Author Topic: 3U Maths Question Thread (Read 1235625 times) Tweet Share

VydekiE · « **Reply #1890 on:** April 23, 2017, 12:39:06 pm »

Hi, could I please get some help on this question,
1. Grain is poured at a constant rate of 0.5 cubic metres per second. It forms a conical pile, with the angle at the apex of the cone equal to 60^o. The height of the pile is h metres and the radius of the base is r metres. You may use the formula V=1/3 pi r² h
a) Show that r= h/root3 (I've proved this)
b) Show that the volume of the pile, V, is given by V=pih³ /9 (I've proved this)
c) Hence, find the rate at which the height of the pile is increasing when the height of the pile is 3 metres.
Thank you!!

Shadowxo · « **Reply #1891 on:** April 23, 2017, 12:47:29 pm »

Quote from: VydekiE on April 23, 2017, 12:39:06 pm

Hi, could I please get some help on this question,
1. Grain is poured at a constant rate of 0.5 cubic metres per second. It forms a conical pile, with the angle at the apex of the cone equal to 60^o. The height of the pile is h metres and the radius of the base is r metres. You may use the formula V=1/3 pi r² h
a) Show that r= h/root3 (I've proved this)
b) Show that the volume of the pile, V, is given by V=pih³ /9 (I've proved this)
c) Hence, find the rate at which the height of the pile is increasing when the height of the pile is 3 metres.
Thank you!!

You need dh/dt, you know dV/dt (0.5) so as dh/dt = dh/dV * dV/dt, you just need dh/dV. You can find this by differentiating part b) (dh/dV = 1 / (dV/dh)

jakesilove · « **Reply #1892 on:** April 23, 2017, 12:47:45 pm »

Quote from: VydekiE on April 23, 2017, 12:39:06 pm

Hi, could I please get some help on this question,
1. Grain is poured at a constant rate of 0.5 cubic metres per second. It forms a conical pile, with the angle at the apex of the cone equal to 60^o. The height of the pile is h metres and the radius of the base is r metres. You may use the formula V=1/3 pi r² h
a) Show that r= h/root3 (I've proved this)
b) Show that the volume of the pile, V, is given by V=pih³ /9 (I've proved this)
c) Hence, find the rate at which the height of the pile is increasing when the height of the pile is 3 metres.
Thank you!!

Hey! Basically, we're looking for

$\frac{dh}{dt}$

ie the rate of change in height. Now, this chain rule conversion becomes natural over time, but for now use the information they have supplied you with to guess that

$\frac{dh}{dt}=\frac{dh}{dV} \times \frac{dV}{dt}$

This is obviously true; you can just cancel out the dVs, and the equation looks like what it should! We already know the rate of change of volume over time; it's 0.5 cubic metres per second. We can quickly find the other term

$V=\frac{\pi h^3}{9}$
$\frac{dV}{dh}=\frac{\pi h^2}{3}$
$\frac{dh}{dV}=\frac{3}{\pi h^2}$

Putting this all together, we can show that

$\frac{dh}{dt}=\frac{3}{\pi h^2} \times 0.5$

Finally, we simply sub h=3 to find our final answer

All of these sorts of questions are identical, and the method is exactly what I've used above. Practice is the only way to perfect this section; do a billion of these questions, and the answers just roll off your pen!

VydekiE · « **Reply #1893 on:** April 23, 2017, 01:19:25 pm »

Thank you!!

itssona · « **Reply #1894 on:** April 24, 2017, 01:12:58 pm »

heey so I was looking at this question and the first thing it says is:
volume of solid is given by: pi integral y^2 dx

how???

I read everything after that and it makes sense, and I understand the integrating perfectly, so my real problem is actually understanding how the volume is given by pi integral y^2 dx

thank you!

RuiAce · « **Reply #1895 on:** April 24, 2017, 01:36:23 pm »

Quote from: sssona09 on April 24, 2017, 01:12:58 pm

heey so I was looking at this question and the first thing it says is:
volume of solid is given by: pi integral y^2 dx

how???

I read everything after that and it makes sense, and I understand the integrating perfectly, so my real problem is actually understanding how the volume is given by pi integral y^2 dx

thank you!

That's a formula that you're taught in 2U integration.

RuiAce · « **Reply #1896 on:** April 25, 2017, 02:05:36 pm »

Quote from: davidss on April 22, 2017, 08:55:45 pm

Would appreciate it if someone helps with 31 and 5c pls. Answers are: 2^n-2 and 108

(Image removed from quote.)

Sorry, forgot to get back to this.

The question's quite bizarre. There has got to be an easier way of doing it. What I did was I tried to invent a method for part a), and then copy it but with the even number restriction for part c.
____________________________________

For part a), we first note that every 5 digit number is greater than 4000. For {0,2,4,5,7}, the number of 5 digit numbers we can make (without repetition) is 4*4! = 96 because the first digit cannot be 0.

This leaves us with 72 possible 4 digit numbers greater than 4000. To check this, we may split the cases based off the issue that the first digit cannot be a 0 or 2.
$\text{Case 1: 0 is picked but not 2.}\\ \text{Then 4, 5, 7 are also picked.}\\ \text{2 can't be first digit, but after the first digit is done the other digits are random}\\ \text{This case yields }3\times 3! = 18\text{ possible outcomes}$
$\text{Case 2: 2 is picked but not 0.}\\ \text{Follows exactly as above, hence }18\text{ possible outcomes}$
$\text{Case 3: 0 and 2 are both picked}\\ \text{We first have the issue that we need to choose 2 out of }\{4,5,7\}\\ \text{This is doable in }\binom32\text{ ways}$
$\text{Once again, the first digit cannot be 0 or 2}\\ \text{But now we only have 2 choices for the first digit, not 3}\\ \text{But the rest are random}\\ \text{Hence }\binom{3}{2}\times 2\times 3! = 36\text{ ways}$
$\text{Recombine: }18+18+36=72$
Note: 0 and 2 cannot BOTH be not picked, because then our numbers are only 3 digits
____________________________________

Nice. Now let's add in the restriction.

For 5 digit numbers that are even, the last digit must be 0, 2 or 4. Split the cases up.
$\text{If }0\text{ is the last digit, then the first four digits can be in any order}\\ \therefore 4!\\\text{If }2\text{ is the last digit, we still have to address that 0 can't be the first digit}\\ \therefore 3\times 3!\\ \text{If 4 is the last digit, we have the same as 2 being the last digit}\\ \therefore 3\times 3!\\ \text{Add: }4!+3\times 3!+3\times 3!=60$
108 - 60 = 48, so apparently there are 48 possible 4 digit numbers. Let's check this.

Problem: We have TWO restrictions on 0 and 2. They can't be the first digit, and they may be forced into being the last digit. This is going to look bizarre.

Instead, consider the complementary event, where the numbers are still greater than 4000 BUT they are ODD. Remember that from above, we know that the total number of 4 digit numbers is 72, so we expect that the total number of odd 4 digit numbers is 24.

Note that we actually don't need to use the complement, and the maths is POTENTIALLY easier (I have not tried it). But it will make the visualisation even harder.
$\text{Case 1: 0 is chosen but 2 is not}\\ \text{Fix 5 or 7 into the last digit: }2\\ \text{Deny 0 from being the first digit and then arrange normally: }2\times 2!\\ \text{So this case yields }2\times 2\times 2! = 8$
$\text{Case 2: 2 is chosen but 0 is not}\\ \text{Exact same as above, so this yields }8$
$\text{Case 3: Both 0 and 2 are chosen}$
Things go haywire, as we need to decide between {4,5,7} which get chosen.

Subcase 3.1: {4,5} chosen
5 has to be the last digit
4 has to be the first digit
Then 0 and 2 can be placed in any order - 2 possible ways

Subcase 3.2: {4,7} chosen
Just like above - 2 possible ways

Subcase 3.3: {5,7} chosen
5 or 7 is the last digit - 2 possible ways
The leftover has to be the first digit
0 and 2 can still be placed in any order - 2 possible ways
So this subcase yields 2*2 = 4 possible ways
$\text{Combine the subcases back into case 3:}\\ 2+2+4=8$
$\text{Combine all cases back to what we wanted}\\ 8+8+8=24$
Try repuzzling everything from here.

legorgo18 · « **Reply #1897 on:** April 26, 2017, 07:27:52 pm »

Yo i cant do these 2 inequalities, well i dont understand the answer for the first one and the second one i just need some direction

1) Solve the inequation 2^2x - 2(2^x) < or equal to -1 [ans is x=0]
2) Solve log base (1/2) (1/x) > log base 2 (3x - 1) for x

Thank you so much!!

jakesilove · « **Reply #1898 on:** April 26, 2017, 07:45:41 pm »

$2^{2x}-2(2^x)-1\le 0$

Now, if we make a clever substitution, we can solve this equation

$u=2^x$
$u^2-2u-1\le0$
$(u-1)^2\le0$

Now, the left hand side will NEVER be less than zero. A real number squared is positive. However, it may be EQUAL to zero. In that case,

$u=1$
$2^{x}=1$
$x=0$

Now,

$log_{\frac{1}{2}}(\frac{1}{x}) > log_2(3x-1)$

First, we can use our log laws to rewrite the left hand side

$-log_{\frac{1}{2}}(x) > log_2(3x-1)$

Finally, I'll rewrite the left hand side to turn everything into base 2

$-\frac{log_2x}{log_2(\frac{1}{2})}>log_2(3x-1)$

Conveniently, the bottom is just equal to -1

$log_2x>log_2(3x-1)$
$x>3x-1$
$-2x>-1$
$x<\frac{1}{2}$

Now, there is actually a bit more that is tough to spot. The value within a log must be positive. Thus,

$3x-1>0$
$x>\frac{1}{3}$

Edit by Rui: Caught some typos

bluecookie · « **Reply #1899 on:** April 27, 2017, 06:00:08 pm »

1. cosinv(3/5)-taninv(-2)=sininv(k)

jakesilove · « **Reply #1900 on:** April 27, 2017, 07:07:52 pm »

So we have

$\cos^{-1}\frac{3}{5}-\tan^{-1}(-2)=\sin^{-1}k$

If we 'sin' both sides, we get

$\sin(\cos^{-1}\frac{3}{5}-\tan^{-1}(-2))=\sin(\sin^{-1}k)=k$

Then, we expand the left hand side out using the standard sin(a-b) rule. Finally, draw yourself two triangles such that

$\cos\alpha=\frac{3}{5}$
and
$\tan\beta=\frac{-2}{1}$

Use Pythag to get the third side, and find cos and sin of the same angles. I don't have time to finish the whole solution, hopefully that sets you on your way! (Or Rui jumps in)

legorgo18 · « **Reply #1901 on:** April 27, 2017, 09:38:03 pm »

Hello, i need some help with the following

1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc

Thank you :c

jamonwindeyer · « **Reply #1902 on:** April 27, 2017, 09:44:57 pm »

Quote from: legorgo18 on April 27, 2017, 09:38:03 pm

Hello, i need some help with the following

1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc

Thank you :c

Might be worth posting this in the 4U thread instead legorgo!! This is definitely beyond 3U, and I'd hate for the 4U crew to miss it

bluecookie · « **Reply #1903 on:** April 29, 2017, 10:11:52 am »

Quote from: jakesilove on April 27, 2017, 07:07:52 pm

So we have

$\cos^{-1}\frac{3}{5}-\tan^{-1}(-2)=\sin^{-1}k$

If we 'sin' both sides, we get

$\sin(\cos^{-1}\frac{3}{5}-\tan^{-1}(-2))=\sin(\sin^{-1}k)=k$

Then, we expand the left hand side out using the standard sin(a-b) rule. Finally, draw yourself two triangles such that

$\cos\alpha=\frac{3}{5}$
and
$\tan\beta=\frac{-2}{1}$

Use Pythag to get the third side, and find cos and sin of the same angles. I don't have time to finish the whole solution, hopefully that sets you on your way! (Or Rui jumps in)

Actually, could you finish it off? That was actually the part I was having trouble with, the rest is all fine

I'm just confused about which side of the triangle is negative (is the -2 and 1 or -1 and 2) and whether we take the negative when calculating another sin/cos values for that triangle. I thought since inverse tan existed only for angles in the 1st and 4th quadrant, that we draw the triangle in either the 1st or 4th quadrant: https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/ebaa19ac-ff8b-43a6-a793-a00d9ac15e86.png and let one of the sides be the negative depending on which one it is. But I tried it, and then did the LHS/RHS with final answer subbed in, if I did it correctly I should have gotten 1, but I didn't, so I don't know where I went wrong ://

RuiAce · « **Reply #1904 on:** April 29, 2017, 11:12:35 am »

Quote from: bluecookie on April 29, 2017, 10:11:52 am

Actually, could you finish it off? That was actually the part I was having trouble with, the rest is all fine I'm just confused about which side of the triangle is negative (is the -2 and 1 or -1 and 2) and whether we take the negative when calculating another sin/cos values for that triangle. I thought since inverse tan existed only for angles in the 1st and 4th quadrant, that we draw the triangle in either the 1st or 4th quadrant: https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/ebaa19ac-ff8b-43a6-a793-a00d9ac15e86.png and let one of the sides be the negative depending on which one it is. But I tried it, and then did the LHS/RHS with final answer subbed in, if I did it correctly I should have gotten 1, but I didn't, so I don't know where I went wrong ://

Screw the negatives. Make everything positive using the following identities.
$\begin{align*}\sin(-x)&=-\sin x\\ \cos(-x)&=\cos x\\ \tan(-x)&=-\tan x\\ \sin^{-1}(-x)&=-\sin^{-1}x\\ \cos^{-1}(-x)&=\pi - \cos^{-1}x\\ \tan^{-1}(-x)&=-\tan^{-1}x\end{align*}$
Seriously, positives are always better to work with than negatives. There's always confusions if you leave negative arguments everywhere.
$\text{So in your question, ignore }-\tan^{-1}(-2)\\ \text{Just replace it with }\tan^{-1}2$

Search the forums now!

We have moved!

Author Topic: 3U Maths Question Thread (Read 1235625 times) Tweet Share

VydekiE

Re: 3U Maths Question Thread

Shadowxo

Re: 3U Maths Question Thread

jakesilove

Re: 3U Maths Question Thread

VydekiE

Re: 3U Maths Question Thread

itssona

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

legorgo18

Re: 3U Maths Question Thread

jakesilove

Re: 3U Maths Question Thread

bluecookie

Re: 3U Maths Question Thread

jakesilove

Re: 3U Maths Question Thread

legorgo18

Re: 3U Maths Question Thread

jamonwindeyer

Re: 3U Maths Question Thread

bluecookie

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

Recent Posts

User:
Password: