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Anonymous

MATlab 2016-01-10 22:41:47 Post No. 7774653

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MATlab 2016-01-10 22:41:47 Post No. 7774653

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Can any of you show me how to solve this? This isnt my homework or anything but I am practicing for a numerical methods class I have this semester.

I checked chegg, but dont really get their process. Thanks

>>

>plugging numbers in

>he can't even do that

Fuck off and get someone else to do your homework.

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>>7774653

define a function that returns 8000 - the right hand side of the volume function and replace b with a/K.

Then all you have to do is use a root finder or optimization routine to solve for a at every K.

Then b then S from the saved a and K values.

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>>7774653

>Recent high school graduate who will be first entering college later this month reporting in.

If my understanding is correct then for any value of K you need only replace a with Kb in the volum formula and then calculation until you have something like

8000 = c*b where c is whatever constants result from multiplying 1/4 times pi squared times all that a b shit going on. Then you can get b. and because a = kb then you can also get a.

Then plug in your values of a and b into the formula for surface area and you are done.

So... from what I can gather Numerical methods is just basic algebra?

Man, I used to be impressed when I read engineering curriculums include 'Numerical methods for engineers'. Now I know that should read 'Basic algebra for people who didn't do enough high school'.

Anyways, please tell me if I am incorrect and also

>inches

for fucks sake man, that is heresy.

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>>7774673

>So... from what I can gather Numerical methods is just basic algebra?

No. This is obviously an intro problem meant to get you used to coding in MATLAB.

You are also wrong and didn't consider that there are several roots and bitching about inches makes you sound like a redditor.

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>>7774699

>You are also wrong

Sad. Maybe I should try to code it in C.

>sound like a redditor

It makes me sound like a person who uses fucking meters and other reasonable units. Why do some countries cling to these ungodly units of measurement? I am almost sure that when doing research or real work you will most likely use SI units. And I am completely sure that if you were ever going to work with someone abroad, you'd have to stick to SI units anyways.

But whatever, no conversions are happening but this also makes me wonder why do they even define the unit, they could just say '8000 units cubed'.

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>>7774710

>that las expression

For fucks sake what the fuck is that shit. How do I simplify that shit? I guess... I can't wait for college

>Erased my proof for matrix addition being equal to addition of linear maps for this

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>>7774710

Units do not matter are all. Only retards bitch when their little brains have to struggle to think in different units.

I grew up in SI and use Imperial when I feel like it.

You don't worry about units in research because it's most often dimensionless quantities. If you really need to do conversions it takes a second to code the conversions and is really not something to complain about unless you're more interested in fanboy memes than real science/engineering.

>>7774747

It's a cubic equation, there is an analytical solution, but the whole point is practice using a root solver and writing routines to choose the appropriate root.

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>>7774710

>>7774673

I hate to pull the "you're just a high school grad out of your element" card, but I mean, you're pretty out of your element. Particularly in engineering, pounds/inches/etc are the standard in the U.S. I know you probably think SI units are some sort of beautifully defined units or whatever it is teens think, but the real world just isn't like that.

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>>7774770

In all honesty, imperial units just look like a pityful attempt at rebelling against Europe's monopoly in scientific progress (back in the day, at least).

Considering that you were ruled over by the brits and assuming that back then the schools taught the brit units I see no practical reason to change units other than to look different and edgy.

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>>7774653

basic algebra, but I also need to practice my skills and I havent done anything productive today

8000=(1/4)(pi^2)(a+b)(b-a)^2

4*8000/(pi^2)=(a+b)(b-a)^2

a=kb, so 4*8000/(pi^2)=(kb+b)(b-kb)^2

expand

(kb+b)(b^2-2kb^2+(k^2)b^2)

(b^3-2kb^3+(k^2)b^3)+k(b^3-2kb^3+(k^2)b^3)

b^3-(k^2)b^3-kb^3+k^3

take out common factor (b^3)(1-k^2-k+k^3)

4*8000/(pi^2)=(b^3)(1-k^2-k+k^3)

32000/((pi^2)(1-k^2-k+k^3))=b^3

b=(32000/((pi^2)(1-k^2-k+k^3)))^(1/3)

>>

[math]

>>7774803

[/math]

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