Doing a calculus course at uni and I am very much not a maths person.
At all.
I can’t remember how to do the integral thingy for the equation in this question. Is it by part or substitution or …. ?
And I can’t figure out how to solve the equations for part B in this one either. It’ll have an x or y where there should be a number. (??)
Is anyone able to help please?
Sibeen?
Michael V said:
Sibeen?
And/or Deevs & Mollwumblegrumble
Spiny Norman said:
Doing a calculus course at uni and I am very much not a maths person.
At all.I can’t remember how to do the integral thingy for the equation in this question. Is it by part or substitution or …. ?
And I can’t figure out how to solve the equations for part B in this one either. It’ll have an x or y where there should be a number. (??)
Is anyone able to help please?
Can you wait until tomorrow? If not, I’m sure Rev D can do it, and esselte and KJW, etc.
It doesn’t look like integration by parts or substitution. It looks like a standard double integral. Solve the inside one first then the outside one second.
mollwollfumble said:
Can you wait until tomorrow? If not, I’m sure Rev D can do it, and esselte and KJW, etc.
It doesn’t look like integration by parts or substitution. It looks like a standard double integral. Solve the inside one first then the outside one second.
Yep tomorrow is okay thanks.
This helps a bit ….
Got the first one I reckon.
Spiny Norman said:
Got the first one I reckon.
Looks good to me.
Spiny Norman said:
Got the first one I reckon.
Billy boy, have a think about your answer. The question asked about a rod with a variable density that started at 2 kg/m and increased as you went along the bar. As the bar is 2 metres long then the answer has to be above 4. Always check your maths with some common sense :)
Multiply that last bit of the equations out.
So you’ll have 0.4x – 0.2x^2, now do the int to get -(0.2/3)x^3 + 0.2x^2.
Put that bit into you equation and you should end up with your actual answer being 4.267 or thereabouts :)
sibeen said:
Spiny Norman said:
Got the first one I reckon.
Billy boy, have a think about your answer. The question asked about a rod with a variable density that started at 2 kg/m and increased as you went along the bar. As the bar is 2 metres long then the answer has to be above 4. Always check your maths with some common sense :)
Multiply that last bit of the equations out.
So you’ll have 0.4x – 0.2x^2, now do the int to get -(0.2/3)x^3 + 0.2x^2.
Put that bit into you equation and you should end up with your actual answer being 4.267 or thereabouts :)
rushes back in
The question asked about a rod with a variable density that started at 2 kg/m and increased as you went along the bar.
That was damn inelegant of me, although up to 2 metres the density does increase, so the above sentence stands, as does my maths :)
Above 2 metres the density per metre drops. In fact, WTF, at 10 metres long this thing works out to have a mass of minus 26.7 kg.
scratches head
is this sorted
no otherwise it would be hist
Well you would, wouldn’t you.
https://i.imgur.com/tPfCcys.mp4
PermeateFree said:
Well you would, wouldn’t you.https://i.imgur.com/tPfCcys.mp4
Oops, wrong fred.
quote=Spiny Norman
Doing a calculus course at uni and I am very much not a maths person.
At all.
And I can’t figure out how to solve the equations for part B in this one either. It’ll have an x or y where there should be a number. (??)
Is anyone able to help please?
/quote
mollwollfumble said:
quote=Spiny Norman
Doing a calculus course at uni and I am very much not a maths person.
At all.And I can’t figure out how to solve the equations for part B in this one either. It’ll have an x or y where there should be a number. (??)
Is anyone able to help please?
/quote
I was going to reply last night, but like a good engineer I followed Sibeen’s advice before he gave it and decided that half a glass of wine and a hard day at the Indian on-line visa site had addled my brain, and I’d better not. I’ll get back to it this afternoon if I can.
For Part 2, to get a picture of what is happening, I suggest drawing a 3D graph of density against position in the XY plane. The volume under that graph will be the weight.
SCIENCE said:
no otherwise it would be hist
clever…
I forgot last night that I have a way of showing equations directly. I’m on holidays, it was late, and I’d drunk way more than the Rev had.
Bill, just showing here where you went wrong with the second part of the equation in question 1. You brought the 0.2 out to the front of the integral as such:
and then you split the integral into its component parts, but made a boo boo by not using the 0.2 for both parts.
Egad – missed a minus sign. Should be:
sibeen said:
Spiny Norman said:
Got the first one I reckon.
Billy boy, have a think about your answer. The question asked about a rod with a variable density that started at 2 kg/m and increased as you went along the bar. As the bar is 2 metres long then the answer has to be above 4. Always check your maths with some common sense :)
Multiply that last bit of the equations out.
So you’ll have 0.4x – 0.2x^2, now do the int to get -(0.2/3)x^3 + 0.2x^2.
Put that bit into you equation and you should end up with your actual answer being 4.267 or thereabouts :)
Okay I see where I went wrong with that one – I didn’t put the x & stuff back into brackets – after I did that it came out as 4 4/15.
mollwollfumble said:
I don’t understand why the 2nd squiggle has the x at the bottom and not at the top ?
Spiny Norman said:
mollwollfumble said:
I don’t understand why the 2nd squiggle has the x at the bottom and not at the top ?
I had it as zero on the bottom and x at the top.
Spiny Norman said:
Spiny Norman said:
mollwollfumble said:
I don’t understand why the 2nd squiggle has the x at the bottom and not at the top ?
I had it as zero on the bottom and x at the top.
Pops in:
Because you are integrating the shaded vertical slice at x, which extends from y=x to y=1.
Pops out.
The Rev Dodgson said:
Spiny Norman said:
Spiny Norman said:I don’t understand why the 2nd squiggle has the x at the bottom and not at the top ?
I had it as zero on the bottom and x at the top.
Pops in:
Because you are integrating the shaded vertical slice at x, which extends from y=x to y=1.
Pops out.
Why not zero to x ?
Spiny Norman said:
The Rev Dodgson said:
Spiny Norman said:I had it as zero on the bottom and x at the top.
Pops in:
Because you are integrating the shaded vertical slice at x, which extends from y=x to y=1.
Pops out.
Why not zero to x ?
Ah never mind, I understand why now. The way I was doing it, it’d be under the line.
Because y=zero to y=x is space, then you get the shaded region.
It’s the vertical slice we are looking at, as in the figure on the left in your first post.
Spiny Norman said:
Spiny Norman said:
The Rev Dodgson said:Pops in:
Because you are integrating the shaded vertical slice at x, which extends from y=x to y=1.
Pops out.
Why not zero to x ?
Ah never mind, I understand why now. The way I was doing it, it’d be under the line.
That’s it :)
Spiny Norman said:
The Rev Dodgson said:
Spiny Norman said:I had it as zero on the bottom and x at the top.
Pops in:
Because you are integrating the shaded vertical slice at x, which extends from y=x to y=1.
Pops out.
Why not zero to x ?
As Rev D says, that’s the wrong triangle. Every point in the triangle has y > x.
If it was zero to x then that would be y < x.
The Rev Dodgson said:
Spiny Norman said:
Spiny Norman said:Why not zero to x ?
Ah never mind, I understand why now. The way I was doing it, it’d be under the line.
That’s it :)
It’d be jolly good if we did like the real world does – With a program of some sort. University of Queensland does that that way, grumble grumble ….
Spiny Norman said:
Spiny Norman said:
The Rev Dodgson said:Pops in:
Because you are integrating the shaded vertical slice at x, which extends from y=x to y=1.
Pops out.
Why not zero to x ?
Ah never mind, I understand why now. The way I was doing it, it’d be under the line.
No worries.
Spiny Norman said:
The Rev Dodgson said:
Spiny Norman said:Ah never mind, I understand why now. The way I was doing it, it’d be under the line.
That’s it :)
It’d be jolly good if we did like the real world does – With a program of some sort. University of Queensland does that that way, grumble grumble ….
Peers over glasses.
If you want to learn how to feed a black box, go to TAFE. At university we learn to understand things.
(But I’ll probably post a spreadsheet to do double integration in a day or two).
Spiny Norman said:
The Rev Dodgson said:
Spiny Norman said:Ah never mind, I understand why now. The way I was doing it, it’d be under the line.
That’s it :)
It’d be jolly good if we did like the real world does – With a program of some sort. University of Queensland does that that way, grumble grumble ….
Jaysus. I remember when Kothos was back doing his degree and he was whinging about the same thing. I’ll tell you what I told him back then, “suck it up, princess” :)
The Rev Dodgson said:
Spiny Norman said:
The Rev Dodgson said:That’s it :)
It’d be jolly good if we did like the real world does – With a program of some sort. University of Queensland does that that way, grumble grumble ….
Peers over glasses.
If you want to learn how to feed a black box, go to TAFE. At university we learn to understand things.
(But I’ll probably post a spreadsheet to do double integration in a day or two).
DGAF how it works, not interested. I’m more than happy to just do it how it’s usually done commercially.
Spiny Norman said:
The Rev Dodgson said:
Spiny Norman said:It’d be jolly good if we did like the real world does – With a program of some sort. University of Queensland does that that way, grumble grumble ….
Peers over glasses.
If you want to learn how to feed a black box, go to TAFE. At university we learn to understand things.
(But I’ll probably post a spreadsheet to do double integration in a day or two).
DGAF how it works, not interested. I’m more than happy to just do it how it’s usually done commercially.
Commercially?
Um. For simple intervals like these I do it by hand, just the way I’ve explained.
More complicated and I write a computer program to do the Riemann sums.
If I feel like getting fancy I look it up in Handbook of Chemistry and Physics or, better, Gradshteyn and Ryzhik.
For a fast result on a difficult integrand I use https://www.wolframalpha.com/calculators/integral-calculator/
The most accurate integration software is Mathematica. Other software will solve integrals but tend to make more mistakes.
mollwollfumble said:
Spiny Norman said:
The Rev Dodgson said:Peers over glasses.
If you want to learn how to feed a black box, go to TAFE. At university we learn to understand things.
(But I’ll probably post a spreadsheet to do double integration in a day or two).
Probably 99.999(lots more 9s)% of integrals are done numerically, in FEA software and the like.
I’ll post something showing how to do it for these examples.
DGAF how it works, not interested. I’m more than happy to just do it how it’s usually done commercially.
Commercially?
Um. For simple intervals like these I do it by hand, just the way I’ve explained.
More complicated and I write a computer program to do the Riemann sums.If I feel like getting fancy I look it up in Handbook of Chemistry and Physics or, better, Gradshteyn and Ryzhik.
For a fast result on a difficult integrand I use https://www.wolframalpha.com/calculators/integral-calculator/
The most accurate integration software is Mathematica. Other software will solve integrals but tend to make more mistakes.
Not sure what happened there! My reply is buried in the middle under BS.
