Author Topic: 3U Maths Question Thread (Read 1524489 times) Tweet Share

Shadowxo · « **Reply #1635 on:** March 10, 2017, 12:16:40 am »

Quote from: jamonwindeyer on March 09, 2017, 01:31:29 am

Nah nah, you could definitely assume that, I suppose hemisphere is the wrong word (though we will need it) - I mean more the volume of a spherical cap - I think to do this we'll need a formula linking the volume of a spherical cap to the height of said cap (and the radius of the sphere), which with a Google is apparently:

$V=\frac{\pi}{3}H^2(3R-H)$

I can't think of a way to do it without this formula, and even with it I'm not sure about the approach. I'll have a proper think and attempt at it tomorrow, I'm thinking we'll need to define $\frac{dH}{dt}=0.3+A$, where $A$ is some function of the volume of the sphere in the water at any given time. Play with derivatives a bit from there?

With regards to this question (I know it's a bit late), the only way I could find to solve it was rotating the curve y=√(4-(x-2)²) aka the first half of the circle about the x axis from 0 to h. You could use this to find The volume of the sphere with respect to height then dV/dh for sphere and go from there to solve the question (ended up with 3/1000 though - fairly close but not the same). Anyway, they shouldn't ask you to do this, at least not all in a single question, just wanted to show a way to do it using year 12 knowledge

Edit: I used VCE specialist maths knowledge so I'm not sure if this method is known for 3U maths

hanaacdr · « **Reply #1636 on:** March 10, 2017, 05:54:33 am »

Quote from: RuiAce on March 09, 2017, 09:54:05 pm

Hint for first part: Consider the following theorems.

1. Alternate segment theorem
2. Three right angles in that diagram, two of which are angle AMQ and angle ALQ

yes thank you,
but can i please confirm that <QBN = <AQM
not 100% sure
thank you

hanaacdr · « **Reply #1637 on:** March 10, 2017, 06:10:33 am »

Also could i please get some help on this inequalities
I am not sure how to approach it
Thanks Hanaa

ellipse · « **Reply #1638 on:** March 10, 2017, 06:54:39 am »

Quote from: hanaacdr on March 10, 2017, 06:10:33 am

Also could i please get some help on this inequalities
I am not sure how to approach it
Thanks Hanaa(Image removed from quote.)

(i)
$a+1\leq 1+b$
as both a and b are positive, you can take the reciprocal and change in the inequality sign
$\frac{1}{1+a}\geq \frac{b}{1+b}$
$\frac{1+a}{1+a}-\frac{a}{1+a}\geq \frac{1+b}{1+b}-\frac{b}{1+b}$
$-\frac{a}{1+a}\geq -\frac{b}{1+b}$
now multiply each side by -1 and reverse the inequality sign
$\frac{a}{1+a}\leq \frac{b}{1+b}$

ellipse · « **Reply #1639 on:** March 10, 2017, 07:03:51 am »

Quote from: hanaacdr on March 10, 2017, 06:10:33 am

Also could i please get some help on this inequalities
I am not sure how to approach it
Thanks Hanaa(Image removed from quote.)

(ii) not the best at inequalities, so appreciate if someone could check over this
from part (i)
$\frac{a}{1+a}-\frac{b}{1+b}\leq 0\\$
But as $c\geq 0, \frac{c}{1+c} \geq 0 \\ \therefore \frac{a}{1+a}-\frac{b}{1+b}\leq \frac{c}{1+c} \\\Rightarrow \frac{a}{1+a}\leq \frac{b}{1+b}+\frac{1}{1+c}$

kiwiberry · « **Reply #1640 on:** March 10, 2017, 07:33:06 am »

Quote from: hanaacdr on March 10, 2017, 05:54:33 am

yes thank you,
but can i please confirm that <QBN = <AQM
not 100% sure
thank you

If you're using the alternate segment theorem, $\angle QBN = \angle MAQ$, can you see it?

Shadowxo · « **Reply #1641 on:** March 10, 2017, 08:25:59 am »

Quote from: hanaacdr on March 10, 2017, 06:10:33 am

Also could i please get some help on this inequalities
I am not sure how to approach it
Thanks Hanaa(Image removed from quote.)

Way I'd do it is, by kind of expanding it out and figuring what I need,
$a \leq b \\ a+ab \leq b+ab \\ a(1+b) \leq b(1+a) \\ \frac{a}{1+a} \leq \frac{b}{1+b}$

Alternatively, you could do it like ellipse,
$a+1 \leq 1+b$
$\frac{1}{1+a}\geq \frac{1}{1+b}$
$\frac{1}{1+a}-\frac{1+a}{1+a} \geq \frac{1}{1+b} - \frac{1+b}{1+b}$
$-\frac{a}{1+a}\geq -\frac{b}{1+b}$ $\frac{a}{1+a}\leq \frac{b}{1+b}$
(fixed some errors)
And your working for ii seems to be right

RuiAce · « **Reply #1642 on:** March 10, 2017, 08:33:55 am »

Quote from: hanaacdr on March 10, 2017, 06:10:33 am

Also could i please get some help on this inequalities
I am not sure how to approach it
Thanks Hanaa(Image removed from quote.)

$\textbf{I do not recall seeing this in 3U.}$
$\textbf{Work backwards on scrap paper}\text{ or else it's too unobvious}\\ \begin{align*}\frac{a}{1+a}&\le \frac{b}{1+b}\\ a(1+b)&\le b(1+a)\\ a+ab& \le b+ab\\ a&\le b\end{align*}$
$\text{Therefore, write it out in the correct way (essentially Shadowxo's method)}\\ \text{And we are done.}\\ \textbf{Note: Working backwards first makes it more obvious though.}$

Quote from: ellipse on March 10, 2017, 07:03:51 am

(ii) not the best at inequalities, so appreciate if someone could check over this
from part (i)
$\frac{a}{1+a}-\frac{b}{1+b}\leq 0\\$
But as $c\geq 0, \frac{c}{1+c} \geq 0 \\ \therefore \frac{a}{1+a}-\frac{b}{1+b}\leq \frac{c}{1+c} \\\Rightarrow \frac{a}{1+a}\leq \frac{b}{1+b}+\frac{1}{1+c}$

$\text{This is fine, however you should make it more explicit}\\ \text{by showing that }\frac{a}{1+a}-\frac{b}{1+b} \le 0\\ \text{whereas }\frac{c}{1+c}\ge 0\text{ because we're dividing a non-negative with a positive}\\ \text{and that we're combining inequalities}$

ellipse · « **Reply #1643 on:** March 10, 2017, 08:42:34 am »

Quote from: RuiAce on March 10, 2017, 08:33:55 am

$\text{This is fine, however you should make it more explicit}\\ \text{by showing that }\frac{a}{1+a}-\frac{b}{1+b} \le 0\\ \text{whereas }\frac{c}{1+c}\ge 0\text{ because we're dividing a non-negative with a positive}\\ \text{and that we're combining inequalities}$

oh yes I forgot to write that out after typing 'from part (i)' Thank you though

legorgo18 · « **Reply #1644 on:** March 10, 2017, 09:41:21 pm »

Hey guys cant do part b of this q

so a is Find the eqn of the tangent to the curve root x + root y = root c at point P(a,b) on the curve

Using implicit diff i got (root a)*y + (root b)*x = a*(root b) + b*(root a)

and b: The tangent meets the x- and y axes at Q and R respectively. Show that OQ + OR = c for all positions of P, where O is the origin

So i made the tangent x and y =0 and i used distance formula from Q to O and R to O then added them together but cant simplify it or am i doing something wrong?

Shadowxo · « **Reply #1645 on:** March 10, 2017, 10:04:16 pm »

Quote from: legorgo18 on March 10, 2017, 09:41:21 pm

Hey guys cant do part b of this q

so a is Find the eqn of the tangent to the curve root x + root y = root c at point P(a,b) on the curve

Using implicit diff i got (root a)*y + (root b)*x = a*(root b) + b*(root a)

and b: The tangent meets the x- and y axes at Q and R respectively. Show that OQ + OR = c for all positions of P, where O is the origin

So i made the tangent x and y =0 and i used distance formula from Q to O and R to O then added them together but cant simplify it or am i doing something wrong?

For part a) I got something different
$\sqrt x + \sqrt y =\sqrt c\\ \frac{1}{2}x^{-1/2}+\frac{1}{2}y^{-1/2} \frac{dy}{dx}=0 \\ \frac{dy}{dx}=\frac{- \sqrt{y}}{\sqrt x} \\ =- \sqrt{\frac{b}{a}} \\ \text{using y-y1=m(x-x1), } y= - \sqrt{\frac{b}{a}}x+\sqrt{ba} +b$

Skipped a couple steps but does this help?
Edit: rearranged your equation to get same as mine, going to do next part

Next Part:
Subbing in x=a, y=b into original equation , √c = √a + √b
via tangent formula we calculated,
x intercept Q (a + √(ab) , 0) y intercept R (0 , b + √(ab) )
OQ = a+√(ab), OR = b+√(ab)
OQ + OR = a + 2√(ab)+b
c= (√a + √b)²= a + 2√(ab)+b = OQ + OR
Sorry if it's not the right notation but here's the general idea

RuiAce · « **Reply #1646 on:** March 10, 2017, 10:50:48 pm »

Quote from: legorgo18 on March 10, 2017, 09:41:21 pm

Hey guys cant do part b of this q

so a is Find the eqn of the tangent to the curve root x + root y = root c at point P(a,b) on the curve

Using implicit diff i got (root a)*y + (root b)*x = a*(root b) + b*(root a)

and b: The tangent meets the x- and y axes at Q and R respectively. Show that OQ + OR = c for all positions of P, where O is the origin

So i made the tangent x and y =0 and i used distance formula from Q to O and R to O then added them together but cant simplify it or am i doing something wrong?

$\textbf{Note that questions requiring implicit differentiation are NOT examinable in 3U.}$
$\text{From }\sqrt{b}x+\sqrt{a}y=a\sqrt{b}+b\sqrt{a}\\ \text{For the coordinates of }Q\text{, set }y=0:\\ x\sqrt{a}=a\sqrt{b}+b\sqrt{a}\implies x = \sqrt{ab}+b\\ \text{For the coordinates of }R\text{, set }x=0:\\ y\sqrt{b}=a\sqrt{b}+b\sqrt{a}\implies y=a+\sqrt{ab}$
$\text{Hence }Q\text{ is }(b+\sqrt{ab},0)\\ R\text{ is }(0,a+\sqrt{ab})\\ \text{So it can be checked that our distances are}\\ \begin{align*}OQ&=b+\sqrt{ab}\\OR&=a+\sqrt{ab}\end{align*}$
$\text{This means that}\\ \boxed{OQ + OR = a+2\sqrt{ab}+b = (\sqrt{a}+\sqrt{b})^2}$
$\text{Now, recall that point }P(a,b)\\ \textbf{lies on the curve }\sqrt{x}+\sqrt{y}=\sqrt{c}$
$\text{Since we KNOW }P\text{ lies on the curve}\\ \text{We should be able to substitute the coordinates in like this:}$
$\text{Substituting in }(x,y)=(a,b)\text{ gives}\\ \boxed{\sqrt{a}+\sqrt{b}=\sqrt{c}}$
$\text{So substituting this into what we have earlier we observe that}\\ OQ+OR = (\sqrt{c})^2=c\text{ as required.}$

shreya_ajoshi · « **Reply #1647 on:** March 10, 2017, 10:58:43 pm »

Hi
Some perms and combs questions I need help with
1) find the number of krrangements of the letters of the word EDUCATION if:
a) C must be next to A
B) I must precede E
C) there must be three letters between D and N

Thanks

RuiAce · « **Reply #1648 on:** March 10, 2017, 11:08:05 pm »

Quote from: shreya_ajoshi on March 10, 2017, 10:58:43 pm

Hi
Some perms and combs questions I need help with
1) find the number of krrangements of the letters of the word EDUCATION if:
a) C must be next to A
B) I must precede E
C) there must be three letters between D and N

Thanks

$\text{Since }C\text{ must be next to }A\\ \text{We treat }CA\text{ as a group of letters.}\\ \text{Note that there are }2!\text{ ways of arranging }CA$
$\text{We group }CA\text{ into }\boxed{CA}\\ \text{and now are interested in arranging the 7 'letters}\\ E, D, U, \boxed{CA}, T, I, O, N\\ \text{Which is done in 8! ways}$
$\therefore\text{Answer: }\boxed{2!\times 8!}$
______________________________________
$\textit{Trick:} \text{ First, note that the total possible arrangements is }9!\\ \text{Now, there are only TWO possibilities:}$
$\text{Either }I\text{ preceeds }E\\ \text{Or }E\text{ preceeds }I$
$\text{But note that for every arrangement where }I\text{ preceeds }E\\ \text{If you just swap them around, you get an arrangement}\\ \text{where }E\text{ preceeds }I$
$\text{By doing this with EVERY scenario where }I\text{ preceeds }E\\ \text{We see that the amount of words with }I\text{ before }E\text{ and }E\text{ before }I\\ \text{are evenly split!}$
$\therefore\text{Answer: }\boxed{\frac{9!}{2}}$
______________________________________
$\_ \,\_ \,\_ \,\_ \,\_ \,\_ \,\_ \,\_ \,\_ \\ \text{When we have a restriction, we }\textbf{deal with the restriction first}\\ \text{First, note that either the }D\text{ or the }N\text{ could come first}\\ \text{So we arrange the }D\text{ and }N\text{, which is done in 2! ways}$
$\text{Now, for explanation's sake (since we did the 2! stuff)}\\ \text{Just assume the }D\text{ comes first}\\ \text{Observe that since }D\text{ must come first, we only have the following 5 cases:}$
$D\, \_ \,\_ \,\_ \,N\,\_ \,\_ \,\_ \,\_ \\ \_ D\, \_ \,\_ \,\_ \,N\, \_ \, \_ \, \_ \, \_\\ \_ \, \_ \, D\, \_ \,\_ \,\_ \,N \_ \, \_ \, \_\\ \_ \, \_ \, \_ \, D\, \_ \,\_ \,\_ \,N \_ \, \_ \\ \_ \, \_ \, \_ \, \_ \, D\, \_ \,\_ \,\_ \,N \_ \\ \_ \, D\, \_ \,\_ \,\_ \,N$
$\text{i.e. there are only 7 possible choices}\\ \text{for where we can put our }D\text{ and }N$
$\text{Lastly, the remaining 7 letters can be arranged in any of the leftover spots}\\ \text{This is done in 7! ways}$
$\therefore\text{Answer: }\boxed{2!\times 7\times 7!}$

shreya_ajoshi · « **Reply #1649 on:** March 10, 2017, 11:26:29 pm »

Quote from: RuiAce on March 10, 2017, 11:08:05 pm

$\text{Since }C\text{ must be next to }A\\ \text{We treat }CA\text{ as a group of letters.}\\ \text{Note that there are }2!\text{ ways of arranging }CA$
$\text{We group }CA\text{ into }\boxed{CA}\\ \text{and now are interested in arranging the 7 'letters}\\ E, D, U, \boxed{CA}, T, I, O, N\\ \text{Which is done in 8! ways}$
$\therefore\text{Answer: }\boxed{2!\times 8!}$
______________________________________
$\textit{Trick:} \text{ First, note that the total possible arrangements is }9!\\ \text{Now, there are only TWO possibilities:}$
$\text{Either }I\text{ preceeds }E\\ \text{Or }E\text{ preceeds }I$
$\text{But note that for every arrangement where }I\text{ preceeds }E\\ \text{If you just swap them around, you get an arrangement}\\ \text{where }E\text{ preceeds }I$
$\text{By doing this with EVERY scenario where }I\text{ preceeds }E\\ \text{We see that the amount of words with }I\text{ before }E\text{ and }E\text{ before }I\\ \text{are evenly split!}$
$\therefore\text{Answer: }\boxed{\frac{9!}{2}}$
______________________________________
I'll do C in a bit unless someone gets there before me.

Thanksssss. Yep I need help with C

I also have another question:
A basketball team of 15 players is to be culled into a starting squad of 5 players. In how many ways can this be done so that:
A) the shortest player is not included
B) the tallest player is included
C) the three eldest people are included and thr three youngest people are not included

I'm confused about it because they've hardly given us any info :/

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Author Topic: 3U Maths Question Thread (Read 1524489 times) Tweet Share

Shadowxo

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