Unsolved Textbook Exercises: Seeking Help and Solutions

[kelvintc]

I encountered many problems while doing exercises in textbooks. And i have stated it down in a word document attached in this post. Hope someone can help and teach me how to solve those problems. Answers are given. I just don't know how to get those answer.

Thanks a lot.

p/s : Emergency = next week test.

(2), (4), (5), (6) solved.. thanks

 

[arildno]

Welcome to PF!

You must show some of your own work here; don't expect your homework to be done for you!
Having said that, let's take a specific question, nr. 4:
Now, what are your problems with this particular exercise?
Make a detailed comment on this.

 

[kelvintc]

r dot grad T should be a scalar ? i wonder y answer's a vector. I used formulas and can't get those answers. Just can't understand why. So seeking help. Thanks.

 

[arildno]

The answer should be a vector, as it is given in the answer (i haven't checked if the answer given is correct)

I hope you know about the grad "vector":
grad=\vec{a}_{x}\frac{\partial}{\partial{x}}+\vec{a}_{y}\frac{\partial}{\partial{y}}+\vec{a}_{z}\frac{\partial}{\partial{z}}

For example, the divergence of a vector \vec{v} is given by:
div\vec{v}=grad\cdot\vec{v}

Are you familiar with this notation?

 

[kelvintc]

actually i don't know what means (r dot grad) ? i know div and grad as well
p/s: how you draw those symbols?

 

[arildno]

You can click on the LATEX code to see how you write it .

OK, so you know the "grad", which I'll henceforth write as \nabla

Let's first review how we get the scalar known as "divergence"
(\nabla\cdot\vec{v})
Let \vec{v}=u\vec{a}_{x}+v\vec{a}_{y}+w\vec{a}_{z}

We then have that:
\nabla\cdot\vec{v}=\vec{a}_{x}\cdot(\frac{\partial}{\partial{x}}\vec{v})+\vec{a}_{y}\cdot(\frac{\partial}{\partial{y}}\vec{v})+\vec{a}_{z}\cdot(\frac{\partial}{\partial{z}}\vec{v})

This simplifies to, in our case:
\nabla\cdot\vec{v}=\frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{y}}+\frac{\partial{w}}{\partial{z}}

(Please comment if this doesn't make sense to you!)

Now, we're ready to tackle \vec{v}\cdot\nabla
This is also a "dot" product (scalar product), and looks like:
\vec{v}\cdot\nabla=u\vec{a}_{x}\cdot\nabla+v\vec{a}_{y}\cdot\nabla+w\vec{a}_{z}\cdot\nabla

Or, simplified:
\vec{v}\cdot\nabla=u\frac{\partial}{\partial{x}}+v\frac{\partial}{\partial{y}}+w\frac{\partial}{\partial{z}}

This is a "scalar" operator which you then apply on T.

 

[arildno]

The answer is correct, BTW

 

[kelvintc]

oic thanks

 

[kelvintc]

in q 5 : i found the surface integral = -2, but line integral = 7/6. it didn't match Stoke's theorem.. i wonder y...

 

[kelvintc]

i never learn that inverse dot product b4.. hehe... thanks

 

[kelvintc]

and q 2 i really have no idea

 

[arildno]

OK, first:
Have you checked that you get no.4 right?

Secondly, try and group together a few questions you think "belong" to each other, which you would like to focus on.

 

[kelvintc]

no.1 to 6 are about vectors. All about integrals.. I　ｊｕｓｔ　ｗｏｎｄｅｒ y can't get answers using formulas. No.2 is really don't know how to start also.

 

[kelvintc]

no.5 curl = -x^2 in z direction.. then integral can't get 7/6

 

[arildno]

OK, we'll look into 2 (but you did check 4, or what?).

2.
Now, you've been given equations for two surfaces.
In general, if you have a surface given on the form S(x,y,z)=0 (or constant),
you know that the normal on that surface at a point (x,y,z) is parallell to the gradient of S (evaluated on the same point).
Post what you get here in some detail.

 

[kelvintc]

can't get no.4 answer

 

[arildno]

Post what you've done. In detail.

 

[kelvintc]

1) since z=0, a_{r} has no a_{x} component , a_{\phi} = -\sin{\phi}{a}_{r} + \cos{\phi}{a}_{\phi} ... and i can't t the answer for (a) and so on

 

[arildno]

I meant on Q4; the one I started with.

 

[kelvintc]

(r dot grad T ) = (\vec{r}\cdot\nabla){T}={{x}a_{x}}\frac{\partial{2zy}}{\partial{x}}+{{y}a_{y}}\frac{\partial{xy^2}}{\partial{y}}+{{z}a_{z}}\frac{\partial{x^2yz}}{\partial{z}}
is it?

 

[kelvintc]

(2), (3) no ideas
(5) i got \nabla\times{F}={-x^2}{a}_{z} and can't get the answer

 

[arildno]

Absolutely not!
We gained:
\vec{r}\cdot\nabla=x\frac{\partial}{\partial{x}}+y\frac{\partial}{\partial{y}}+z\frac{\partial}{\partial{z}}

We then have:
(\vec{r}\cdot\nabla)\vec{T}=x\frac{\partial\vec{T}}{\partial{x}}+y\frac{\partial\vec{T}}{\partial{y}}+z\frac{\partial\vec{T}}{\partial{z}}

 

[arildno]

2) Take your first surface (given as an equation)
a)Rewrite that equation into a form S(x,y,z)=0,(introduce S(x,y,z) for the expression in x,y,z)

b) Calculate the gradient of S

 

[kelvintc]

ooops.. finally understand and get it.. thanks for patience... hehe

 

[kelvintc]

2) Take your first surface (given as an equation)
a)Rewrite that equation into a form S(x,y,z)=0,(introduce S(x,y,z) for the expression in x,y,z)

can't get it... for an example?

 

[arildno]

In question 2, the coordinates for the intersection doesn't make sense

Either it should be (1,2,1) (not (1-2,1)), or there is some other wrong troubling it.

Question 3 doesn't seem to make any sense at all..

 

[kelvintc]

oh typed it wrong.. should be (-1, 2, 1)

(3) it wants \oint{V}{dS}

 

[kelvintc]

that means S1(x,y,z)={x^2}{y}+{z-3}, S2(x,y,z)={x}{\log}{z}-{y^2}+{4} ?
but {\nabla}{S2} i never learn b4 (the log one)

 

[arildno]

Allright, I'll give you an example of what I mean:
(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=R^{2}

Clearly, this equation describes the shell of a sphere with center at (a,b,c).
We may rewrite the equation as:
S(x,y,z)=0
where in this case, we have:
S(x,y,z)=(x-a)^{2}+(y-b)^{2}+(z-c)^{2}-R^{2}

That is, the spherical shell is composed of those points on which the function S is zero(right?)

The gradient of S is easily found:
\nabla{S}=2(x-a)\vec{a}_{x}+2(y-b)\vec{a}_{y}+2(z-c)\vec{a}_{z}

The unit normal at a given point (x,y,z) is parallell to \nabla{S} there, but of unit length.
See if this helps you along.

 

[arildno]

You have the correct expressions for S1 and S2.
Now differientiate with the gradient; log(z) is the natural logarithm to z, if you are familiar with that concept.

 

[kelvintc]

{S1}{(x,y,z)}={x^2}{y}+{z-3}
{\nabla}{S1}={2xy}a_{x}+{x^2}a_{y}+a_{z}
S2(x,y,z)={x}{\log}{z}-{y^2}+{4}
{\nabla}{S2}={\log}{z}a_{x}-{2y}a_{y}+a_{z}\frac{{x}{\log}{e}}{z} ?
{\cos}{\phi}=\frac{{\nabla}{S1}{\cdot}{\nabla}{S2}}{{\mid}{\nabla}{S1}{\cdot}{\nabla}{S2}{\mid}} ?

 

[arildno]

Precisely!
Now, you must figure out the unit normals at the intersection point, and calculate the angle between them. (log(e)=1, BTW)

 

[kelvintc]

log e is not in base 10? it's in base e ? i get the answer correct if log e = 1..

 

[arildno]

Sure:
Now, the natural logarithm of a number "a" is a number b=log(a), so that:
e^{b}=a (right?)
If a=e, we have:
e^{b}=e=e^{1}\to{b}=log(e)=1

 

[kelvintc]

ic.. thanks...

 

[kelvintc]

oic... about no.5 it asks me to prove the Stoke's Theorem.
i get \nabla\times{F}={-x^2}{a}_{z}
to find \int(\nabla\times{F})\cdot{dS} , do i need to separate the triangle to 2 pieces, one is x=0 to x=1, y=0 to y=1 and another one is x=1 to x=2, y=1 to y=0 ?
I can't get 7/6 with both ways i did.
But, \oint{F}\cdot{dl} i got 7/6.. curious..

 

[kelvintc]

i'm taking electrical engineering...

 

[arildno]

1.We agree on the expression on the curl
2. I don't see the need to split up the triangle (I'll get into that)
3. I took a master's in fluid mechanics some time ago..

 

[arildno]

My choices for the two triangles:
0<=x<=1,0<=y<=x triangle 1

1<=x<=2, 0<=y<=2-x triangle 2

 

[kelvintc]

yet, i can't get the 7/6. my answer for triangle 1 = -1/4, then triangle 2 = -19/40... sums up can't get 7/6.. anything wrong?

master! oh, it's incredible..

 

[kelvintc]

about (1) since z=0, a_{r} has no a_{x} component , a_{\phi} = -\sin{\phi}{a}_{x} + \cos{\phi}{a}_{y}
e^{-2}\sin\frac{\phi}{2}*-\sin\phi=-0.102

 

[kelvintc]

no.6 i translate all x,y,z into cylindrical coordinates.
x = r\cos\phi
y = r\sin\phi
z = z
a_{x} = \cos{\phi}{a}_{r} - \sin{\phi}{a}_{\phi}
a_{y} = \sin{\phi}{a}_{r} + \cos{\phi}{a}_{\phi}
a_{x} = a_{z}
dS = rd\phi{d}{r}a_{r}
Hence,
F_{r} = r^2\cos^3\phi + r^2\sin^3\phi
integrate i can't get the answer also..

 

[arildno]

Hi again!
I got 7/6 in 5); I'll take that first.

1. 7/6 is the number you get by the line integral going around the surface in one direction; -7/6 if you go in the opposite direction

2. We let the normal vector in the surface integral be \vec{n}=-\vec{a}_{z}

3. Hence, we must calculate I:
I=\int_{0}^{1}\int_{0}^{x}x^{2}dydx+\int_{1}^{2}\int_{0}^{2-x}x^{2}dydx

The first integral is: I_{1}=\frac{1}{4}
The second integral is: I_{2}=\frac{11}{12}

This yields 7/6 in total..

 

[kelvintc]

thanks again.. hehe... oh... i really made too much mistakes in calculation.. sad

 

[arildno]

No wonder you're stuck by 6)!

As far as I can see, the surface integral (the flux) on the cylindrical shell is 0.

However, if you consider the fluxes through the top and bottom disks (z=0 and z=2) as well, you'll get for the total flux 16\pi which seems to agree with the answer..

 

[arildno]

Please post some work on 1)

 

[kelvintc]

replace z=0 and z=2 to get (z^2-1) ?
then find \int{rd\phi}{dr} ?
by this i get 16\pi...
i also get the surface integral in r direction = 0

 

[kelvintc]

(1) a_{r} = \cos{\phi}{a}_{x} + \sin{\phi}{a}_{y}
a_{\phi} = -\sin{\phi}{a}_{x} + \cos{\phi}{a}_{y}
a_{z} = a_{z}
then z=0, so vector in x direction only depends on phi .
-e^{-2}\sin\frac{\phi}{2}\sin\phi=-0.0586

 

[arildno]

I'm not sure if your post 47 is a cry for help, or if you got my meaning.
I'll look into 1) later

 

[kelvintc]

i think i got the answer for (6). just not sure whether the method is wrong or not