HSC 2013 Maths Marathon (archive) (1 Viewer)

[Menomaths]

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]

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]

Re: HSC 2013 2U Marathon

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

 

[Badsiii]

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]

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]

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?

 

[spanky125]

Re: HSC 2013 2U Marathon

No you are right, you can not use

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

When you have the gradient of the entire function, not just a tangent.

 

[RealiseNothing]

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]

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]

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]

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]

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]

Re: HSC 2013 2U Marathon

Because the gradient isn't 6 it's 2.. Plus you can use that way.. Just with the gradient of 2 not 6

If you are referring to using then no, you can't use that way.

 

[RealiseNothing]

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]

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.

 

[spanky125]

Re: HSC 2013 2U Marathon

Find the derivative of:

 

[Badsiii]

Re: HSC 2013 2U Marathon

You can only use when finding the equation of a straight line, whether that be a tangent to a curve or the function itself.

Yes true

 

[andybandy]

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>

 

[Badsiii]

Re: HSC 2013 2U Marathon

Find the derivative of:

Tan3x + 3xsec^2(3x)

 

[Menomaths]

Re: HSC 2013 2U Marathon

Find the derivative of:

uv'+u'v

x.3sec^2(3x)+tan(3x)

= 3xsec^2(3x)+tan(3x)