Author Topic: VCE Specialist 3/4 Question Thread! (Read 2579384 times) Tweet Share

~T · « **Reply #1725 on:** May 30, 2013, 08:05:34 pm »

... and further to my point, in the working for the second question, you have a moment where:

$x=\frac{-1}{2}e^{1-2y}+c$ which suggests that $x<c$ because $\frac{-1}{2}e^{1-2y}<0$

Meaning that at the end, $y=-ln\left|2c-2x\right|+\frac{1}{2}$ can become $y=-ln\left(2c-2x\right)+\frac{1}{2}$

Sorry for spamming the forum, but I think I've figured it out for myself now $:-\$

e^1 · « **Reply #1726 on:** May 30, 2013, 08:58:17 pm »

Explain why the function with the equation $f(x) = \sin(\cos^{-1}(x))$ is not defined for $x < 0$ .

Not sure how to approach this question. Hints?

Alwin · « **Reply #1727 on:** May 30, 2013, 09:09:54 pm »

Quote from: e^1 on May 30, 2013, 08:58:17 pm

Explain why the function with the equation $f(x) = \sin(\cos^{-1}(x))$ is not defined for $x < 0$ .

Not sure how to approach this question. Hints?

Hint: It's a composite function. Firstly assuming you meant Sin(x) not sin(x), think of range of $\cos^{-1}(x)$ and domain of $\sin(x)$

ikhalid · « **Reply #1728 on:** May 30, 2013, 09:22:00 pm »

Quote from: Alwin on May 30, 2013, 07:49:03 pm

Yeah, i realised. I just had no better way of explaining it and I took that example out of context of the log_e function and that the constant A can be positive or negative. sorry, i just sort of smudged over it. I'll go back and change it now

hint: cot^2(x) + 1 = cosec^2(x)

I think you can do it from here

but its not ct^2(x) its just cot (x)

b^3 · « **Reply #1729 on:** May 30, 2013, 09:25:43 pm »

$\begin{alignedat}{1}\cot^{2}\left(\theta\right)+1 & =\text{cosec}^{2}\left(\theta\right) \\ 2\text{cosec}^{2}\left(\theta\right)+\cot\left(\theta\right) & =5 \\ 2\left(\cot^{2}\left(\theta\right)+1\right)+\cot\left(\theta\right) & =5 \\ 2\cot^{2}\left(\theta\right)+\cot\left(\theta\right)-3 & =0 \\ \text{Let }u=\cot\left(\theta\right) \\ 2u^{2}+u-3 & =0 \end{alignedat} $
I think you can take it from there

random_person · « **Reply #1730 on:** May 30, 2013, 09:44:48 pm »

Can someone please help me with part b of this question?
Heinemann 7.5 Extended answer question 9

A flexible beam of length $D$ is supported at the ends, which are at the same horizontal level. The deflection $y$ of the beam, measured downwards from the horizontal, satisfies the differential equation
$14\frac{d^2y}{dx^2}=D-4x, 0<x<D$

(a) Find an expression for $\frac{dy}{dx}$
(b) Find an expression for the deflection $y$

Answer to part (a):

Spoiler

$\frac{dy}{dx}=\frac{Dx}{14}-\frac{x^2}{7}+c$

Answer to part (b):

Spoiler

$y=\frac{Dx^2}{28}-\frac{x^3}{21}+\frac{D^2x}{84}$

I got $y=\frac{Dx^2}{28}-\frac{x^3}{21}+cx+d$
How do you get from my answer to the correct answer?

Kanye East · « **Reply #1731 on:** May 30, 2013, 10:10:39 pm »

Could someone help me with this? Notsureifbadengstudentortiredofstudying
The region bounded by the line y=5 and the curve y=x^2 +1 is rotated about the x-axis. Find the volume generated.

Phy124 · « **Reply #1732 on:** May 30, 2013, 10:34:50 pm »

Quote from: random_person on May 30, 2013, 09:44:48 pm

Can someone please help me with part b of this question?
Heinemann 7.5 Extended answer question 9

...

How do you get from my answer to the correct answer?

The deflection at the ends is zero.

So you have have:
$y(0)=y(D)=0$

The gradient of the beam at the centre is zero.

So you have:
$\frac{dy}{dx}=0$ at $x=\frac{D}{2}$

Put in one of the first two conditions and the third condition in your equations and solve for $c$ and $d$ .

You will find that $c=\frac{D^2}{84}$ and $d=0$

Conic · « **Reply #1733 on:** May 30, 2013, 10:36:05 pm »

Find when the curves are equal:

$5=x^2+1$

$4=x^2$

$x=\pm 2$

The top curve is y=5 and the bottom curve is $y=x^2+1$ and the bounds are x=-2 and x=2, so you can substitute into this formula:

$V=\pi\int_{lower\:bound}^{upper\:bound}[top\:curve]^2-[bottom\:curve]^2dx$

$V=\pi\int_{-2}^{2}((5)^2-(x^2+1)^2)dx$

It's symmetrical around the y axis so you can write it as:

$V=2\pi\int_{0}^{2}((5)^2-(x^2+1)^2)dx$

It ends up being $\frac{1088\pi}{15}$

zvezda · « **Reply #1734 on:** May 31, 2013, 07:08:10 pm »

Hey,
With ex 9D q16d in essentials, i get to a point where t=-100log|25-m|+c. However, i dont know whether, when transposed, e is positive or negative.
How would i go about figuring this out?
Thanks

BubbleWrapMan · « **Reply #1735 on:** May 31, 2013, 07:22:28 pm »

Use the initial condition.

Henreezy · « **Reply #1736 on:** May 31, 2013, 07:45:52 pm »

can anyone help me integrate cos^3(x)/sin^(x)?
thanks in advance!!

zvezda · « **Reply #1737 on:** May 31, 2013, 07:54:31 pm »

Quote from: Timmeh on May 31, 2013, 07:22:28 pm

Use the initial condition.

Yeah yeah.
Thanks

e^1 · « **Reply #1738 on:** May 31, 2013, 08:45:56 pm »

Quote from: Henreezy on May 31, 2013, 07:45:52 pm

can anyone help me integrate cos^3(x)/sin^(x)?
thanks in advance!!

Spoiler

$\begin{align} \small{} \int{\frac{\cos^3(x)}{\sin(x)}} dx &= \int{\frac{\cos(x)\cdot \cos^2(x)}{\sin(x)}} dx \\ &= \int{\frac{\cos(x)\cdot (1- \sin^2(x))}{\sin(x)}} dx \\\\ \text{Let } u = \sin(x) \\ \therefore \frac{du}{dx} = \cos(x) \\ \implies dx = \frac{du}{\cos(x)} \\\\ \text{Substitute u and dx:} \\ \int{\frac{(1- u^2)}{u}} du &= \int{(\frac{1}{u} - u}) du \\ &= \log_{e}(|u|) - \frac{u^2}{2} + c \\\\ \therefore \int{\frac{\cos^3(x)}{\sin(x)}} dx &= \log_{e}(|\sin(x)|) - \frac{\sin^2(x)}{2} + c \end{align}$

EDIT after your correction:

try it yourself first

$\begin{align} \small{} \int{\frac{\cos^3(x)}{\sin^2(x)}} dx &= \int{\frac{\cos^2(x)}{\sin^2(x)} \cdot \cos(x)} dx \\ &= \int{\frac{1 - \sin^2(x)}{\sin^2(x)} \cdot \cos(x)} dx \\\\ \text{Let } u = \sin(x) \\ \therefore \frac{du}{dx} = \cos(x) \\ \implies dx = \frac{du}{\cos(x)} \\\\ \text{Substitute u and dx:} \\ \int{\frac{1 - \sin^2(x)}{\sin^2(x)} \cdot \cos(x)} dx &= \int{\frac{1 - u^2}{u^2}} du \\ &= \int{\frac{1}{u^2} - 1} du \\\\ \therefore \int{\frac{\cos^3(x)}{\sin^2(x)}} dx &= -\frac{1}{\sin(x)} - \sin(x) + C \end{align}$

Henreezy · « **Reply #1739 on:** May 31, 2013, 09:16:51 pm »

ahh! I made a mistake in my question, I meant to write dy/dx = cos^3(x)/sin^2(x)
so sorry about that D:

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Author Topic: VCE Specialist 3/4 Question Thread! (Read 2579384 times) Tweet Share

~T

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