Intergration calculating centroid of an area

Hedgecutter · 01-01-2015 11:16PM

TheBody wrote: »

No, it's not partial fractions. DIFFERENTIATE the top line and see if you notice anything......

12x-5 all over 12x-5

TheBody · 01-01-2015 11:39PM

So I got a pen and paper to do this problem and it's a lot harder than it looks. I initially thought it was substitution. Based on the standard of problem you have shown me so far, I am wondering if it is an error. I think they probably wanted to have:

[latex]\int \frac{12x-5}{6x^2-5x+3}dx[/latex].

This would be much more in keeping with the style of problem you have been doing so far.

Hedgecutter · 01-01-2015 11:46PM

TheBody wrote: »

So I got a pen and paper to do this problem and it's a lot harder than it looks. I initially thought it was substitution. Based on the standard of problem you have shown me so far, I am wondering if it is an error. I think they probably wanted to have:

[latex]\int \frac{12x-5}{6x^2-5x+3}[/latex].

This would be much more in keeping with the style of problem you have been doing so far.

Ill find that out next week when I'm back. I have heard of the odd type o on some exam papers.

TheBody · 01-01-2015 11:52PM

You might like this website.

http://patrickjmt.com/

Scroll down the page to the calculus section. REALLY good videos on everything you need to know!!

Hedgecutter · 02-01-2015 12:02AM

TheBody wrote: »

You might like this website.

http://patrickjmt.com/

Scroll down the page to the calculus section. REALLY good videos on everything you need to know!!

Cheers. Look very good. Looks to have a hell of a lot of information there.

TheBody · 02-01-2015 12:05AM

Hedgecutter wrote: »

Cheers. Look very good. Looks to have a hell of a lot of information there.

Personally, I think it's better than khan academy (https://www.khanacademy.org/) which you may know about already. Use the search function if you can't spot the topic you are looking for.

Hedgecutter · 02-01-2015 12:11AM

TheBody wrote: »

Personally, I think it's better than khan academy (https://www.khanacademy.org/) which you may know about already. Use the search function if you can't spot the topic you are looking for.

Ya sometimes I find Khan methods slightly different to the way I'm being taught.

TheBody · 02-01-2015 12:15AM

Hedgecutter wrote: »

Ya sometimes I find Khan methods slightly different to the way I'm being taught.

The problem I have with khans videos is that by the time he is finished, the screen is a mess of scribbles and stuff all over the place.

Hedgecutter · 02-01-2015 12:19AM

TheBody wrote: »

The problem I have with khans videos is that by the time he is finished, the screen is a mess of scribbles and stuff all over the place.

Like the guy who teaches me applied mechanics nightmare to take notes down.

TheBody · 02-01-2015 12:24AM

Time for bed but keep the questions coming and I'll get back to you on them as soon as I can.

Hedgecutter · 02-01-2015 12:27AM

TheBody wrote: »

Time for bed but keep the questions coming and I'll get back to you on them as soon as I can.

Ya I'm hitting the hay too. I have part (2) attempted so I'll post it tomorrow.

Mathrew · 08-01-2015 11:41AM

I just wish I could help with math, but I couldn't, I love math, but math doesn't love me. haha

Hedgecutter · 30-03-2015 09:00PM

Two past paper questions i'am attempting. Am I using the right rules?

TheBody · 30-03-2015 09:07PM

Hedgecutter wrote: »

Two past paper questions i'am attempting. Am I using the right rules?

I presume you are asked to differentiate both? If so, they look good to me. The only suggestion I would make is to tidy up the last lines in each problem.

Well done!

Hedgecutter · 30-03-2015 09:17PM

TheBody wrote: »

I presume you are asked to differentiate both? If so, they look good to me. The only suggestion I would make is to tidy up the last lines in each problem.

Well done!

Ya, in most past papers we are told "by rule". this paper just told differentiate both.
Ill tidy them up alright.

TheBody · 30-03-2015 09:18PM

How's the course going this term?

Hedgecutter · 30-03-2015 09:37PM

TheBody wrote: »

How's the course going this term?

On the home run now, exams in june. Thermodynamics taking up a huge amount of time.

TheBody · 30-03-2015 09:38PM

Hedgecutter wrote: »

On the home run now, exams in june. Thermodynamics taking up a huge amount of time.

Great. Keep putting in the hours and you will do fine. Good luck!

Hedgecutter · 30-03-2015 09:54PM

Attempting Q6(I) not sure how to factories the denominator.
Intergration by partial fractions right?

TheBody · 30-03-2015 10:05PM

You asked this problem before. At the time, I suggested, based on the level of questions I have seen you post in the past, that this may be a typo. I think it may be upside down. I think they may have wanted:

[latex]\int\frac{12x-5}{6x^2-5x+3}dx[/latex].

Perhaps you can clarify this with your lecturer?

Hedgecutter · 30-03-2015 10:17PM

TheBody wrote: »

You asked this problem before. At the time, I suggested, based on the level of questions I have seen you post in the past, that this may be a typo. I think it may be upside down. I think they may have wanted:

[latex]\int\frac{12x-5}{6x^2-5x+3}dx[/latex].

Perhaps you can clarify this with your lecturer?

YA, I saw a note to ask about it alright but I couldn't remember the issue.

TheBody · 30-03-2015 10:20PM

If it's the way I suggest, it's just regular substitution. Let [latex]u=6x^2-5x+3[/latex].

Hedgecutter · 30-03-2015 10:27PM

TheBody wrote: »

If it's the way I suggest, it's just regular substitution. Let [latex]u=6x^2-5x+3[/latex].

Ya if it's typo should be straight forward enough.

Hedgecutter · 31-03-2015 09:46PM

Finding these second derivative's a bit tricky, wondering if I'm on the right track?

TheBody · 31-03-2015 09:58PM

Hedgecutter wrote: »

Finding these second derivative's a bit tricky, wondering if I'm on the right track?

In both cases your first derivatives are correct but the second derivatives are incorrect.

For the first one:

[latex]y=3\cos(2\theta) [/latex]. Differentiating we get

[latex]\frac{dy}{dx}=-6\sin(2\theta) [/latex]. Differentiating again we get:

[latex]\frac{d^2y}{dx^2}=-12\cos(2\theta) [/latex].

For the second one:

[latex]y=2x^2-4\ln(x)[/latex]. Differentiating we get:

[latex]\frac{dy}{dx}=4x-\frac{4}{x}=4x-4x^{-1}[/latex]. Differentiating again we get:

[latex]\frac{d^2y}{dx^2}=4+4x^{-2}=4+\frac{4}{x^2}[/latex]

Hedgecutter · 31-03-2015 10:03PM

TheBody wrote: »

In both cases your first derivatives are correct but the second derivatives are incorrect.

For the first one:

[latex]y=3\cos(2\theta) [/latex]. Differentiating we get

[latex]\frac{dy}{dx}=-6\sin(2\theta) [/latex]. Differentiating again we get:

[latex]\frac{d^2y}{dx^2}=-12\cos(2\theta) [/latex].

For the second one:

[latex]y=2x^2-4\ln(x)[/latex]. Differentiating we get:

[latex]\frac{dy}{dx}=4x-\frac{4}{x}=4x-4x^{-1}[/latex]. Differentiating again we get:

[latex]\frac{d^2y}{dx^2}=4+4x^{-2}=4+\frac{4}{x^2}[/latex]

Thanks
For the second one I forgot to bring what under the line up before deriving.

TheBody · 31-03-2015 10:05PM

Not to worry. You know for the future!

Hedgecutter · 31-03-2015 10:15PM

Sticky one here again. Think I not on the right track with the denominator

TheBody · 31-03-2015 10:17PM

Hedgecutter wrote: »

Sticky one here again. Think I not on the right track with the denominator

No point in multiplying out the denominator. Just use regular substitution. Let [latex]u=4x^2-3[/latex]

Hedgecutter · 31-03-2015 10:22PM

TheBody wrote: »

No point in multiplying out the denominator. Just use regular substitution. Let [latex]u=4x^2-3[/latex]

intergration by parts?

Intergration calculating centroid of an area

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