Author Topic: 4U Maths Question Thread (Read 663333 times) Tweet Share

ellipse · « **Reply #1035 on:** March 29, 2017, 09:11:07 pm »

Quote from: saloni99.95 on March 29, 2017, 08:58:17 pm

Heey
I got part a by subbing z= i.
How do u solve part b?

You guys are awesome!!

let the other root be 'a.'

By doing the sum of roots, a+i=(1+i)/(2-i)
By realising the RHS, we get 1/5+3i/5=a+i.
Now solving for 'a,' you get a=1/5(1-2i)

ellipse · « **Reply #1036 on:** March 29, 2017, 09:42:53 pm »

Quote from: chelseam on March 29, 2017, 08:10:43 pm

Hi! Can someone please help me with part b of this question? Thank you!

jakesilove · « **Reply #1037 on:** March 30, 2017, 08:17:36 am »

Quote from: ellipse on March 29, 2017, 09:42:53 pm

You absolute Mathematical legend

chelseam · « **Reply #1038 on:** March 30, 2017, 06:43:06 pm »

Thank you so much!

Quote from: ellipse on March 29, 2017, 09:42:53 pm

qwertycc · « **Reply #1039 on:** March 31, 2017, 09:19:08 pm »

Hi can someone please help me find the time taken for this resisted motion question. thanks very much!!

RuiAce · « **Reply #1040 on:** April 01, 2017, 10:35:16 am »

Quote from: qwertycc on March 31, 2017, 09:19:08 pm

Hi can someone please help me find the time taken for this resisted motion question. thanks very much!!

In the future, please provide any progress.
$\text{From }v^2=\frac{2gR^2}{x}-gR\\ \text{noting that we need the positive square root for the upwards journey}$
$\begin{align*}v^2&=gR\left(\frac{2R}{x}-1\right)\\ &= gR \frac{2R-x}{x}\\ \frac{dx}{dt}&=\sqrt{gR}\sqrt{\frac{2R-x}{x}}\\ \sqrt{\frac{x}{2R-x}}dx&=\sqrt{gR}\,dt\end{align*}$
$\text{Note that}\\ \begin{align*}\int \sqrt{\frac{x}{2R-x}}dx &= \int \frac{x}{\sqrt{x(2R-x)}}dx\\ &= \int \frac{R-x}{\sqrt{2Rx-x^2}}dx + \int \frac{R}{\sqrt{R^2-(x-R)^2}}dx\\ &= -\sqrt{2Rx-x^2} + R \sin^{-1}\left(\frac{x-R}{R}\right)+C\end{align*}$
$\text{When }t=0, x=R\\ \text{When }t=T, x=2R\\ \begin{align*}\therefore \int_R^{2R} \sqrt{\frac{x}{2R-x}}dx&=\int_0^T\sqrt{gR}\,dt \\ \left[R \sin^{-1}\left(\frac{x-R}{R}\right) - \sqrt{2Rx-x^2} \right]_R^{2R}&= \sqrt{gR}T\\ \frac{R\pi}{2}-R &=\sqrt{gR}T\\ \sqrt{\frac{R}g}\left(\frac{\pi}{2}-1\right)&=T\end{align*}$

chelseam · « **Reply #1041 on:** April 01, 2017, 10:04:17 pm »

Hi! Is there a way to recognise when to draw a major/minor arc that's above or below the x-axis for the type of question
arg(z±(...)/z±(...)) = π/(...)?

Thank you

RuiAce · « **Reply #1042 on:** April 01, 2017, 10:14:12 pm »

Quote from: chelseam on April 01, 2017, 10:04:17 pm

Hi! Is there a way to recognise when to draw a major/minor arc that's above or below the x-axis for the type of question
arg(z±(...)/z±(...)) = π/(...)?

Thank you

$\text{I always write it as}\\ \arg(z-z_1) - \arg(z-z_2) = \theta$
$\text{Once it's in this form, I always remember it as}\\ \text{you're drawing in a counter-clockwise direction from }z_1\text{ to }z_2$
I keep forgetting why so I'll link to one of Eddie Woo's videos

chelseam · « **Reply #1043 on:** April 01, 2017, 10:27:52 pm »

That's so helpful! Thank you so much!

Quote from: RuiAce on April 01, 2017, 10:14:12 pm

$\text{I always write it as}\\ \arg(z-z_1) - \arg(z-z_2) = \theta$
$\text{Once it's in this form, I always remember it as}\\ \text{you're drawing in a counter-clockwise direction from }z_1\text{ to }z_2$
I keep forgetting why so I'll link to one of Eddie Woo's videos

qwertycc · « **Reply #1044 on:** April 02, 2017, 09:33:20 am »

Rui Ace, thanks for your help!

Kle123 · « **Reply #1045 on:** April 03, 2017, 01:02:31 pm »

can anybody help me out with this question. thank you

Kle123 · « **Reply #1046 on:** April 03, 2017, 01:14:00 pm »

In this question i got pi as the answer but i did it through sketching on the argand diagram. Is the answer pi? and is this the simplest method?

ellipse · « **Reply #1047 on:** April 03, 2017, 01:16:33 pm »

Quote from: Kle123 on April 03, 2017, 01:02:31 pm

can anybody help me out with this question. thank you

ellipse · « **Reply #1048 on:** April 03, 2017, 01:21:02 pm »

Quote from: Kle123 on April 03, 2017, 01:14:00 pm

In this question i got pi as the answer but i did it through sketching on the argand diagram. Is the answer pi? and is this the simplest method?

you could do it using argument rules

Kle123 · « **Reply #1049 on:** April 03, 2017, 02:00:32 pm »

Thank you so much

Search the forums now!

We have moved!

Author Topic: 4U Maths Question Thread (Read 663333 times) Tweet Share

ellipse

Re: 4U Maths Question Thread

ellipse

Re: 4U Maths Question Thread

jakesilove

Re: 4U Maths Question Thread

chelseam

Re: 4U Maths Question Thread

qwertycc

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

chelseam

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

chelseam

Re: 4U Maths Question Thread

qwertycc

Re: 4U Maths Question Thread

Kle123

Re: 4U Maths Question Thread

Kle123

Re: 4U Maths Question Thread

ellipse

Re: 4U Maths Question Thread

ellipse

Re: 4U Maths Question Thread

Kle123

Re: 4U Maths Question Thread

Recent Posts

User:
Password: