Maths ruinded my life...

Dr_Noswad · Sep 9, 2004

im not sure if ive set that out right either :$

Ikerous · Sep 9, 2004

Close.. but no.

Dr_Noswad · Sep 9, 2004

Dr_Noswad said:
im not sure if this is correct but is it sumwhere along the lines of:
(x+21^2)-(x+{x+2})
i think thats right, ish...

i c now, (x+21^2)-((x)x{x+2})

Ikerous · Sep 9, 2004

use * instead of x to muliply.. and thats a lot closer. I think theres still one thing wrong. I might be mistaken though because i havent actually looked at the attached image to figure out the answer. (I think the ^2 is in the wrong place though...)

SLH · Sep 9, 2004

Yep, but don't forget to put brackets round 'x+21' (because raising something to a power has a high precedence than adding).

The next thing you could do is introduce a variable called 'L' (for example) which is the length of the T shape. (you've done a formula where L=3). You could then develop a formula that used 2 variables, the one you've got currently, and this new one called 'L'.

To do this use the same method as before. For different values of X and L try and workout what the numbers in the corners and the bottom of the T would be. To do this easily, only change one variable at a time - i.e. start with something like X=6, L=3 then move on to X=6, L=4.

Once you've done that you could try making a formula for if you rotated the T shape.

Anyway i'm off to bed now, good luck!

Dr_Noswad · Sep 9, 2004

i c im being a noob once again here

so its deff this
(x+21)^2-(x*{x+2})

Ikerous · Sep 9, 2004

; Dear goodness you must have had the longest dang answers on your math hw... why would he spend his time coming up with all that :-/ I'm pretty sure its not expect of him...

And yea, i think you've got it.

Neutrino · Sep 9, 2004

Ikerous said:
; Dear goodness you must have had the longest dang answers on your math hw... why would he spend his time coming up with all that :-/ I'm pretty sure its not expect of him...

Probably not expected, but it can be fun.

Oh and ya you have the answer. Now you can simplify it down into your original equation.

Dr_Noswad · Sep 9, 2004

ah well never hurt to do a bit better...but this is odd i start off not having a clue wot to even do and now im onto that

PvtRyan · Sep 9, 2004

Dr_Noswad said:
i c now, (x+21^2)-((x)x{x+2})

Almost

Another test for any mathlovers out there:

Find the formula for tangent that touches the graph of the formula
Y = sqrt(2x²+x) * x² in X = 2

(see attachment, find the formula of the line with the red arrow

)

SLH · Sep 9, 2004

Dr_Noswad said:
i c now, (x+21^2)-((x)x{x+2})

Looking at this part:

x+21^2

As it stands it means x+441 because you do the squaring before the adding. To make the formula correct place brackets as below:

(x+21)^2

Here is a better version of the formula:

(x+21)^2-x(x+2)

That only includes brackets when needed (i removed the multiplication sign as it's implied when written as i did).
NOTE: this is in it's simplified form.

Dr_Noswad · Sep 9, 2004

So is this correct? :S
(x+21)^2-(x*{x+2})

PvtRyan · Sep 9, 2004

Dr_Noswad said:
So is this correct? :S
(x+21)^2-(x*{x+2})

Yes that is correct.

SLH · Sep 9, 2004

Ikerous said:
; Dear goodness you must have had the longest dang answers on your math hw... why would he spend his time coming up with all that :-/ I'm pretty sure its not expect of him...

And yea, i think you've got it.

In my GCSE maths class a problem very similar to this was given. It was to be done over a few days, and was open-ended so pupils could show how good at maths problems they were. Doing what i suggested should get him high marks.

Dr_Noswad · Sep 9, 2004

oh cool

is there n e thing else u would recomend to include? i mean like workings out etc or simply sumthing as obv as it may seem....

SlagOps · Sep 9, 2004

Check this out, I'm going into the army, but im so horrible at math, I acually have to study for the ASVAB test. haha I'm great in english class tho.

Dr_Noswad · Sep 9, 2004

hey thats sweet, u just finished ure gcse's?

SLH · Sep 9, 2004

Dr_Noswad said:
oh cool is there n e thing else u would recomend to include? i mean like workings out etc or simply sumthing as obv as it may seem....

Not sure what you mean, just make sure to show that you worked out the relationship between where the T was and the values in the corners/bottom.
Give examples of where you tested it.

If you're not sure ask your teacher(s).

Dr_Noswad · Sep 9, 2004

right im new to this forum as u can tell

is there a way i could end this post or delete it as its taking up server space....

but thanks u guys for the help :thumbs:

SLH · Sep 9, 2004

Dr_Noswad said:
hey thats sweet, u just finished ure gcse's?

lol, no i did my A-Levels 2 years ago (one was Maths), i'm at uni now (not doing maths).

SLH · Sep 9, 2004

Dr_Noswad said:
right im new to this forum as u can tell is there a way i could end this post or delete it as its taking up server space....

Nope, only mods can do that, and they will if they need to, so don't worry about it.

Glad I/we could help.

Dr_Noswad · Sep 9, 2004

SLH said:
Nope, only mods can do that, and they will if they need to, so don't worry about it.

Glad I/we could help.

Well SLH it was just u rly

other guys were trying to confuse me :S but hey i got it after much guidance

and now im off 2 bed so night and thanks again! :thumbs:

Ikerous · Sep 9, 2004

the derivative of: Y = sqrt(2x²+x) * x² in X = 2
is 12*srqt(2) right?
which would be the slope of that tangent line, right?

PvtRyan · Sep 9, 2004

Ikerous said:
the derivative of: Y = sqrt(2x²+x) * x² in X = 2
is 12*srqt(2) right?
which would be the slope of that tangent line, right?

You're going in the right direction with the derivative, but the answer is incorrect

I'm off to bed too

Neutrino · Sep 9, 2004

Dr_Noswad said:
So is this correct? :S
(x+21)^2-(x*{x+2})

You might as well show the process of simplifying it though.

Dr_Noswad · Sep 9, 2004

yeah gd idea thanks...

Ikerous · Sep 9, 2004

wtf is the derivitive of that function >_> This is gunna bug me. I havent done this in a really long time.

Now i'm getting 17*sqrt(2) -_-
(which is wrong)

Anyone else workin on that...?

blahblahblah · Sep 9, 2004

So you guys will help me with my homework now, right? :angel:

Great!!! I'll have you start doing some tax research followed up by some more accounting research. To top it off you can collect some data for my research project. kthxbye.

You guys were to nice to him with his homework problem. :O

Ikerous · Sep 9, 2004

PvtRyan said:
Another test for any mathlovers out there:

Find the formula for tangent that touches the graph of the formula
Y = sqrt(2x²+x) * x² in X = 2

(see attachment, find the formula of the line with the red arrow )

o.< Anyone gonna answer this...? Its gunna bother me if i dont find out the right answer...

Neutrino · Sep 9, 2004

Ikerous said:
wtf is the derivitive of that function >_> This is gunna bug me. I havent done this in a really long time.

Now i'm getting 17*sqrt(2) -_-

Anyone else workin on that...?

I believe the derivative of

y=sqrt(2x^2+x)*x^2

is

y'=[(x^2)(4x+1)(2x^2+x)^-1/2]/2 +(2x)(2x^2+x)^1/2

Then @ x=2

y'= (29*10^1/2)/5

But I might have messed up.

Ikerous · Sep 9, 2004

And might i ask how you did that...

JimmehH · Sep 9, 2004

Ikerous said:
And might i ask how you did that...

Binomial expansion, man. Christ. I hoped I'd never have to think about that again.

Ikerous · Sep 9, 2004

How do you use binomial expansion to solve for the derivitve..?
(Now im getting 4+sqrt(2) ) -_-

I've tried f(x) - f(c) over x-c which is hard cuz its limits which are lame. And just simplifying the equation then bringing the exponent down and dropping it by one.

JimmehH · Sep 9, 2004

I'd use binomial expansion to simplify the sqrt(2x^2+x) then muliply what I get for that by x^2 and find the dirivitive. Then solve for x=2

Too much work for 3am!

Edit: Not quite sure if the binomial theorem will work if they're both terms of x though...

Or you could do what Neutrino did \/
Like I said - 3am

Neutrino · Sep 9, 2004

Just use the product rule for derivatives.

So you end up with:

y' = (29*10^1/2)/5

@ x=2

y = 4*10^1/2 = the y coordinate
y' = (29*10^1/2)/5 = the slope

So you know the coordinate: (2, 4*10^1/2)

Now use the formula: y-y1 = m(x-x1)

y-4*10^1/2 = (29*10^1/2)/5(x-2)

Which simplifies into:

y = [(29*10^1/2)/5]*x - (38*10^1/2)/5

or

y = 18.34x - 24.03 = The equation of the tangent line.

I could be wrong as I didn't recheck my calculations, but I think that's right as I did graphed it.

Ikerous · Sep 9, 2004

I'm going to take your word on that cuz i have no clue how you arrived at that for the derivitve.

Neutrino · Sep 9, 2004

Ikerous said:
I'm going to take your word on that cuz i have no clue how you arrived at that for the derivitve.

If you have:

h(x) = f(x) g(x)

Then the derivative is:

h'(x) = [f(x)g(x)]' = f'(x) g(x) + f(x) g'(x)

Does that help? If not I can try to explain better.

C4-Explosive · Sep 9, 2004

Actually you dont even have to use the product rule..
Y = sqrt(2x²+x) * x²
= sqrt(2x³+x²)

then just differentiate that...

Neutrino · Sep 9, 2004

C4-Explosive said:
Actually you dont even have to use the product rule..
Y = sqrt(2x²+x) * x²
= sqrt(2x³+x²)

then just differentiate that...

Well, that would be convenient, but it doesn't work that way sorry.

Prince of China · Sep 9, 2004

Maths ruinded my life...

I don't think it's math who ruined your life, who ever taughted you English did a bad job.

Maths ruinded my life...

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