Mathematics Question Thread

[SSSS]

Ok then, thanks for the help!! 

[Thebarman]

Here is the method.

Do a free-hand sketch clearly showing the focus and the directrix. Then, figure out the orientation of the parabola (does it concave upwards, downwards or sideways). Then work out the focal length, and finally work out the vertex (which you can write its equation afterwards).

Note: The focal length is the distance from the vertex to the focus.
The (perpendicular) distance from the focus to the directrix is twice the focal length


If it doesn't come out, post up your working

Thank you so much, everything is good now!

Mind if I ask another question?
The price of shares in a particular company is falling by an average of 2% each day. How many days would it take fo rhte shares to half in value, and after how many days will the shares be worth 25% of their value?
Thanks again 

[RuiAce]

Thank you so much, everything is good now!

Mind if I ask another question?
The price of shares in a particular company is falling by an average of 2% each day. How many days would it take fo rhte shares to half in value, and after how many days will the shares be worth 25% of their value?
Thanks again 

[katnisschung]

find any stationary points on the curve y=(x-3) (4-x)^1/2
the second part is all under a square root....

[RuiAce]

find any stationary points on the curve y=(x-3) (4-x)^1/2
the second part is all under a square root....

[katnisschung]

ahaha i differentiated it wrong thanks ruiace

[RuiAce]

ahaha i differentiated it wrong thanks ruiace

These things happen haha. When in doubt, use WolframAlpha

[Wales]

Just a handy tip my tutor told me. Not sure if it's been here yet but

When you differentiate a Root it ALWAYS :

dy/dx = u'/2root(u)

eg:

SQRT(3x^2+2x+3)

dy/dx = 6x+2/2SQRT(3x^2+2x+3)
          =3x+1/SQRT(3x^2+2x+3)
Saves the time writing it all out. Just remember u'/2rootu

[RuiAce]

Just a handy tip my tutor told me. Not sure if it's been here yet but

When you differentiate a Root it ALWAYS :

dy/dx = u'/2root(u)

eg:

SQRT(3x^2+2x+3)

dy/dx = 6x+2/2SQRT(3x^2+2x+3)
          =3x+1/SQRT(3x^2+2x+3)
Saves the time writing it all out. Just remember u'/2rootu

It's better if you can remember it as a mnemonic or some catchy rhyme
_________________

Personally I never showed working out for any chain rule. I just smashed it in one step for every single thing.

[Happy Physics Land]

Just a handy tip my tutor told me. Not sure if it's been here yet but

When you differentiate a Root it ALWAYS :

dy/dx = u'/2root(u)

eg:

SQRT(3x^2+2x+3)

dy/dx = 6x+2/2SQRT(3x^2+2x+3)
          =3x+1/SQRT(3x^2+2x+3)
Saves the time writing it all out. Just remember u'/2rootu

I still work everything out through first principles. It's safer that way.

[Wales]

It's better if you can remember it as a mnemonic or some catchy rhyme
_________________

Personally I never showed working out for any chain rule. I just smashed it in one step for every single thing.

Yeah the explanation is on point. I personally found it much easier to just remember it, 1 over 2rootu is pretty catchy in my head aha. Each to their own though.

I still work everything out through first principles. It's safer that way.

HUH... Wouldn't that literally take a entire page? I hated First Principals. It was pointless to me and never made any sense why we needed to know

[jakesilove]

HUH... Wouldn't that literally take a entire page? I hated First Principals. It was pointless to me and never made any sense why we needed to know

Yeah. He's weird. Ignore him.

[Happy Physics Land]

Yeah. He's weird. Ignore him.

Stfu Jake.

By first principle I meant like I like to differentiate it using the normal way, not using any tricks.
Like Im not talking about the f'(x) = f(x-h) - f(x) / h thing.

[RuiAce]

[Wales]

Stfu Jake.

By first principle I meant like I like to differentiate it using the normal way, not using any tricks.
Like Im not talking about the f'(x) = f(x-h) - f(x) / h thing.

Bit misleading aye :^)

That aside... Don't ever want to see First Principals again   As Rui Said, building blocks