projection required for maximum range is $\frac{\pi}{4}$ radians, assuming that the projectile is launched from the real-life equivalent of the origin, (0,0) (not that it really matters).

Oh. and a fun question. True or false, without using calculus -
$\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} (tanx-cotx)dx = 0$

Drongoski · Jul 16, 2011

bleakarcher said:
prairiewood high school

Thought you were from Ruse or something.

bleakarcher · Jul 16, 2011

post the solution to that mechanics question. i dont much about maths extension II projectile motion

benji_10 · Jul 16, 2011

Lol it's 3U, not 4U projectile motion. I didn't mean 4U mechanics when I posted the question. My bad. You just apply the formulae for the x and y directions in creative ways then it basically turns into a max/min problem.

I'll give you a hint bleak, as it's not fun spilling the beans on someone so quick, especially since you're supposed to be owning us:

$y=\frac{-g}{2}t^2+Vtsin\alpha$ , where $y$ is horizontal displacement, $\alpha$ is the angle of projection, $g$ is the acceleration due to gravity, $t$ is time, and $V$ initial velocity. Let y=0. Remember, you're trying to find the angle such that it gives maximum range.

bleakarcher · Jul 16, 2011

y=(-1/2)gt^2+Vtsinθ
Range occurs when y=0,
t=(2Vsinθ)/g
Now, x=Vtcosθ
Range occurs when y=0 ie when t=(2Vsinθ)/g,
x=V[(2Vsinθ)/g]cosθ
x=[2V^2sinθcosθ]/g
x=[V^2sin(2θ)]/g
dx/dθ=[2V^2cos(2θ)]/g
Maximum range occurs when dx/dθ=0,
2V^2cos(2θ)/g=0
cos(2θ)=0
2θ=90
θ=45=pi/4 radians

benji_10 · Jul 16, 2011

bleakarcher said:
y=(-1/2)gt^2+Vtsinθ
Range occurs when y=0,
t=(2Vsinθ)/g
Now, x=Vtcosθ
Range occurs when y=0 ie when t=(2Vsinθ)/g,
x=V[(2Vsinθ)/g]cosθ
x=[2V^2sinθcosθ]/g
x=[V^2sin(2θ)]/g
dx/dθ=[2V^2cos(2θ)]/g
Maximum range occurs when dx/dθ=0,
2V^2cos(2θ)/g=0
cos(2θ)=0
2θ=90
θ=45=pi/4 radians

One problem: You didn't answer the question properly. You forgot to prove that the stationary point at θ=45 is a max T.P
It's incredibly important that you prove that the stationary point is either a max or a min T.P. If you don't prove it, then you've just stated a value which could either give a max, min or an invalid answer, depending on whether there were limits placed. Hence, not giving what the markers want.

bleakarcher · Jul 16, 2011

ah ok thanks for that

bleakarcher · Jul 16, 2011

how do u do the second one? graph?

benji_10 · Jul 16, 2011

I did it with a graph. I don't think there's any other way to do it.

cutemouse · Jul 16, 2011

benji_10 said:
Oh. and a fun question. True or false, without using calculus -
$\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} (tanx-cotx)dx = 0$

True, using the result $\int^b_a f(x) dx = \int^b_a f(a+b-x)dx$

benji_10 · Jul 16, 2011

cutemouse said:
True, using the result $\int^b_a f(x) dx = \int^b_a f(a+b-x)dx$

Can that result be quoted in exams, or do you have to prove it first?

deterministic · Jul 16, 2011

Not quite right... try using the fact $cot(\theta)=tan(\frac{\pi}{2}-\theta)$ and an appropriate substitution

bleakarcher · Jul 16, 2011

next question

Shadowdude · Jul 16, 2011

cutemouse said:
True, using the result $\int^b_a f(x) dx = \int^b_a f(a+b-x)dx$

I can't immediately see how that result fits the question though.

Next question? Okay, let's delve away from integration then:

$\textup{Show\;that\;the\;distance\;from\;the\;origin\;to\;the\;line}\;Ax+By+C=0\;\textup{is\;given\;by:} d(0,l)=\frac{|C|}{\sqrt{A^{2}+B^{2}}}$

cutemouse · Jul 16, 2011

benji_10 said:
Can that result be quoted in exams, or do you have to prove it first?

I don't know. Depends on your teacher.

In the HSC you probably wont get this type of question exactly though. There'll be a step before it or give you a hint...

cutemouse · Jul 16, 2011

Shadowdude said:
I can't immediately see how that result fits the question though.

Because 3pi/8 + pi/8 = pi/2

So the second term becomes cot (pi/2 -x) = tan x

So you're integrating tanx -tanx ie. 0 so integral is 0.

benji_10 · Jul 16, 2011

cutemouse said:
Because 3pi/8 + pi/8 = pi/2

So the second term becomes cot (pi/2 -x) = tan x

So you're integrating tanx -tanx ie. 0 so integral is 0.

=_=

If only I knew that. The I wouldn't have had to waste 5 min graphing that damn graph. I could've done another question... Probably failed the test >_>
And shadow, could you perhaps rewrite what you were trying to write in tex? I don't understand what you wrote.

And I'm gong to chuck random questions from the test I did today until bleak folds.

$Use\ calculus\ to\ prove\ that\ the\ inequality\ lnx<\sqrt{x}\ is\ true\ for\ any\ real\ positive\ number\ x.$

This one was a 7 marker

Shadowdude · Jul 16, 2011

benji_10 said:
=_=

If only I knew that. The I wouldn't have had to waste 5 min graphing that damn graph. I could've done another question... Probably failed the test >_>
And shadow, could you perhaps rewrite what you were trying to write in tex? I don't understand what you wrote.

And I'm gong to chuck random questions from the test I did today until bleak folds.

$Use\ calculus\ to\ prove\ that\ the\ inequality\ lnx<\sqrt{x}\ is\ true\ for\ any\ real\ positive\ number\ x.$

This one was a 7 marker

1. That was in tex, here it is again:

$\textup{Show\;that\;the\;distance\;from\;the\;origin\;to\;the\;line}\;Ax+By+C=0\;\textup{is\;given\;by:}$
$d(0,l)=\frac{|C|}{\sqrt{A^{2}+B^{2}}}$

2. I could cheat via the Mean Value Theorem =P

benji_10 · Jul 16, 2011

Isn't that the perpendicular distance formula?

$P(0,0)\ l:Ax+By+C$
$\therefore\ x_1=0,\ y_1=0,\ a=A,\ b=B,\ c=C$
$\begin{align*}d &=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\ \therefore\ d &=\frac{|A(0)+B(0)+C|}{\sqrt{A^2+B^2}}\\ &=\ \frac{|C|}{\sqrt{A^2+B^2}}\end{align*}$

For some stupid reason there are stupid br/ tags in my post. Solution to that would be nice.

And no cheating shadow, it's moar fun this way =)

how big a change is yr 11 maths to yr 10 maths (1 Viewer)

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