Author Topic: 3U Maths Question Thread (Read 1464347 times) Tweet Share

jamonwindeyer · « **Reply #1440 on:** February 08, 2017, 09:20:05 am »

Quote from: bsdfjn;lkasn on February 08, 2017, 06:24:11 am

Thank you Rui. I understand the first part but was wondering where the 1 from the cos2x formula went when you were solving part b. How are you able to just equate the two parts of the expression with the angle?

Hey hey, which bit specifically? The first line of Part B comes directly from the end of Part A, is that what you mean?

legorgo18 · « **Reply #1441 on:** February 08, 2017, 04:51:42 pm »

Hi guys need some help with part c in this projectile question

An aircraft is flying with constant velocity U at a constant height h above horizontal ground. When the plane is at A it is directly over a gun at O. When the plane is at B, a shell is fired from the gun at the aircraft along OB. The shell is fired with an initial velocity V at an angle of elevation alpha.

a) If x and y are the horizontal and vertical displacements of the shell from O at time t seconds, show that if g is the acceleration due to gravity, x = VT cos alpha and y = VT sin alpha - 1/2 gt^2

I did: x''= 0, y''=-g

x'=c, t=0, c=0 so x'= v cos alpha
y'=-gt + c, t=0, c=0 so y'=v sin alpha

x= Vt cos alpha + c t=0, c=0 so x= Vt cos alpha + c
y= Vt sin alpha - 1/2gt^2 + c t=0, c=0 so y= Vtsinalpha - 1/2gt^2

b) Show that if the shell hits the aircraft at time T at point C, then VT cos alpha= h/ tan alpha + UT
i did:
x= AB + BC
tan alpha = h/AB
AB = h /tan alpha

BC= UT
so x = h/tan alpha + UT so VT cos alpha = h/tan alpha + UT

c) Show that if the shell hits the aircraft then 2U(V cos alpha - U) tan^2 alpha = gh

no idea... any help is appreciated

RuiAce · « **Reply #1442 on:** February 08, 2017, 05:29:21 pm »

Quote from: legorgo18 on February 08, 2017, 04:51:42 pm

Hi guys need some help with part c in this projectile question

An aircraft is flying with constant velocity U at a constant height h above horizontal ground. When the plane is at A it is directly over a gun at O. When the plane is at B, a shell is fired from the gun at the aircraft along OB. The shell is fired with an initial velocity V at an angle of elevation alpha.

a) If x and y are the horizontal and vertical displacements of the shell from O at time t seconds, show that if g is the acceleration due to gravity, x = VT cos alpha and y = VT sin alpha - 1/2 gt^2

I did: x''= 0, y''=-g

x'=c, t=0, c=0 so x'= v cos alpha
y'=-gt + c, t=0, c=0 so y'=v sin alpha

x= Vt cos alpha + c t=0, c=0 so x= Vt cos alpha + c
y= Vt sin alpha - 1/2gt^2 + c t=0, c=0 so y= Vtsinalpha - 1/2gt^2

b) Show that if the shell hits the aircraft at time T at point C, then VT cos alpha= h/ tan alpha + UT
i did:
x= AB + BC
tan alpha = h/AB
AB = h /tan alpha

BC= UT
so x = h/tan alpha + UT so VT cos alpha = h/tan alpha + UT

c) Show that if the shell hits the aircraft then 2U(V cos alpha - U) tan^2 alpha = gh

no idea... any help is appreciated

The working out in the previous parts isn't really needed; just the final result of b)
$\text{I take that the result proven was}\\ \boxed{VT\cos \alpha = \frac{h}{\tan \alpha}+UT}$
$\text{Observe that }T\text{ vanishes in the result of c)}\\ \text{This means we should make }T\text{ the subject and prepare to eliminate it.}\\ \boxed{T=\frac{h}{\tan \alpha(V\cos \alpha-U)}}$
$\text{The only piece of information not used is the equation in }y\\ \text{Keep in mind that we want to make }gh\text{ the subject.} \\ \text{Note that when }t=T, \quad y=h$
$\begin{align*}\therefore h&=VT\sin \alpha - \frac{gT^2}{2}\\ \therefore h&=V\sin \alpha \left(\frac{h}{\tan \alpha(V\cos \alpha-U)}\right)-\frac{g}{2}\left(\frac{h}{\tan \alpha(V\cos \alpha-U)}\right)^2\\ \tan^2\alpha&=\frac{V\sin \alpha\tan \alpha}{V\cos \alpha-U}-\frac{gh}{2}\left(\frac{1}{V\cos \alpha-U}\right)^2\\ \frac{gh}{(V\cos \alpha-U)^2}&= \frac{2V\sin \alpha\tan \alpha}{V\cos \alpha-U}-2\tan^2\alpha\\ gh&=(V\cos \alpha-U)(2V\sin \alpha\tan \alpha-2\tan^2\alpha (V\cos \alpha-U))\\ &= 2(V\cos \alpha-U)(V\sin \alpha\tan \alpha-V\cos \alpha\tan^2\alpha-U\tan^2\alpha)\\ &= 2(V\cos \alpha-U)(U\tan^2\alpha)\tag{*}\\ &= 2U(V\cos \alpha-U)\tan^2\alpha\end{align*}$
$(*)-\text{Note that }\\ \cos \alpha \tan^2\alpha=\cos \alpha \tan \alpha\tan \alpha = \cos \alpha\left(\frac{\sin \alpha}{\cos \alpha}\right)\tan \alpha=\sin \alpha\tan \alpha\\ \text{so the two adjacent terms cancel out.}$

legorgo18 · « **Reply #1443 on:** February 08, 2017, 06:53:04 pm »

Hi, just got 1 more question

Its just the last bit of this projectile motion question of subbing (dcos thet, dsin thet) into y = x tan alpha - (gx^2 sec^2 alpha)/2v^2
Answer is d = 2v^2 cos alpha sin(alpha - thet)/ gcos^2thet but i got an extra cos thet on top, anyone mind playing spot the mistake?

I did: d sin thet= d cos thet tan alpha - g(d^2cos^2thet) sec^2 alpha/ 2V^2

g(d^2cos^2alpha) sec^2alpha/ 2V^2 = d(costhet tan alpha - sinthet)

g(dcos^2alpha) sec^2alpha = 2V^2(cos alpha tan alpha - sinthet)

d= [2V^2 (costhet tan alpha - sinthet)]/ gcos^2 thet sec^2 alpha

d= 2v^2 (costhet cos alpha sin alpha - sinthet)/gcos^2thet

d=2V^2 costhet cos alpha sin(alpha- thet)/gcos^2thet

Sorry if this is hard to read, idk how to do the formatting. Any help is appreciated

RuiAce · « **Reply #1444 on:** February 08, 2017, 07:14:01 pm »

Quote from: legorgo18 on February 08, 2017, 06:53:04 pm

Hi, just got 1 more question

Its just the last bit of this projectile motion question of subbing (dcos thet, dsin thet) into y = x tan alpha - (gx^2 sec^2 alpha)/2v^2
Answer is d = 2v^2 cos alpha sin(alpha - thet)/ gcos^2thet but i got an extra cos thet on top, anyone mind playing spot the mistake?

I did: d sin thet= d cos thet tan alpha - g(d^2cos^2thet) sec^2 alpha/ 2V^2

g(d^2cos^2alpha) sec^2alpha/ 2V^2 = d(costhet tan alpha - sinthet)

g(dcos^2alpha) sec^2alpha = 2V^2(cos alpha tan alpha - sinthet)

d= [2V^2 (costhet tan alpha - sinthet)]/ gcos^2 thet sec^2 alpha

d= 2v^2 (costhet cos alpha sin alpha - sinthet)/gcos^2thet

d=2V^2 costhet cos alpha sin(alpha- thet)/gcos^2thet

Sorry if this is hard to read, idk how to do the formatting. Any help is appreciated

Ignoring LaTex, could use θ for theta and

Code: [Select]

[sup][/sup] for superscripts to indicate powers. But the easy way out is to just post your working out on paper.
$\begin{align*}d\sin \theta &= d \cos \theta \tan \alpha - \frac{gd^2\cos^2\theta\sec^2\alpha}{2V^2}\\ \frac{gd\cos^2\theta\sec^2\alpha}{2V^2}&=\cos \theta\tan \alpha-\sin \theta\\ d&=\frac{2V^2(\cos \theta\tan \alpha-\sin \theta)}{\cos^2\theta\sec^2\alpha}\end{align*}$
$\text{And then there's confusion. The next line should read}\\ \boxed{d=\frac{2V^2(\cos \theta \cos \alpha\sin \alpha-\boxed{\sin \theta\cos^2\alpha})}{\cos^2\theta}}\\ \text{The line after that is also wrong despite the error carry through}$

legorgo18 · « **Reply #1445 on:** February 08, 2017, 07:29:27 pm »

Omg thx rui, i got it!

it was just the sin sum thing after simplifying...

queenie09 · « **Reply #1446 on:** February 08, 2017, 08:45:01 pm »

Hey!
Can someone please help me with these questions?
1. A game is played in which two coloured dice are rolled once. The six faces of the black die are numbered 5, 7, 8, 10, 11, 14. The six faces of the white die are numbered 3, 6, 9, 12, 13, 15. The player wins if the number on the black die is bigger than the number on the white die.Calculate the probability that a player will lose at least once in two consecutive games.

2. In a game, Anna and Ingrid take turns at drawing, and immediately replacing, a ball from an urn containing three blue and four green balls. The game is won when Anna draws a blue ball, or when Ingrid draws a green ball. Anna goes ﬁrst. Find the probability that:
(c) Anna wins in less than four of her turns
(d) Anna wins the game.

Thank you in advance!!

RuiAce · « **Reply #1447 on:** February 08, 2017, 09:05:34 pm »

Quote from: queenie09 on February 08, 2017, 08:45:01 pm

Hey!
Can someone please help me with these questions?
1. A game is played in which two coloured dice are rolled once. The six faces of the black die are numbered 5, 7, 8, 10, 11, 14. The six faces of the white die are numbered 3, 6, 9, 12, 13, 15. The player wins if the number on the black die is bigger than the number on the white die.Calculate the probability that a player will lose at least once in two consecutive games.

2. In a game, Anna and Ingrid take turns at drawing, and immediately replacing, a ball from an urn containing three blue and four green balls. The game is won when Anna draws a blue ball, or when Ingrid draws a green ball. Anna goes ﬁrst. Find the probability that:
(c) Anna wins in less than four of her turns
(d) Anna wins the game.

Thank you in advance!!

Draw a table of outcomes for Q1 first, to find the probability of the player winning just 1 game.
__________________________________________________
$\text{Note that there is }\textbf{replacement.}$
Presumably parts a) and b) would've been useful in building up to c).
$\text{Since Anna goes first, the probability of Anna winning in the first turn is }\frac{3}{7}$
$\text{Anna can NOT win the second turn, however if Ingrid wins}\\ \text{the game terminates.}\\ \text{So Anna and Ingrid must both fail to draw the right ball}\\ \text{The probability of this occurring is }\frac{4}{7}\times \frac{3}{7}$
$\text{If Anna wins in the third turn (which is her 2nd turn), then after the failures above}\\ \text{Anna draws the right ball. So this occurs with probability}\\ \left(\frac47\times\frac37\right)\times \frac37$
$\text{Now, in a similar way, for there to be a 5th turn (i.e. Anna's 3rd)}\\ \text{They must both draw the wrong balls TWICE: }\left(\frac47\times\frac37\right)^2\\ \text{Then Anna draws the right ball: }\frac37\\ \text{So this can occur with probability }\left(\frac47\times\frac37\right)\times \frac37$
$\text{So if Anna wins in less than 4 of her turns, we simply add the cases up}\\ \begin{align*}Pr(\text{Anna wins before 4th})&=\frac37 + \left(\frac47\times\frac37\right)\times \frac37 + \left(\frac47\times\frac37\right)^2\times \frac37\end{align*}$
__________________________________________
$\text{And once again, for there to be a 7th turn (i.e. Anna's 4th)}\\ \text{We find the probability to be }\left(\frac47\times\frac37\right)^3\times \frac37$
$\text{The idea is that this keeps going. There is a pattern forming}\\ \text{Ultimately, we arrive at a G.P.}$
$\begin{align*}Pr(\text{Anna wins})&=Pr(\text{Anna wins on 1st}) + Pr(\text{Anna wins on 2nd})+Pr(\text{Anna wins on 3rd})+\dots\\ &= \frac37 + \left(\frac47\times\frac37\right)\times \frac37 + \left(\frac47\times\frac37\right)^2\times \frac37 + \dots\\ &= \frac{3}{7}\left(1+\frac{12}{49}+\left(\frac{12}{49}\right)^2+\dots\right)\\ &= \frac37 \left(\frac{1}{1-\frac{12}{49}}\right)\\ &= \frac{21}{37}\end{align*}$
Subject to some minor inaccuracy

queenie09 · « **Reply #1448 on:** February 08, 2017, 09:24:42 pm »

Hey Jake,
I am struggling to answer this questions:
There are two white and three black discs in a bag. Two players A and B are playing a game in which they draw a disc alternately from the bag and then replace it. Player A must draw a white disc to win and player B must draw a black disc. Player A goes ﬁrst. Find the probability that: player A wins in less than four of her draws.

RuiAce · « **Reply #1449 on:** February 08, 2017, 09:29:00 pm »

Quote from: queenie09 on February 08, 2017, 09:24:42 pm

Hey Jake,
I am struggling to answer this questions:
There are two white and three black discs in a bag. Two players A and B are playing a game in which they draw a disc alternately from the bag and then replace it. Player A must draw a white disc to win and player B must draw a black disc. Player A goes ﬁrst. Find the probability that: player A wins in less than four of her draws.

$\text{This follows the precise same working as the above question}\\ \text{The answer will be}\\ \frac{2}{5} + \left(\frac{3}{5}\times \frac{2}{5}\right)\times \frac25 + \left(\frac35\times \frac25\right)^2\times \frac25$

queenie09 · « **Reply #1450 on:** February 08, 2017, 09:36:07 pm »

Hey, sorry for the spam. For the following questions, I am confused as to how the probabilities are different for each

Find how many arrangements of the letters of the word UNIFORM are possible: if the M is somewhere to the right of the U.
Find how many arrangements of the letters of the word BEHAVING: have three vowels occurring together.

Thank you in advance!

RuiAce · « **Reply #1451 on:** February 08, 2017, 09:42:35 pm »

Quote from: queenie09 on February 08, 2017, 09:36:07 pm

Hey, sorry for the spam. For the following questions, I am confused as to how the probabilities are different for each

Find how many arrangements of the letters of the word UNIFORM are possible: if the M is somewhere to the right of the U.
Find how many arrangements of the letters of the word BEHAVING: have three vowels occurring together.

Thank you in advance!

$\text{Note that all letters of }UNIFORM\text{ are different}\\ \text{So without any condition, the total number of arrangements is }7!=5040$
$\textbf{Trick: }\text{If you think about it}\\ \text{The }M\text{ must be either somewhere to the right, or to the left of }U$
$\text{e.g. For a random arrangement like }UFOMRNI\text{ the }M\text{ is to the right of }U\\ \boxed{\text{But if you SWAP the }M\text{ and the }U}\\ \text{you have }M\text{ being to the left of }U: \quad MFOURNI$
$\text{The idea is that exactly HALF the cases will be what we want: }M\text{ to right of }U\\ \text{And the other half are what we don't want}\\ \text{So our answer is just }\frac{5040}{2}=2520$
_____________________________________
$\text{Note that all letters of }BEHAVING\text{ are different}$
$\text{Group the vowels: }EAI\\ \text{We must first pick an ordering for these vowels. This can be done in }3!\text{ ways}$
$\text{Then, treat the vowels as one single letter}\\ \text{We are basically arranging the 6 letters:}\\ B, H, V, N, G, \boxed{EAI}\\ \text{Which is done in }6!\text{ ways}$
$\text{So the number of arrangements is }3!\times 6!=4320$

bsdfjnlkasn · « **Reply #1452 on:** February 08, 2017, 10:59:14 pm »

Quote from: jamonwindeyer on February 08, 2017, 09:20:05 am

Hey hey, which bit specifically? The first line of Part B comes directly from the end of Part A, is that what you mean?

Hey, sorry I wasn't very clear.

I'm a confused by the third line of part b) and how we're just able to equate the two inverse functions (also the conclusion doesn't make much sense)
Sorry this is all new to me so i'm still working on what I don't know

RuiAce · « **Reply #1453 on:** February 08, 2017, 11:01:20 pm »

Quote from: bsdfjn;lkasn on February 08, 2017, 10:59:14 pm

Hey, sorry I wasn't very clear.

I'm a confused by the third line of part b) and how we're just able to equate the two inverse functions (also the conclusion doesn't make much sense)
Sorry this is all new to me so i'm still working on what I don't know

$\theta = \sin^{-1}\frac35\text{ was given at the very start}\\ \text{All that happens in line 3 is that I subbed }\theta\text{ back in for }\sin^{-1}\frac35$

yasmineturner · « **Reply #1454 on:** February 09, 2017, 05:20:54 pm »

Hello,

I was wondering if someone could please help me with the following question:

Sand falling from a funnel forms a conical pile such that the height of the pile is one and a half times the radius.
a) Show that the volume of the pile is given by V=(pi)r^3/2.
b) If the sand is falling at the rate of (pi)/10m^3 per minute, find the rate at which the height is increasing when the pile of sand is 3m.

I have answered (a) and got the right answer, but I do not get the right answer in (b). The answer of (b) should be 1/40.

Thankyou so much

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