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June 26, 2019, 12:07:51 pm

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RuiAce

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« Reply #2250 on: April 08, 2019, 03:13:57 pm »
+1
hey how do you do this one:

h/(x+h) < ln(x+h) - ln(x) <h/x . (it's meant to be greater than and equal to btw)
Hint: You can use a sketch to verify the following inequality.
$h \times \frac{1}{x+h} < \int_x^{x+h} \frac{1}{t}dt < h\times \frac{1}{x}$

david.wang28

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« Reply #2251 on: April 13, 2019, 10:01:55 pm »
0
Hello,
I am stuck on Example 7 part a) in the link below. I have also attached my working out. Can anyone please help me out? Thanks
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RuiAce

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« Reply #2252 on: April 13, 2019, 10:40:29 pm »
+2
Hello,
I am stuck on Example 7 part a) in the link below. I have also attached my working out. Can anyone please help me out? Thanks
Two things:
$\text{Firstly, you've correctly noted that }A = \frac12 x^2 \tan \alpha.\\ \text{So your volume of a cross section should just be }\boxed{\delta V = \frac12 x^2 \tan \alpha\, \delta y}\\ \text{i.e. }\boxed{\delta V = \frac12 (r^2-y^2)\tan \alpha\, \delta y}$
Not sure why you multiplied $x^2$ to another $r^2-y^2$ term in your expression. By doing so, because of the fact that $x^2+y^2=r^2$, you're basically trying to deal with $x^4$, or equivalently $(r^2-y^2)^2$
$\text{Secondly, what is }a?\\ \text{The boundaries of integration (and also the sum) are just }-r\text{ and }r$

david.wang28

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« Reply #2253 on: April 15, 2019, 03:25:52 pm »
0
Two things:
$\text{Firstly, you've correctly noted that }A = \frac12 x^2 \tan \alpha.\\ \text{So your volume of a cross section should just be }\boxed{\delta V = \frac12 x^2 \tan \alpha\, \delta y}\\ \text{i.e. }\boxed{\delta V = \frac12 (r^2-y^2)\tan \alpha\, \delta y}$
Not sure why you multiplied $x^2$ to another $r^2-y^2$ term in your expression. By doing so, because of the fact that $x^2+y^2=r^2$, you're basically trying to deal with $x^4$, or equivalently $(r^2-y^2)^2$
$\text{Secondly, what is }a?\\ \text{The boundaries of integration (and also the sum) are just }-r\text{ and }r$
I realised where I went wrong; thanks
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david.wang28

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« Reply #2254 on: April 24, 2019, 09:49:37 pm »
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Hello,
I am stuck on a volumes question in the attachment below (especially the proving part). Can anyone please help me out? Thanks
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RuiAce

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« Reply #2255 on: April 25, 2019, 06:17:54 pm »
+1
Hello,
I am stuck on a volumes question in the attachment below (especially the proving part). Can anyone please help me out? Thanks
Sorry about the delay with this one. Been ages since I used the linear equation method for the volume of a frustrum so I wanted to be more alert when revising it. I just attached the full solution here.

It's essentially based off the exact same method you used. Just with some arbitrary variables, instead of given constants.

david.wang28

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« Reply #2256 on: April 25, 2019, 09:08:04 pm »
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Sorry about the delay with this one. Been ages since I used the linear equation method for the volume of a frustrum so I wanted to be more alert when revising it. I just attached the full solution here.

It's essentially based off the exact same method you used. Just with some arbitrary variables, instead of given constants.

(Image removed from quote.)

(Image removed from quote.)
Wow, this seems top-notch. Thank you!
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diggity

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« Reply #2257 on: April 26, 2019, 10:27:38 pm »
0
Hey Rui! Following the MX2 lecture today at UTS (was awesome btw!), I've decided to try this thread for help. I think my worst area is Applications of Polynomials (locuses and all that). Here's my question from that topic (from textbook):

a) Show that the equation of the normal to the parabola y = x^2 at the point P(t, t^2) may be written as
$t^3 + (\frac{1-2y}{2})t + \frac{-x}{2}=0$

b) Hence, if there exist three normals passing through P(x_0, y_0) then using the result of part (b) in question 7 above show that $y_0 > 3(\frac{x_0}{4})^{\frac{2}{3}} + \frac{1}{2}$.

Question a) is easy, just added it for context. The real head-scratcher is part b), and I don't even know where to start. (side note; is there any way to do LaTeX inline?)
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RuiAce

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« Reply #2258 on: April 26, 2019, 10:56:14 pm »
+3
Hey Rui! Following the MX2 lecture today at UTS (was awesome btw!), I've decided to try this thread for help. I think my worst area is Applications of Polynomials (locuses and all that). Here's my question from that topic (from textbook):

a) Show that the equation of the normal to the parabola y = x^2 at the point P(t, t^2) may be written as
$t^3 + (\frac{1-2y}{2})t + \frac{-x}{2}=0$

b) Hence, if there exist three normals passing through P(x_0, y_0) then using the result of part (b) in question 7 above show that $y_0 > 3(\frac{x_0}{4})^{\frac{2}{3}} + \frac{1}{2}$.

Question a) is easy, just added it for context. The real head-scratcher is part b), and I don't even know where to start. (side note; is there any way to do LaTeX inline?)

I just had a look at this question. I'm hesitant to comment right now though because I actually got it to work easily with the cubic discriminant formula, Will probably spend more time on it later, but for my reference which textbook is this from?

diggity

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« Reply #2259 on: April 26, 2019, 11:05:05 pm »
0
Terry Lee's "Advanced mathematics; A complete HSC MX2 Course" (exercise 3.4 q8). I've never heard of the cubic discriminant formula before? Edit: and that's okay! take all the time you need
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diggity

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« Reply #2260 on: April 26, 2019, 11:09:17 pm »
0
I just noticed that I did not include the result mentioned in the question! It's; if f(x) is in the form:

$f(x)=x^3+ax+b$
then,
$27b^2 + 4a^3 < 0$

(I'm not actually sure how much information from the previous question is needed, but this is the result obtained)
« Last Edit: April 26, 2019, 11:14:00 pm by diggity »
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RuiAce

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« Reply #2261 on: April 26, 2019, 11:38:44 pm »
+2
I just noticed that I did not include the result mentioned in the question! It's; if f(x) is in the form:

$f(x)=x^3+ax+b$
then,
$27b^2 + 4a^3 < 0$

(I'm not actually sure how much information from the previous question is needed, but this is the result obtained)
Ah cool, the question is definitely doable now aha! That above result is quite important because it holds only when $f(x)$ has 3 distinct $x$-intercepts.
$\text{The idea is the following.}\\ \text{Firstly, suppose that the normal at }P\text{ passes through }(x_0,y_0).\\ \text{Then, from subbing in we have}\\ \boxed{t^3 + \left( \frac{1-2y_0}{2}\right)t + \frac{-x_0}{2} =0}.$
$\text{But then if }\textbf{three}\text{ normals pass through }(x_0,y_0)\\ \text{that means we have }\textbf{three distinct choices}\text{ for the point }P\\ \text{that will make the boxed equation hold true.}$
$\text{Which is then equivalent to saying that}\\ \text{there are three distinct solutions for }t\\ \text{that make the equation true.}$
It is because of that why we're able to use the result of the previous question. (The tough part of this question was translating what they tell you from words into mathematics.)
$\text{So using the previous question's result, here we have }\\\boxed{27\left(- \frac{x_0}{2} \right)^2 + 4\left( \frac{1-2y_0}{2} \right) ^3< 0}$
\text{Now we just do battle with algebra.}\\ \begin{align*} 4\left( \frac{2y_0-1}{2} \right)^3 &> 27\left( \frac{x_0}{2} \right)^2\\ \left( y_0-\frac12 \right)^3 &> 27 \left( \frac{x_0}{4} \right)^2\\ y_0 - \frac12 &> 3\left( \frac{x_0}{4} \right)^{2/3}\\ y_0 &= 3\left( \frac{x_0}{4} \right)^{2/3}+\frac12\end{align*}
Code: (Also, in regards to inline LaTeX, it works like this) [Select]
$stuff goes here$
« Last Edit: April 26, 2019, 11:40:15 pm by RuiAce »

diggity

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« Reply #2262 on: April 27, 2019, 12:15:17 am »
+1
OH! That makes so much sense! I didn't recognise the equation of the normal was in that form, and then being able to use that inequality to achieve the answer. You're right, a huge part of this topic is the translation from words to mathematics. I see now that this question is quite simple when boiled down to its fundamental components. Major props!
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diggity

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« Reply #2263 on: April 27, 2019, 02:28:39 pm »
0
Okay, I have another question (from the same exercise):

6 Assume $cos3\theta = 4cos^3\theta - 3cos\theta$, find in terms of $\pi$ the exact values of the solutions of
a) $x^3-3x+1=0$

The answer in the book begins quite simply, with $x=kcos\theta$, $k^3cos^3\theta - 3kcos\theta + 1 = 0$. They then mention that:
"We want $\frac{k^3}{3k} = \frac{4}{3}$, ∴ $k^2=4$, ∴ $k=2$."
From this, they attain the equation:
$8cos^3\theta - 6cos\theta + 1 = 0\\
2cos3\theta + 1 = 0$

... and being able to solve for theta. What I'm confused at is why they wanted that original equation that gave them k. What I originally tried to do was something like:
$k^3 = 4\\
3k = 3$

Why did they divide the two equations? I see that there is no solution for my two equations, which is where I got stuck. Help!
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HSC 2019 discord. Minimal work is done, but more people to do that work would be nice. Feel free to join.

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RuiAce

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« Reply #2264 on: April 27, 2019, 03:00:30 pm »
+2
Okay, I have another question (from the same exercise):

6 Assume $cos3\theta = 4cos^3\theta - 3cos\theta$, find in terms of $\pi$ the exact values of the solutions of
a) $x^3-3x+1=0$

The answer in the book begins quite simply, with $x=kcos\theta$, $k^3cos^3\theta - 3kcos\theta + 1 = 0$. They then mention that:
"We want $\frac{k^3}{3k} = \frac{4}{3}$, ∴ $k^2=4$, ∴ $k=2$."
From this, they attain the equation:
$8cos^3\theta - 6cos\theta + 1 = 0\\
2cos3\theta + 1 = 0$

... and being able to solve for theta. What I'm confused at is why they wanted that original equation that gave them k. What I originally tried to do was something like:
$k^3 = 4\\
3k = 3$

Why did they divide the two equations? I see that there is no solution for my two equations, which is where I got stuck. Help!
Your equations are essentially a bit too restrictive.

Your claim is that the coefficient of $\cos^3\theta$ must be $4$, and the coefficient of $\cos\theta$ must be $3$. But we do not need to go that far.

For example, the coefficients could've been, say, $16$ and $12$. Then we could've still use the formula, except we would've just had $16\cos^3\theta - 12\cos\theta+1 = 0$, which can still be simplified into $4\cos3\theta+1=0$ anyway!
$\text{The idea is, only the }\textbf{ratio}\text{ of the coefficients need}\\ \text{to be }4:3.\\ \text{The actual coefficients themselves don't have to follow that rule}\\ \text{because we don't }\textbf{need}\text{ to aim for }\cos3\theta\text{ by itself.}\\ \text{Rather, we can aim for a constant multiplied to }\cos3\theta\text{ instead.}$
$\text{This is where they get }\frac{k^3}{3k} = \frac{4}{3}\text{ from.}$
It just so turns out that $16$ and $12$ weren't correct, but rather $8$ and $6$ were. Which of course, are still in the ratio 4:3, but don't actually equal 4 and 3 respectively!
« Last Edit: April 27, 2019, 03:02:40 pm by RuiAce »