HSC 2013 Maths Marathon (archive) (4 Viewers)

Status
Not open for further replies.

Menomaths

Exaı̸̸̸̸̸̸̸̸lted Member
Joined
Jul 9, 2013
Messages
2,373
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

Why bother? Just integrate it then sub in the points to find the value of C
Yeah but what if in the exam I think about using the equation, then I'll have two different answers so I must know why we can't use the equation
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

Yeah but what if in the exam I think about using the equation, then I'll have two different answers so I must know why we can't use the equation
What's stopping you?

Edit: Ok I realised this sounded rhetorical. Something actually is stopping you, I'm asking what is it.
 

Menomaths

Exaı̸̸̸̸̸̸̸̸lted Member
Joined
Jul 9, 2013
Messages
2,373
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

different answers...I think...I have a feeling I'm doing something wrong (like usual)
 

Badsiii

Member
Joined
Oct 19, 2013
Messages
33
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

Yeah but what if in the exam I think about using the equation, then I'll have two different answers so I must know why we can't use the equation
Use that only if you have the points and gradient but not the equation. For example finding tangent to the line or the normal equation. Here they give you the derivative of the curve. So just integrate find C by subbing the points and that's your equation; y=2x^3-6x-5
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

different answers...I think...I have a feeling I'm doing something wrong (like usual)
No you are right, you can not use

Think about what this equation actually means. i.e. when can you use it?
 

spanky125

Just waitin' for a mate
Joined
Sep 7, 2012
Messages
285
Location
Classified
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

It's so much easier with the 3U method, wish we could use it



For this question why can't we use
Is this correct, btw how do I make it show the image in the thread?

Solution.jpg
 
Last edited:

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

Use that only if you have the points and gradient but not the equation. For example finding tangent to the line or the normal equation. Here they give you the derivative of the curve. So just integrate find C by subbing the points and that's your equation; y=2x^3-6x-5
He's asking why does it give you a different answer, not necessarily which method is better.
 

Menomaths

Exaı̸̸̸̸̸̸̸̸lted Member
Joined
Jul 9, 2013
Messages
2,373
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

Ahh I get it, thanks =)

yeah that's the correct answer spanky

You can show images by typing [ img]xxx[/ img] without the spaces
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

When you have the gradient of the entire function, not just a tangent.
That is true. But on a more general level, what does using that equation imply about the curve that may not be true? (in this case it isn't true, which is why you can't use it).
 

Badsiii

Member
Joined
Oct 19, 2013
Messages
33
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

He's asking why does it give you a different answer, not necessarily which method is better.
Because the gradient isn't 6 it's 2.. Plus you can use that way.. Just with the gradient of 2 not 6
 

spanky125

Just waitin' for a mate
Joined
Sep 7, 2012
Messages
285
Location
Classified
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

He's asking why does it give you a different answer, not necessarily which method is better.
It gives a different answer because the derivative of the equation only gives you the gradient at a point, if you try to put that gradient into the equation: y-y1=m(x-x1) then you will find the equation of the tangent to the curve at that point.
Whereas if you integrate and sub in any points on the curve then you will find the equation of the function.
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

It gives a different answer because the derivative of the equation only gives you the gradient at a point, if you try to put that gradient into the equation: y-y1=m(x-x1) then you will find the equation of the tangent to the curve at that point.
Whereas if you integrate and sub in any points on the curve then you will find the equation of the function.
Pretty much. A better explanation would be that using implies that the curve is a straight line. But this isn't always the case, which is why you can't always use it.
 

spanky125

Just waitin' for a mate
Joined
Sep 7, 2012
Messages
285
Location
Classified
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

Pretty much. A better explanation would be that using implies that the curve is a straight line. But this isn't always the case, which is why you can't always use it.
You can only use when finding the equation of a straight line, whether that be a tangent to a curve or the function itself.
 

andybandy

Member
Joined
Sep 1, 2012
Messages
294
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\partial&space;}{\partial&space;x}[xtan3x]&space;=&space;x.3sec^{2}3x&plus;1.tan3x=&space;3xsec^{2}3x&plus;tan3x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\partial&space;}{\partial&space;x}[xtan3x]&space;=&space;x.3sec^{2}3x&plus;1.tan3x=&space;3xsec^{2}3x&plus;tan3x" title="\frac{\partial }{\partial x}[xtan3x] = x.3sec^{2}3x+1.tan3x= 3xsec^{2}3x+tan3x" /></a>
 
Last edited:
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 4)

Top