Author Topic: VCE Specialist 3/4 Question Thread! (Read 2813847 times) Tweet Share

b^3 · « **Reply #3225 on:** June 11, 2014, 08:36:53 pm »

Quote from: Zezima. on June 11, 2014, 08:09:41 pm

Hey can someone please explain how (in the attachment) my teacher knew which domain had which derivative?

As in for x > pi, for instance, how did he know the derivative was negative?

Thanks!!

It's because of the modulus function, we have
$\begin{alignedat}{1}|x| & =\begin{cases} -x & \:\text{if }x<0 \\ x & \:\text{if }x>0 \end{cases}\end{alignedat}$
But $x$ isn't defined for $-\pi<x<\pi$ , so the modulus above in regard to the question becomes
$\begin{alignedat}{1}|x| & =\begin{cases} -x & \:\text{if }x<\pi \\ x & \:\text{if }x>\pi \end{cases}\end{alignedat}$
Then pushing that through the function gives the above.

keltingmeith · « **Reply #3226 on:** June 11, 2014, 09:24:23 pm »

Zezima - that logic isn't correct for two reasons.

1. The first operation is wrong...

The derivative of tan is sec^2.
2. You need a constant of integration there.

Say - f(x) = sin[2arccos(x)], dom_sin(x) = [-pi/2, pi/2]
So, for this, we're going to say that -pi/2 <= 2arccos(x) <= pi/2
.: -pi/4 <= arccos(x) <= pi/4
-sqrt(2)/2 <= x <= sqrt(2)/2

So, we now have the implied domain of x. The range isn't as hard as you might think - sin(theta) will only ever oscillate between -1 and 1. -sqrt(2)/2 to sqrt(2) is a pretty big range, so if we guess that the range is from -1 to 1, you'd probably be right. A quick graph seems to indicated that we'd be right.

To clarify for future, if we have g(x) = circ[f(x)], where circ is some circular function, and we know that normally a <= x <= b, then the implied domain is given by a <= f(x) <= b and solving for x.

Mieow · « **Reply #3227 on:** June 11, 2014, 09:57:27 pm »

Can someone please help me with these questions:
1) Find the value of $\int_{0}^{a}2sin^{2}(3x)cos^{2}(3x)dx$
2) Find the value of $\int_{0}^{a}\frac{2x + 1}{\sqrt{5-2x}}dx$

Zealous · « **Reply #3228 on:** June 11, 2014, 10:19:23 pm »

Quote from: Mieow on June 11, 2014, 09:57:27 pm

Can someone please help me with these questions:
1) Find the value of $\int_{0}^{a}2sin^{2}(3x)cos^{2}(3x)dx$
2) Find the value of $\int_{0}^{a}\frac{2x + 1}{\sqrt{5-2x}}dx$

1: Use trigonometric identities:

$\begin{eqnarray*}\\\int_{0}^{a}2\sin^{2}(3x)\cos^{2}(3x)dx & = & \int_{0}^{a}2(\sin(3x)\cos(3x))^{2}dx\\\\\ use\ \sin(ax)\cos(ax) & = & \frac{1}{2}\sin(2ax)\\\ hence\ \sin(3x)\cos(3x) & = & \frac{1}{2}\sin(6x)\\\\\int_{0}^{a}2(\sin(3x)\cos(3x))^{2}dx & = & \int_{0}^{a}2(\frac{1}{2}\sin(6x))^{2}dx=\int_{0}^{a}\frac{1}{2}\sin^{2}(6x)dx\\\ use\ \sin^{2}(x) & = & \frac{1}{2}(1-\cos(2x))\\\ hence\ \sin^{2}(6x) & = & \frac{1}{2}(1-\cos(12x))\\\int_{0}^{a}\frac{1}{2}\sin^{2}(6x)dx & = & \int_{0}^{a}\frac{1}{2}[\frac{1}{2}(1-\cos(12x))]dx=\frac{1}{4}\int_{0}^{a}(1-\cos(12x))dx\\\end{eqnarray*}$

I'm quite sure you can go from here.

2: Make use of a linear substitution:

$\begin{eqnarray*}\\let\ u & = & 5-2x\implies\frac{du}{dx}=-2\ \ and\ \ dx=\frac{du}{-2}\\x & = & \frac{5-u}{2}\\when\ x=a\implies u & = & 5-2a \ (\text{adjusting terminals!}) \\when\ x=0\implies u & = & 5 \\ \\\therefore\int_{0}^{a}\frac{2x+1}{\sqrt{5-2x}}dx & = & \int_{5}^{5-2a}\frac{2(\frac{5-u}{2})+1}{\sqrt{u}}(\frac{du}{-2})\\ & = & -\frac{1}{2}\int_{5}^{5-2a}(6-u)(u^{-\frac{1}{2}})du=-\frac{1}{2}\int_{5}^{5-2a}(6u-u^{\frac{1}{2}})du\\\end{eqnarray*}\\$

Again, I'm quite sure you'll be able to go from here.

Sayf44 · « **Reply #3229 on:** June 11, 2014, 10:20:26 pm »

Thanks Euler

Ok this is the last question I'll be asking for a while because my SAC is tomorrow.

I'm not sure how to prove questions b and c (in the attachment). Someone please explain, thanks

e^1 · « **Reply #3230 on:** June 11, 2014, 11:26:10 pm »

Quote from: Sayf44 on June 11, 2014, 10:20:26 pm

Thanks Euler

Ok this is the last question I'll be asking for a while because my SAC is tomorrow.

I'm not sure how to prove questions b and c (in the attachment). Someone please explain, thanks

This is quite long and I hope I didn't make any errors. You can do part (c) by using the working out in it.

Working out in here

$\begin{alignedat}{1} \small \sin(a+b) &= \sin(a)\cos(b) + \sin(b)\cos(a) \\\\ & \text{Let } a = \sin^{-1}(x), b= \sin^{-1}(y), \;\;\; \left(|x| \leq 1, |y| \leq 1\right) \\\\ \implies \sin(\sin^{-1}(x)+\sin^{-1}(y)) &= \sin(\sin^{-1}(x))\cos(\sin^{-1}(y)) + \sin(\sin^{-1}(y))\cos(\sin^{-1}(x)) \\ &= x\cos(\sin^{-1}(y)) + y\cos(\sin^{-1}(x)) \end{alignedat} $

$\begin{alignedat}{1} \small \\\\ \cos^2(\sin^{-1}(y)) + \sin^2(\sin^{-1}(y)) = 1 \\ \implies \cos^2(\sin^{-1}(y)) + y^2 = 1 \\ \implies \cos(\sin^{-1}(y)) = \pm \sqrt{1 - y^2} \;\;\; \left(|y| \leq 1\right) \\\\ \text{Since } -\frac{\pi}{2} \leq \sin^{-1}(y) \leq \frac{\pi}{2}, \text{ then } \cos(\sin^{-1}(y)) \geq 0. \\ \text{Thus } \cos(\sin^{-1}(y)) = \sqrt{1 - y^2} \text{ only}. \\ \text{Similarly } \cos(\sin^{-1}(x)) = \sqrt{1 - x^2} \;\;\; \left(|x| \leq 1\right). \end{alignedat} $

$\begin{alignedat}{1} \small \therefore \sin(\sin^{-1}(x)+\sin^{-1}(y)) &= x\sqrt{1 - y^2} + y\sqrt{1 - x^2} \\ \implies \sin^{-1}(x)+\sin^{-1}(y) &= \sin^{-1}(x\sqrt{1 - y^2} + y\sqrt{1 - x^2}) \;\;\;\; \left( -\frac{\pi}{2} \leq \sin^{-1}(x)+\sin^{-1}(y) \leq \frac{\pi}{2} \right) \end{alignedat} $

From there you can assume that it is also true for $0\leq x \leq 1, 0 \leq y \leq 1$ since it is still within the restrictions set in our working out. Also, for the $\cos(\sin^{-1}(y)) \geq 0$ bit, you know cos(x) is positive for -pi/2 <= x <= pi/2 (ie. the forth and first quadrant when drawn on a graph), thus the statement afterwards. If you want, you can do the 'triangle method' to find $\cos(\sin^{-1}(x))$ and $\cos(\sin^{-1}(y))$ instead.

EDIT: Forgot a restriction that occurred in the last step.

Sayf44 · « **Reply #3231 on:** June 12, 2014, 07:50:21 am »

Thanks!!!

M_BONG · « **Reply #3232 on:** June 12, 2014, 06:56:05 pm »

Ok how on earth do you solve this.

It was on the MHS Spesh SAC today tech free; worth 5 marks.

My working was only 2 lines so I figured this is harder than it seems..

EVALUATE:

$\frac{d}{dx}(cosec^{-1}x)$

I applied the simple quotient rule... but I don't think that's right?

kinslayer · « **Reply #3233 on:** June 12, 2014, 07:25:35 pm »

Quote from: Zezima. on June 12, 2014, 06:56:05 pm

Ok how on earth do you solve this.

It was on the MHS Spesh SAC today tech free; worth 5 marks.

My working was only 2 lines so I figured this is harder than it seems..

EVALUATE:

$\frac{d}{dx}(cosec^{-1}x)$

I applied the simple quotient rule... but I don't think that's right?

Quote from: Zezima. on June 12, 2014, 06:56:05 pm

Ok how on earth do you solve this.

It was on the MHS Spesh SAC today tech free; worth 5 marks.

My working was only 2 lines so I figured this is harder than it seems..

EVALUATE:

$\frac{d}{dx}(cosec^{-1}x)$

I applied the simple quotient rule... but I don't think that's right?

Could it be the inverse cosecant?

$\begin{alignedat}1 y &= \csc^{-1} x \\ \csc y &= x \\ \Rightarrow \frac{\mbox{d}y}{\mbox{d} x} &= \left( \frac{\mbox{d}}{\mbox{d}y} \csc y\right)^{-1} \\ &= \ldots \end{alignedat}$

Just a thought

nhmn0301 · « **Reply #3234 on:** June 12, 2014, 07:33:42 pm »

Quote from: Zezima. on June 12, 2014, 06:56:05 pm

Ok how on earth do you solve this.

It was on the MHS Spesh SAC today tech free; worth 5 marks.

My working was only 2 lines so I figured this is harder than it seems..

EVALUATE:

$\frac{d}{dx}(cosec^{-1}x)$

I applied the simple quotient rule... but I don't think that's right?

This would probably easier if I know how to use Latex...
if you let y=cosec^(-1) (x)
=> cosec (y) = x
Using implicit differentiation
-cosec (y) cot (y) dy/dx = 1
dy/dx = -1 / (cosec (y) x cot (y) )
dy/dx = -sin(y) tan(y)
Now try to find sin (y) and tan (y) in terms of x
cosec (y) = 1/sin(y) => sin (y) = 1/cosec (x) = 1/ x (using above fact that cosec (y) = x )
Imagine drawing a triangle, if sin (y) = 1/x, then tan (y)= 1/( sqrt (x^2 -1) )
Hence, dy/dx = 1/ ( x . sqrt (x^2 -1) )
Let me know if there is any errors.

lzxnl · « **Reply #3235 on:** June 12, 2014, 08:59:42 pm »

Quote from: nhmn0301 on June 12, 2014, 07:33:42 pm

This would probably easier if I know how to use Latex...
if you let y=cosec^(-1) (x)
=> cosec (y) = x
Using implicit differentiation
-cosec (y) cot (y) dy/dx = 1
dy/dx = -1 / (cosec (y) x cot (y) )
dy/dx = -sin(y) tan(y)
Now try to find sin (y) and tan (y) in terms of x
cosec (y) = 1/sin(y) => sin (y) = 1/cosec (x) = 1/ x (using above fact that cosec (y) = x )
Imagine drawing a triangle, if sin (y) = 1/x, then tan (y)= 1/( sqrt (x^2 -1) )
Hence, dy/dx = 1/ ( x . sqrt (x^2 -1) )
Let me know if there is any errors.

I love how you immediately went from derivative of csc y is -csc y cot y (which is correct btw)
Only issue is, what about the sign of tan y? If y is in the fourth quadrant, tan y is negative.
As it turns out, tan y has the same sign as sin y which is 1/x. You want tan y to be negative when 1/x is negative and you want tan y to be positive when 1/x is positive. See how tan y / x, or the negative of your derivative, is always positive? This means you need a minus sign outside absolute value bars around your expression as the derivative is always negative. Think about it. dy/dx = - sin y tan y. For y ranging from (-pi/2, pi/2), dy/dx<0

Bestie · « **Reply #3236 on:** June 13, 2014, 09:16:32 pm »

just a quick question, will we ever be asked be asked to draw y = xe^x without the aid of the calculator?
If yes, how would u go about solving it? rather than just subbing in x values?
thank you

keltingmeith · « **Reply #3237 on:** June 13, 2014, 09:20:10 pm »

You shouldn't be, no. The best way to do it is by treatment of concavity which they explicitly state in the study design is not covered in the course.

Plus, there are a bunch more graphs you do more obviously do which would be better - like the trig graphs, reciprocal graphs or graphs with oblique asymptotes.

lzxnl · « **Reply #3238 on:** June 13, 2014, 09:24:58 pm »

I've been asked to sketch this by hand in a SAC. It's not undoable.

y=xe^x
dy/dx = e^x + xe^x = e^x (x+1)
So you have a local minimum at x = -1
Then, d^2/dx^2 = e^x + e^x (x+1) = e^x(x+2)
So at x=-1, d^2y/dx^2 > 0 => local minimum
Intercept at origin obviously, y<0 for x<0
So your graph should go down below the axis, stop at x=-1 and then just increase

The only dodgy bit is recognising that exponentials diminish much faster than x, so there is actually a horizontal asymptote at y=0

keltingmeith · « **Reply #3239 on:** June 13, 2014, 09:51:21 pm »

Surely not in an exam, though? I mean, I always thought you should check for some gradients = 0, their nature, then some concavity natures either side, which as VCAA say, isn't in the curriculum.

Search the forums now!

We have moved!

Author Topic: VCE Specialist 3/4 Question Thread! (Read 2813847 times) Tweet Share

b^3

Re: Specialist 3/4 Question Thread!

keltingmeith

Re: Specialist 3/4 Question Thread!

Mieow

Re: Specialist 3/4 Question Thread!

Zealous

Re: Specialist 3/4 Question Thread!

Sayf44

Re: Specialist 3/4 Question Thread!

e^1

Re: Specialist 3/4 Question Thread!

Sayf44

Re: Specialist 3/4 Question Thread!

M_BONG

Re: Specialist 3/4 Question Thread!

kinslayer

Re: Specialist 3/4 Question Thread!

nhmn0301

Re: Specialist 3/4 Question Thread!

lzxnl

Re: Specialist 3/4 Question Thread!

Bestie

Re: Specialist 3/4 Question Thread!

keltingmeith

Re: Specialist 3/4 Question Thread!

lzxnl

Re: Specialist 3/4 Question Thread!

keltingmeith

Re: Specialist 3/4 Question Thread!

Recent Posts

User:
Password: