Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

SmeagAL · « **Reply #1005 on:** March 21, 2017, 02:01:16 pm »

Hi. It would be great if I could get some help on this question.

RuiAce · « **Reply #1006 on:** March 21, 2017, 03:14:58 pm »

Quote from: SmeagAL on March 21, 2017, 02:01:16 pm

Hi. It would be great if I could get some help on this question.

This question involves a Riemann sum. (Which is not 'explicitly' in the syllabus.)

The whole foundation of integration is the idea of Riemann sums. By reducing the width of the rectangles, we assert that the area of the rectangles will eventually be EQUAL to the area under the curve. So since the limit is actually representative of an (infinite, aka. limiting) Riemann sum, it is equivalent to the ACTUAL area under the curve.

Do mention if my handwriting was too messy.

lsong · « **Reply #1007 on:** March 21, 2017, 09:04:33 pm »

Hi, I can't seem to do part (i) by using the given statement. Any help would be great thanks!

SmeagAL · « **Reply #1008 on:** March 21, 2017, 10:10:24 pm »

THANKS! Extremely helpful.

Never heard of the Riemann Sum before.

RuiAce · « **Reply #1009 on:** March 22, 2017, 04:38:37 am »

Quote from: lsong on March 21, 2017, 09:04:33 pm

Hi, I can't seem to do part (i) by using the given statement. Any help would be great thanks!

$\text{All you have to show is that }RHS\ge 0\text{ really}\\ \text{where by }RHS\text{ I mean the }RHS\text{ of what you were given}$
$\text{Clearly }x+y+z\ge 0\text{ as }x\ge 0\text{ etc.}$
$\begin{align*}x^2+y^2+z^2-xy-xz-yz&= (x^2+y^2+z^2)-2(xy-xz-yz)+(xy+xz+yz)\\ &= (x+y+z)^2 + xy+xz+yz\\ &\ge 0\end{align*}\\ \text{Exercise: How come?}$
$\text{So since }x+y+z>0 \text{ and }x^2+y^2+z^2-xy-xz-yz \ge 0\\ \text{The product of two non-negative numbers is still non-negative}\\ \text{So }RHS=(x+y+z)(x^2+y^2+z^2-xy-xz-yz) > 0$
$\text{Since }RHS > 0\\ \text{We also have }LHS > 0\text{ (because obviously }RHS=LHS)\\ \text{So }x^3+y^3+z^3-3xyz \ge 0 \implies x^3+y^3+z^3 \ge 3xyz\\ \text{and the result is proven}$
Note: The question actually says x>0 and not x≥0. However, I have assumed otherwise because if x>0 then for part i), we should have > as well instead of ≥.

lsong · « **Reply #1010 on:** March 22, 2017, 10:24:37 am »

Quote from: RuiAce on March 22, 2017, 04:38:37 am

$\text{All you have to show is that }RHS\ge 0\text{ really}\\ \text{where by }RHS\text{ I mean the }RHS\text{ of what you were given}$
$\text{Clearly }x+y+z\ge 0\text{ as }x\ge 0\text{ etc.}$
$\begin{align*}x^2+y^2+z^2-xy-xz-yz&= (x^2+y^2+z^2)-2(xy-xz-yz)+(xy+xz+yz)\\ &= (x+y+z)^2 + xy+xz+yz\\ &\ge 0\end{align*}\\ \text{Exercise: How come?}$
$\text{So since }x+y+z>0 \text{ and }x^2+y^2+z^2-xy-xz-yz \ge 0\\ \text{The product of two non-negative numbers is still non-negative}\\ \text{So }RHS=(x+y+z)(x^2+y^2+z^2-xy-xz-yz) > 0$
$\text{Since }RHS > 0\\ \text{We also have }LHS > 0\text{ (because obviously }RHS=LHS)\\ \text{So }x^3+y^3+z^3-3xyz \ge 0 \implies x^3+y^3+z^3 \ge 3xyz\\ \text{and the result is proven}$
Note: The question actually says x>0 and not x≥0. However, I have assumed otherwise because if x>0 then for part i), we should have > as well instead of ≥.

Thanks for the help!

VydekiE · « **Reply #1011 on:** March 22, 2017, 11:43:52 am »

Hi,
It would be great if someone could help me with this question
1) The curve on the Argand diagram for which ||z-3| - |z+5||=4 is a hyperbola.
a) Find the eccentricity of this hyperbola
Thank you!!

theblackswan · « **Reply #1012 on:** March 22, 2017, 02:50:19 pm »

Hey,
I'm kinda stuck on questions that require you to find the tangents to the ellipse or hyperbola that are parallel to a line. What's the best method for those types of questions? For example, find the tangents to the ellipse 16x^2 + 25y^2 =400 which are parallel to the line y = x+2. Thanks!!

jakesilove · « **Reply #1013 on:** March 22, 2017, 06:51:15 pm »

Quote from: VydekiE on March 22, 2017, 11:43:52 am

Hi,
It would be great if someone could help me with this question
1) The curve on the Argand diagram for which ||z-3| - |z+5||=4 is a hyperbola.
a) Find the eccentricity of this hyperbola
Thank you!!

Hey! First, recall that |PS-PS'|=2a. It seems sort of intuitive that the focus of the hyperbola will be z=3 and z=5, right? If not, try think about some values for z that hold true, and the shape kinda pops out on its own. You can use the 'location' of the focus', plus the knowledge that a=2, to find the eccentricity.

jakesilove · « **Reply #1014 on:** March 22, 2017, 06:52:34 pm »

Quote from: theblackswan on March 22, 2017, 02:50:19 pm

Hey,
I'm kinda stuck on questions that require you to find the tangents to the ellipse or hyperbola that are parallel to a line. What's the best method for those types of questions? For example, find the tangents to the ellipse 16x^2 + 25y^2 =400 which are parallel to the line y = x+2. Thanks!!

If you are just given points, the implicit differentiation is the way to go (you can do this as normal, sovling for dy/dx and plugging your points in). If a given line is parallel (ie. tangents are involved etc.) then you can try to use the quadratic discriminant. Both Rui and I are seriously busy tonight, so hopefully we can expand on our answers either later or sometime tomorrow!

VydekiE · « **Reply #1015 on:** March 23, 2017, 10:00:57 am »

Quote from: jakesilove on March 22, 2017, 06:51:15 pm

Hey! First, recall that |PS-PS'|=2a. It seems sort of intuitive that the focus of the hyperbola will be z=3 and z=5, right? If not, try think about some values for z that hold true, and the shape kinda pops out on its own. You can use the 'location' of the focus', plus the knowledge that a=2, to find the eccentricity.

Thank you!!

RuiAce · « **Reply #1016 on:** March 23, 2017, 11:28:36 am »

Quote from: theblackswan on March 22, 2017, 02:50:19 pm

Hey,
I'm kinda stuck on questions that require you to find the tangents to the ellipse or hyperbola that are parallel to a line. What's the best method for those types of questions? For example, find the tangents to the ellipse 16x^2 + 25y^2 =400 which are parallel to the line y = x+2. Thanks!!

$\text{Very quickly expanding on what Jake and I discussed}$
$\text{Once we have a }\textbf{line}\text{ and not a }\textbf{point}\\ \text{We use the following theorem}$
$\text{For any conic section (graphed onto a number plane)}\\ \boxed{\text{A tangent to a conic section only ever intersects it ONCE}}\\ \text{The point being, of course, where it actually is tangential to the curve.}$
$\text{So given a line }y=x+b\text{ which is parallel to }y=x+2\\ \text{and the ellipse }16x^2+25y=400\\ \text{Suppose we wanted to find the point of intersection. Then by simultaneous equations}\\ 16x^2+25(x+a)^2=400$
$\text{This is a quadratic in }x\\ \text{and from the above theorem it }\textbf{only has one solution}\\ \text{Hence, set }\Delta = 0\text{ to solve for }a$
Please find a sample question in post #813, and one solution in post #816

Edit: Typo

smile123 · « **Reply #1017 on:** March 24, 2017, 08:47:15 am »

PLEASE HELP

RuiAce · « **Reply #1018 on:** March 24, 2017, 09:04:27 am »

Quote from: smile123 on March 24, 2017, 08:47:15 am

PLEASE HELP

Three-dimensional curves are not a part of the HSC
$\text{The radius can simply be found by the generalised Pythagoras' theorem:}\\ r=\sqrt{2^2+4^2+3^2}=\sqrt{29}$
$\text{So the equation is just }(x-2)^2+(y-4)^2+(z-3)^2=29$

smile123 · « **Reply #1019 on:** March 24, 2017, 03:42:08 pm »

Thankyou

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