Author Topic: 3U Maths Question Thread (Read 1463304 times) Tweet Share

bsdfjnlkasn · « **Reply #1470 on:** February 11, 2017, 08:23:59 am »

Hey there,

Could someone please help me identify where I went wrong? I thought we had to try and get the general form for an inverse tan integral but I don't know why they got 1/3 as a constant out the front when I thought we should be isolating 1/9 to get 1 (in the denominator) - I hope that this makes sense. I'm really new to this and understand that it's an easy question but any help would be appreciated!

bsdfjnlkasn · « **Reply #1471 on:** February 11, 2017, 08:45:37 am »

How would you solve d) without directly applying the formula?

This is my working and as you can see I got different answers both times (I don't want to have to rely on applying formulas) the longer method was integration by substitution

jamonwindeyer · « **Reply #1472 on:** February 11, 2017, 09:05:20 am »

Quote from: bsdfjn;lkasn on February 11, 2017, 08:23:59 am

Hey there,

Could someone please help me identify where I went wrong? I thought we had to try and get the general form for an inverse tan integral but I don't know why they got 1/3 as a constant out the front when I thought we should be isolating 1/9 to get 1 (in the denominator) - I hope that this makes sense. I'm really new to this and understand that it's an easy question but any help would be appreciated!

(Image removed from quote.)

Hey hey! All good, it's not the easiest and I remember struggling with this very thing when I first learned these

Easy Answer: Check your reference sheet, this is actually on there and yields the answer that they give (I think its on the right hand side of the Extension Page)

More Complete Answer: You've not taken into account the chain rule. Let's differentiate your answer:

$f(x)=\frac{1}{9}\tan^{-1}{\frac{x}{3}}\\f'(x)=\frac{1}{9}\times\frac{1}{1+\frac{x^2}{9}}\times\frac{1}{3}$

That one third out the back is what you've missed - It comes from the chain rule of differentiating $\tan^{-1}\frac{x}{3}$ instead of just $\tan^{-1}x$. You multiply by the derivative of $\frac{x}{3}$ to compensate. So since you've got the 1/9 there already, the answer is out!

So in that regard, you've not quite integrated the function correctly. Instead, you need to leave $\frac{1}{3}$ in the integral and only take $\frac{1}{3}$ out the front, to get the answer you need

hope that helps!

jamonwindeyer · « **Reply #1473 on:** February 11, 2017, 09:10:37 am »

Quote from: bsdfjn;lkasn on February 11, 2017, 08:45:37 am

How would you solve d) without directly applying the formula?

(Image removed from quote.)

This is my working and as you can see I got different answers both times (I don't want to have to rely on applying formulas) the longer method was integration by substitution

(Image removed from quote.)

Hey, good on you for wanting to know the long way and not just learn formula, you'll get a better understanding that way

Little error in the first line of that longer working, should be:

$=\int\frac{dx}{\sqrt{\frac{4}{9}\left(1-\frac{9}{4}x^2\right)}}$

The coefficient on your $x^2$ was the wrong way around - Carry that through and it SHOULD work as expected

bsdfjnlkasn · « **Reply #1474 on:** February 11, 2017, 09:22:27 am »

Thank you so much Jamon!! At least they were easy fixes and nothing majorly wrong in my understanding
I'll make sure to ask more questions as they come up - thanks again

jamonwindeyer · « **Reply #1475 on:** February 11, 2017, 10:02:26 am »

Quote from: bsdfjn;lkasn on February 11, 2017, 09:22:27 am

Thank you so much Jamon!! At least they were easy fixes and nothing majorly wrong in my understanding
I'll make sure to ask more questions as they come up - thanks again

No worries! Yep, there are always small little things like this that pop up, they fade with practice!

bsdfjnlkasn · « **Reply #1476 on:** February 11, 2017, 11:12:13 am »

Could someone please help me with Q23 because even though it's a challenge question I'd like to consolidate my skills. I've got it wrong twice and I think it's because I don't know how to find the area in the first quadrant (how to integrate inverse sinx). My method mostly worked around subtracting easy triangle and squares areas to isolate the shaded curve but alas have been unsuccessful.

RuiAce · « **Reply #1477 on:** February 11, 2017, 11:22:41 am »

Quote from: bsdfjn;lkasn on February 11, 2017, 11:12:13 am

Could someone please help me with Q23 because even though it's a challenge question I'd like to consolidate my skills. I've got it wrong twice and I think it's because I don't know how to find the area in the first quadrant (how to integrate inverse sinx). My method mostly worked around subtracting easy triangle and squares areas to isolate the shaded curve but alas have been unsuccessful.

(Image removed from quote.)

$\text{I don't have much time to explain everything right now but here's my suggestion:}\\ \text{Area of triangle} - \int_{0}^{\frac\pi2}\sin y \,dy$

bsdfjnlkasn · « **Reply #1478 on:** February 11, 2017, 11:56:17 am »

Quote from: RuiAce on February 11, 2017, 11:22:41 am

$\text{I don't have much time to explain everything right now but here's my suggestion:}\\ \text{Area of triangle<br />} - \int_{0}^{\frac\pi2}\sin y \,dy$

Cool thanks for the help! That's all I need (I think) to try again ☺

queenie09 · « **Reply #1479 on:** February 12, 2017, 09:54:17 pm »

Can you please help me with these:

1. (a) Integers are formed from the digits 2, 3, 4 and 5, with repetitions not allowed. (i) How many such numbers are there? (ii) How many of them are even? (b) Repeat the two parts to this question if repetitions are allowed.

2. Numbers less than 4000 are formed from the digits 1, 3, 5, 8 and 9, without repetition.How many of them are divisible by 3?

RuiAce · « **Reply #1480 on:** February 12, 2017, 10:14:02 pm »

Quote from: queenie09 on February 12, 2017, 09:54:17 pm

Can you please help me with these:

1. (a) Integers are formed from the digits 2, 3, 4 and 5, with repetitions not allowed. (i) How many such numbers are there? (ii) How many of them are even? (b) Repeat the two parts to this question if repetitions are allowed.

2. Numbers less than 4000 are formed from the digits 1, 3, 5, 8 and 9, without repetition.How many of them are divisible by 3?

$\text{If we just place the 4 digits in any order (without repetition)}\\ \text{Then clearly we're just arranging them like words in a letter.}\\ \therefore 4!$
$\text{If the number is even, the last digit must be a 2 or a 4: }2\text{ outcomes}\\ \text{And then the remaining three digits get arranged anywhere: }3!\\ \therefore 2\times 3!$
_____________________________________
$\text{Well now if repetition is allowed, for every place value we have 4 possible choices}\\ 4\times 4\times 4\times 4=4^4$
$\text{Then, the last digit must be a 2 or a 4, so again there's only 2 choices}\\ \text{But then all 4 digits can be used everywhere else}\\ \therefore 2\times4\times4\times4 = 2\times 4^3$
_____________________________________
$\textbf{This is insanely hard. Especially for a 3U student.}\\ \text{Recall that a number is divisible by 3}\\ \text{if and only if the sum of its digits is divisible by 3}\\ \text{e.g. 74289 is divisible by 3 as}\\ 7+4+2+8+9=30\text{ is divisible by 3}$
$\text{I will assume the numbers must be 4 digit numbers}$
$\text{Among our 5 numbers, we need to pick 4 numbers}\\ \text{There will be }\binom54=5\text{ different combinations}\\ \text{We will test to see which combinations lead to a number divisible by 3}\\ \textit{Note: If the numbers can be less than 4 digits, then this is ridiculous.}$
$\begin{align*}1+3+5+8&=17&&\text{NO}\\ 1+3+5+9&=18&&\text{YES}\\ 1+5+8+9&=23&&\text{NO}\\ 1+3+8+9&=21&&\text{YES}\\ 3+5+8+9&=20&&\text{NO}\end{align*}$
$\text{So it appears as though if we make numbers out of }1,3,5,9\\ \text{Or }1,3,8,9\\ \text{Then our 4-digit number will be divisible by 3}$
$\textit{Case 1: }1,3,5,9\\ \text{The number has to be less than 4000.}\\ \text{Hence the first number must be a 1 or a 3} \implies 2\\ \text{Then the other 3 digits are just arranged} \implies 3!\\ \therefore 2\times 3!\text{ for this case}$
$\textit{Case 2: }1,3,8,9\\ \text{Happens to be functionally identical to the above.}\\ \text{Hence also }2\times 3!$
$\therefore \text{Final answer: }\boxed{2\times 3!+2\times 3!}\\ \text{from combining the cases}$

yasmineturner · « **Reply #1481 on:** February 13, 2017, 07:24:10 pm »

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

A particle is moving in a straight line with a constant acceleration of 2m/s^2. If it starts from the origin with a velocity of -6m/s find: an expression for the velocity in terms of the displacement.

Thankyou so much,
Yasmine

RuiAce · « **Reply #1482 on:** February 13, 2017, 07:31:50 pm »

Quote from: yasmineturner on February 13, 2017, 07:24:10 pm

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

A particle is moving in a straight line with a constant acceleration of 2m/s^2. If it starts from the origin with a velocity of -6m/s find: an expression for the velocity in terms of the displacement.

Thankyou so much,
Yasmine

$\text{If we want }v\text{ in terms of }x\text{ then we will use the formula }\boxed{\ddot{x}=\frac{d}{dx}\left(\frac12v^2\right)}$
$\begin{align*}\frac{d}{dx}\left(\frac12v^2\right)&=2\\ \frac12v^2&=2x+\frac{C}2\tag{*}\\ v^2&=4x+C\end{align*}\\ (*)-\text{ it makes no difference, as }\frac{C}2\text{ is another constant.}$
$\text{When }x=0, v=-6\\ \begin{align*}\therefore 36&=C\\ \therefore v^2&=4x+36\\ \implies v&=\sqrt{4x+36}\end{align*}$
$\text{We know that when }x=0, v=6\\ \text{Hence when }x=0,\text{ we have a positive velocity }v\ge 0\\ \text{So we take the positive root:}\\ v=\sqrt{4x+36}$

yasmineturner · « **Reply #1483 on:** February 13, 2017, 07:43:09 pm »

Quote from: RuiAce on February 13, 2017, 07:31:50 pm

$\text{If we want }v\text{ in terms of }x\text{ then we will use the formula }\boxed{\ddot{x}=\frac{d}{dx}\left(\frac12v^2\right)}$
$\begin{align*}\frac{d}{dx}\left(\frac12v^2\right)&=2\\ \frac12v^2&=2x+\frac{C}2\tag{*}\\ v^2&=4x+C\end{align*}\\ (*)-\text{ it makes no difference, as }\frac{C}2\text{ is another constant.}$
$\text{When }x=0, v=-6\\ \begin{align*}\therefore 36&=C\\ \therefore v^2&=4x+36\\ \implies v&=\sqrt{4x+36}\end{align*}$
$\text{We know that when }x=0, v=6\\ \text{Hence when }x=0,\text{ we have a positive velocity }v\ge 0\\ \text{So we take the positive root:}\\ v=\sqrt{4x+36}$

Thankyou, I understand where I went wrong now

RuiAce · « **Reply #1484 on:** February 13, 2017, 09:26:00 pm »

Quote from: queenie09 on February 13, 2017, 09:22:04 pm

Quote from: RuiAce on February 12, 2017, 10:14:02 pm

$\text{If the number is even, the last digit must be a 2 or a 4: }2\text{ outcomes}\\ \text{And then the remaining three digits get arranged anywhere: }3!\\ \therefore 2\times 3!$
_____________________________________
Hey, I did the same thing for this part, is the answer just wrong or is there another way of solving?

If the answer given was 32 (like your presumably deleted post said) then that's definitely wrong, because 32 is larger than the answer for part a), which is 24.

There should never be more outcomes with a restriction than without.

I'm sticking with my answer, so if you got the same thing then yes the answer is wrong.

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