Integration! (2 Viewers)

[hscishard]

Therefore x = [log(2x)]/2 ??

The question that involved this



WTF is wrong with my example?

 

[hscishard]

Is it because dy is on the bottom?

 

[Tofuu]

Therefore x = [log(2x)]/2 ??

i don't think you can do that. You're integrating 1/2x in respect to y not x

 

[hscishard]

i don't think you can do that. You're integrating 1/2x in respect to y not x

But then how can you get a derivative of something in terms of itself? [2nd example]

 

[kcqn93]

therefore y = x^2 + C

LOL

EDIT: ohh man LOL, thanks to the guy below!

 

[MetroMattums]

No, y = x^2 + C

 

[random-1006]

Therefore x = [log(2x)]/2 ??

The question that involved this



WTF is wrong with my example?

for your example to be correct , you are integrating (1/2x) with respect to y, NOT x !

TO GO FROM dx/dy= 1 / (2x) , you must integrate both sides with respect to y, NOT x

so its x= integral 1/ (2x) dy ( very important its dy), and i dont know you do that, think its some uni crap

 

[hscishard]

therefore y = x + C

LOL

x^2, doesn't really answer my question

 

[Tofuu]

But then how can you get a derivative of something in terms of itself? [2nd example]

not sure what you mean, but in the first example this is what you're doing



you cant integrate 1/2x if its dy


the second example is pretty much correct

 

[kcqn93]

x^2, doesn't really answer my question

for the second example




dt = 2 pi r^2 dr

integrate both sides

t = 2/3 pi r^3 + c

 

[MetroMattums]

x^2, doesn't really answer my question

Notice how the dy is on the bottom? You can move it to the RHS of the equation and move the 2x on the otherside to the LHS and chuck integration operators in front to get an integration.

 

[hscishard]

Okay. I get it.
So..



x = logy + C

And

Is the final answer?

 

[random-1006]

Okay. I get it.
So..



x = logy + C

And

Is the final answer?

what is the actual question, i dont think anyone actually knows

 

[Tofuu]

Okay. I get it.
So..



x = logy + C

And

Is the final answer?

the first part is correct

the second part is not the answer
you can only integrate if its 1/2y, not 1/2x

 

[hscishard]

the second part is not the answer
you can only integrate if its 1/2y, not 1/2x

Shit. Now I'm confused.

WTF is wrong with my example?

In this one, you can't find r= (blah blah blah), can you?

 

[Tofuu]

You can do this


you cannot do this because its integrating in respect to y and yet you have x

 

[hscishard]

what is the actual question, i dont think anyone actually knows

The dy/dx = y I just amde up myself.
The real example was in the first post and was from a rates of change question.

 

[random-1006]

Shit. Now I'm confused.



In this one, you can't find r= (blah blah blah), can you?

why not, they would give an initial condition to find the C, then rearrange, take cubed root

 

[hscishard]

why not, they would give an initial condition to find the C, then rearrange, take cubed root

But it's respect to t

 

[Tofuu]

In this one, you can't find r= (blah blah blah), can you?

you can, you just have to rearrange the equation you have for t