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Hard maths question can you help?

km
(edited 8 years ago)

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Reply 1
Avatar for TeeEm
TeeEm
19
Original post by opi453
1.Calculate the divergence of the vector field a=(x,y^2,z^3) at the point P(-2,4,5).

2.Calculate the divergence of the vector field a=(z^2,x^2,y^2) at the point P(1,2,3).


There is nothing hard about this question ...

Look at the definition of the divergence in your notes and it is 1 line each.

(They are so simple I cannot even give you a hint)
Original post by opi453
Please im really struggling
What, exactly, have you done so far?
Original post by opi453
Please im really struggling


divF=∇⋅F\displaystyle \text{div} {F} = \nabla \cdot F

So calculate the div for both fields and then sub in values.

As said above 1 line working.
Original post by opi453
Still struggling cant you just give me the answer?
No. This is not a "do my homework for me" service.

Please read the guide to posting stickied at the top of the forum.
Reply 5
Avatar for TeeEm
TeeEm
19
Original post by opi453
Please im really struggling


divergence of a vector field F is defined as ∇. F where ∇ operator is defined as [∂/∂x, ∂/∂y, ∂/∂z]

Example

divergence of [x2-3y, y2-z2-y, x+2y] at (1,1,1)

[∂/∂x, ∂/∂y, ∂/∂z] . [x2-3y, y2-z2-y, x+2y] =

∂/∂x(x2-3y) + ∂/∂y(y2-z2-y) + ∂/∂z(x+2y) =

2x +2y-1+0=

2x+2y-1

at the required point 3
Original post by opi453
This isent my homework:mad:
Then why do you need the answer so badly?
Original post by opi453
Need the answer so I can understand the method not that's its any of you ****ing business


Wow that's how you're going to treat people trying to help you.
Reply 8
Avatar for TeeEm
TeeEm
19
Original post by opi453
Thanks understand it now


no worries
Original post by opi453
Need the answer so I can understand the method not that's its any of you ****ing business
Well, the answer isn't going to help you learn the method, is it? I mean, if I say the divergence is 6, how is that going to help you understand anything?

Spoiler



Anyhow, thanks for removing the last sliver of doubt about whether it's worth wasting any time trying to help you.
Reply 10
Avatar for TeeEm
TeeEm
19
Original post by DFranklin
Well, the answer isn't going to help you learn the method, is it? I mean, if I say the divergence is 6, how is that going to help you understand anything?

Spoiler



Anyhow, thanks for removing the last sliver of doubt about whether it's worth wasting any time trying to help you.


sorry I did not see his comment as I was busy typing an example for him, so I missed the whole conversation
Original post by TeeEm
sorry I did not see his comment as I was busy typing an example for him, so I missed the whole conversation
Not a problem. If you managed to help him that's great.
Original post by opi453
Need the answer so I can understand the method not that's its any of you ****ing business


Understanding a method will teach you nothing about the mathematics.

You can spot method-people a mile off. Give them an integral like ∫−111x2 dx\displaystyle\int_{-1}^{1} \dfrac{1}{x^2}\ \textrm{d}x and when they come back with -2, you know they don't know what they're really doing.
Original post by FireGarden
Understanding a method will teach you nothing about the mathematics.

You can spot method-people a mile off. Give them an integral like ∫−111x2 dx\displaystyle\int_{-1}^{1} \dfrac{1}{x^2}\ \textrm{d}x and when they come back with -2, you know they don't know what they're really doing.


Is it 0 or does it just not converge? I thought the former but wolfram outputs the latter.
Reply 14
Avatar for Zacken
Zacken
22
Original post by FireGarden
Understanding a method will teach you nothing about the mathematics.

You can spot method-people a mile off. Give them an integral like ∫−111x2 dx\displaystyle\int_{-1}^{1} \dfrac{1}{x^2}\ \textrm{d}x and when they come back with -2, you know they don't know what they're really doing.


You can't integrate over the discontinuity at x=0x=0, right?
Original post by keromedic
Is it 0 or does it just not converge? I thought the former but wolfram outputs the latter.


Check out the graph. It's non-negative everywhere it's defined on that interval, so it couldn't possibly be zero! (let alone negative - all the more reason the method-answer of -2 is completely ridiculous). The issue is it's not defined at zero, and the integral of the function does not converge if 0 is one of the boundaries.

Original post by Zacken
You can't integrate over the discontinuity at x=0x=0, right?


Sort of. Just having discontinuities is not a problem in itself - you can just add together the integrals of the continuous pieces. The problem is that this particular discontinuity blows up "badly enough" that if it's a boundary of an integral, the integral doesn't converge. So adding the continuous pieces doesn't work, because the integrals of the continuous pieces don't exist.
(edited 8 years ago)
Original post by FireGarden
Check out the graph. It's non-negative everywhere it's defined on that interval, so it couldn't possibly be zero! (let alone negative - all the more reason the method-answer of -2 is completely ridiculous). The issue is it's not defined at zero, and the integral of the function does not converge if 0 is one of the boundaries.

Isn't the non-thinking answer the one I provided? I.e. 0. As in F(1)−F(−1)=−1+1=0F(1)-F(-1)=-1+1=0. The graph is symmetrical and because definite integrals are the signed areas, I thought you'd have a net result of 0 because the bits above and below the x-axis would cancel. My bad.

Sort of. Just having discontinuities is not a problem in itself - you can just add together the integrals of the continuous pieces. The problem is that this particular discontinuity blows up "badly enough" that if it's a boundary of an integral, the integral doesn't converge. So adding the continuous pieces doesn't work, because the integrals of the continuous pieces don't exist.

I shall entertain myself by googling some more about discontinuities.
Reply 17
Avatar for Zacken
Zacken
22
Original post by keromedic
I shall entertain myself by googling some more about discontinuities.


f(x)=1x2f(x) = \frac{1}{x^2}, we have ∀x∈R,x2>0⇒1x2>0⇒f(x)>0\forall x \in \mathbb{R}, x^2 > 0 \Rightarrow \frac{1}{x^2} > 0 \Rightarrow f(x) > 0 for all real xx, so the graph always lies above the xx-axis, not quite sure what you're saying about the negative cancelling out the positive, etc...

Furthermore, ∫f(x) dx=F(x)=−1x=−x−1\int f(x) \, \mathrm{d}x = F(x) = -\frac{1}{x} = -x^{-1}, so F(1)−F(−1)=−11−−1−1=−2F(1) - F(-1) = -\frac{1}{1} - \frac{-1}{-1} = -2.

I think you're confusing the function we are integrating, it's 1/x^2. Probably just an oversight on your part. :smile:
(edited 8 years ago)
Reply 18
Avatar for Zacken
Zacken
22
Original post by FireGarden

Sort of. Just having discontinuities is not a problem in itself - you can just add together the integrals of the continuous pieces. The problem is that this particular discontinuity blows up "badly enough" that if it's a boundary of an integral, the integral doesn't converge. So adding the continuous pieces doesn't work, because the integrals of the continuous pieces don't exist.



Well, yes, that's what I was implying by "you can't integrate over the discontinuity" - I'm aware that you can integrate over other, better behaved discontinuities (such as some integrands with trigonometric functions in the denominator), which I learned from hanging around TPIT back in 2013. :rofl:
Original post by Zacken
f(x)=1x2f(x) = \frac{1}{x^2}, we have ∀x∈R,x2>0⇒1x2>0⇒f(x)>0\forall x \in \mathbb{R}, x^2 > 0 \Rightarrow \frac{1}{x^2} > 0 \Rightarrow f(x) > 0 for all real xx, so the graph always lies above the xx-axis, not quite sure what you're saying about the negative cancelling out the positive, etc...

Furthermore, ∫f(x) dx=F(x)=−1x=−x−1\int f(x) \, \mathrm{d}x = F(x) = -\frac{1}{x} = -x^{-1}, so F(1)−F(−1)=−11−−1−1=−2F(1) - F(-1) = -\frac{1}{1} - \frac{-1}{-1} = -2.

I think you're confusing the function we are integrating, it's 1/x^2. Probably just an oversight on your part. :smile:


I made a mistake :tongue:

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