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...