Intergration calculating centroid of an area

Hedgecutter

Be handy if I put up an image

Think part 1 is ok but can't figure out part 2.

TheBody

In the first part, you should have [latex]\frac{d\theta}{dt} [/latex] instead of [latex]\frac{dx}{dt} [/latex].

With regards to the second part, in the first part you found a formula for the angular acceleration to be: [latex]\frac{d^2\theta}{dt^2}=20-12t [/latex].

So just let this equal 0 and solve for t. In other words solve:

[latex]20-12t=0 [/latex] for t.

Hedgecutter

TheBody wrote: »

In the first part, you should have [latex]\frac{d\theta}{dt} [/latex] instead of [latex]\frac{dx}{dt} [/latex].

With regards to the second part, in the first part you found a formula for the angular acceleration to be: [latex]\frac{d^2\theta}{dt^2}=20-12t [/latex].

So just let this equal 0 and solve for t. In other words solve:

[latex]20-12t=0 [/latex] for t.

Handy question if it comes up. Cheers.

TheBody

Hi Hedgecutter.

Just wondering if you passed your exams in the end?

Hedgecutter

TheBody wrote: »

Hi Hedgecutter.

Just wondering if you passed your exams in the end?

Still waiting for results. Should get them this week. Found exam very tough. If I don't pass I'll give it another go.

Scuid Mhór

Hedgecutter I think you may owe TheBody a pint or two when you find out you've passed...

TheBody

Hedgecutter wrote: »

Still waiting for results. Should get them this week. Found exam very tough. If I don't pass I'll give it another go.

Hopefully you passed. Let us know how you got on when you get the results.

TheBody

Scuid Mhór wrote: »

Hedgecutter I think you may owe TheBody a pint or two when you find out you've passed...

I was planning to send him an invoice when he passes!!!

Hedgecutter

A sun holiday I'd say ;D

TheBody

So did you pass?

Hedgecutter

Still no results. Ridiculous at this stage, nearly a month late.

TheBody

Hedgecutter wrote: »

Still no results. Ridiculous at this stage, nearly a month late.

That's very odd. Sure college is starting back soon!! Are you sure it's not just yours that are missing?

Hedgecutter

TheBody wrote: »

That's very odd. Sure college is starting back soon!! Are you sure it's not just yours that are missing?

Everyone in the same boat. Email afew of the lads in my class, they have received nothing. Guys at work who are a year behind have received some but not all. I email the exam office and they said that the delay is at city and guilds side.

Part time course I'm due to start begins in a couple of weeks. I have no idea if i qualify yet. Crazy.

Hedgecutter

Good news passed my exams. Starting day course next week. Maths module Friday morning, so I'm sure it won't be long before I'm back on here looking for help

TheBody

Ah that's great news!! Well done. Congrats!!

Hedgecutter

Stuck with these two turning points questions. Can't even start.

(I) find dy/dx then find X, once you have X you can find Y, then your max and min points.

I can't get of the mark to find X

Hedgecutter

Got number 6 sorted but still stuck on 8



I know this is basic algebra but cant remember how to find X, Is my work correct so far?

Finding turning points

Y=5X-2lnX

dy/dx=5-2/x

dy/dx=5-2x^-1
dydx=0

5-2x^-1=0
-2x^-1=-5

TheBody

Hedgecutter wrote: »

Stuck with these two turning points questions. Can't even start.

(I) find dy/dx then find X, once you have X you can find Y, then your max and min points.

I can't get of the mark to find X

Can you elaborate a little on what yor problem is? With regards the first one:

Rather than using the product rule to differentiate, I'd begin by expanding the brackets.

[latex]x=\theta(6-\theta)=6\theta-\theta^2 [/latex].

You can just differentiate this as normal then. In other words,

[latex]\frac{dx}{d\theta}=6-2\theta [/latex].

Can you finish it from there?

With regrads the second:

[latex]y=5x-2\ln(x)[/latex]. Differentiating we get:

[latex]\frac{dy}{dx}=5-\frac{2}{x}[/latex]. Can you finish it from here?

TheBody

I see you replied while I was writing the previous post!! We have:

[latex]5-\frac{2}{x}=0[/latex]
[latex]\implies\frac{2}{x}=5[/latex]
[latex]\implies 5x=2[/latex]
[latex]x=\frac{2}{5}[/latex].

Hedgecutter

Do I bring X above the line like I did in post 198?

Find the square root of 5/2 using second function on calculator putting a -1 before root symbol?

I got 0.4

Hedgecutter

Ok what rules are you using in post 200. Indices?

Hedgecutter

sorry you just cross multiplied

TheBody

Hedgecutter wrote: »

sorry you just cross multiplied

Yes!! (Although I don't like the term cross multiplying!!)

No need for fancy things like indices here!!

Hedgecutter

How did you get rid of the minus in front of the 2/x?

TheBody

If [latex]5-\frac{2}{x}=0 [/latex] then subtracting 5 from both sides gives:

[latex]-\frac{2}{x}=-5 [/latex]. Multiply both sides by -1 then gives:

[latex]\frac{2}{x}=5 [/latex]

Hedgecutter

Cheers, out of curiosity what is the correct term for cross multiplying?

TheBody

Terms and phrases such as "cross multiplying", "bring it accross the equals sign and change the sign/operation" are widespread in the teaching of maths. Indeed, when I was in school, that was the way I was taught.

The reality is that (in my humble opinion) they are bad ways to think about manipulating equations and formulae. Nothing is moving anywhere!!

An equals sign means what is on both sides of it has EXACTLY the same value or worth. Therefore, we can do what we like (within reason!!) provided we do it to both sides. So, for example, say we had something simple like:

[latex]x-4=5 [/latex]. Instead of saying something like "we bring the 4 accross the equals sign and make it positive". (Which seems like a strange statement when you think about it to get [latex]x=5+4[/latex]we should be saying something like "adding 4 to both sides of the equation", to get same thing but it is much more natural.

Similarily say we had (as we had above) [latex]\frac{2}{x}=5[/latex]. Instead of "cross multiplying" (which again seems like a strange thing to do), we can multiply both sides by x to get [latex]2=5x[/latex].

I never use phrases such as "cross multiply" or "move stuff accross the equals sign" in my lesson. You won't find such nonsense in any text book.

Hedgecutter

Wondering if I'm on the right track here, do I bring 4t^-3 back under the line before I solve for t?

TheBody

Hedgecutter wrote: »

Wondering if I'm on the right track here, do I bring 4t^-3 back under the line before I solve for t?

Yes, I would. Also, you should have [latex]\frac{dx}{dt}[/latex]. The function is called x with a variable of t.

Hedgecutter

Stuck with the finish. Not a 100% sure how to finish this. But I gave it a go. Answer in book is (0.5,6) I'm well off.