nav[aria-label="Primary Navigation"] { padding: 0; & ul { list-style: none; width: 100%; display: flex; flex-direction: row; justify-content: start; align-items: start; gap: 30px; padding: 0; & li { margin: 0; } & ul li { list-style: none; } } }

Child Item

Intergration calculating centroid of an area

Hedgecutter

I'm revising at the moment and I have hit a wall with one question. I have to find the centroid of the area of a curve Y=3x^2 the x axis and ordinates x=0,y=2

looking over my notes and I dont understand how we found y bar.

I don't understand how we went from 3x^2 to 9x^4,

if we integrate it should be 3x^3/3?

I have attached my notes for y bar

any help would be appreciated

Find more posts tagged with

Quick Links

Comments

Hedgecutter

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

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

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

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

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

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

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

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

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

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

Hedgecutter

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

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

Hedgecutter

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

TheBody

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

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

How's the course going this term?

Hedgecutter

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

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

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

TheBody

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

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

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

Hedgecutter

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

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

TheBody

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

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

Not to worry. You know for the future!

Hedgecutter

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

TheBody

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

TheBody wrote: »

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

intergration by parts?