What are some vector calculus questions about gradient, divergence, and curl?

  • Thread starter misogynisticfeminist
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In summary: I really appreciate the help. In summary, the conversation discusses various mathematical equations and operators such as the grad operator, divergence of a vector, and the curl operator. The conversation also includes questions and clarifications on how to derive the equations and formulas. Through the use of product and chain rule, as well as the Leibniz rule, the equations and formulas are simplified and solved.
  • #1
misogynisticfeminist
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1. I was reading on the geometric interpretation of the grad operator. I've did until the point where this particular relation was given.

[tex]d\varphi=0=C_1-C_1=\Delta C=(\nabla \varphi)\bullet d \vec r [/tex]

This is when we permit [tex] \vec r [/tex] to take us from the surface [tex] \varphi (x,y,z)=C_1[/tex] to another adjacent surface [tex]\varphi (x,y,z)=C_2 [/tex] where c are constants.

Why is it that in the first equation we have [tex] C_1-C_1[/tex] ?? Also, why is it that the consequence of the first relation shows that,

for a given [tex]\vert {d\vec r}\vert[/tex], the change of [tex] \varphi [/tex], [tex] d \varphi [/tex] is maximum when [tex]\vert {d\vec r}\vert[/tex] is parellel to [tex] \nabla\varphi [/tex] when [tex] \nabla\varphi [/tex] is normal to the surface.

2. While evaluating the divergence of vector, [tex] \nabla \bullet \vec r f(r) = \frac {\partial}{\partial x} (xf(r))+ \frac {\partial}{\partial y}(yf(r))+ \frac{\partial}{\partial z} (zf(r)) [/tex],

why is it equivalent to,

[tex] 3 f(r)+\frac {x^2}{r} \frac {df}{dr}+\frac {y^2}{r} \frac {df}{dr}+\frac {z^2}{r} \frac {df}{dr} [/tex] ?

I've tried manipulating the partial differentials using chain rules and all but don't seem to get it. Can someone show me the steps how? Also,

3. I was trying to simplify[tex]\nabla \times f \vec v \vert _x = \frac {\partial}{\partial y} (f V_z)-\frac {\partial}{\partial z} (f V_y)[/tex]. Also, how does it reduce to the final answer,

[tex] f \nabla \times \vec V \vert _x +\nabla f \times \vec V \vert _x [/tex]??

I have also tried manipulating the partial derivatives to no avail. Can someone help? thanks alot...

: )
 
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  • #2
misogynisticfeminist said:
2. While evaluating the divergence of vector, [tex] \nabla \bullet \vec r f(r) = \frac {\partial}{\partial x} (xf(r))+ \frac {\partial}{\partial y}(yf(r))+ \frac{\partial}{\partial z} (zf(r)) [/tex],

why is it equivalent to,

[tex] 3 f(r)+\frac {x^2}{r} \frac {df}{dr}+\frac {y^2}{r} \frac {df}{dr}+\frac {z^2}{r} \frac {df}{dr} [/tex] ?
Here you should use the product and chain rule. Also: [itex]r=\sqrt{x^2+y^2+z^2}[/itex]

So for example:

[tex]\frac{\partial}{\partial x}(xf(r))=f(r)+x\frac{\partial}{\partial x}f(r)[/tex]
This explains where the 3f(r) comes from.
Now:

[tex]\frac{\partial}{\partial x}f(r(x,y,z))=\frac{df}{dr}\frac{\partial r}{\partial x}[/tex]
 
  • #3
"This is when we permit to take us from the surface [tex] \varphi (x,y,z)=C_1[/tex]
to another adjacent surface [tex] \varphi (x,y,z)=C_1[/tex] where c are constants."

No, I don't think that's what that particular formula is intended to show. What that shows is that the derivative in a direction tangent to a level surface is 0 so [tex](\nabla \varphi)\bullet d \vec r= 0 [/tex]: that is, the gradient is perpendicular to all level surfaces.
What is true is that, since [itex]\vec u \bullet \vec v= |\vec u||\vec v|cos(\theta)[/tex], [tex](\nabla \varphi)\bullet d \vec r [/tex] is largest when cos(θ) is 1: i.e. θ= 0 and the two vectors are parallel.

For 2, I surprised you didn't see it while looking at the chain rule.
[tex]r= (x^2+ y^2+ z^2)^{\frac{1}{2}}[/tex] so [tex]\frac{\partial r}{\partial x}= \frac{1}{2}(x^2+ y^2+ z^2)^\frac{-1}{2}(2x)= \frac{x}{r}[/tex].
Of course, the same is true of the derivative with respect to y and z.
Now, [tex]\frac{\partial}{\partial x}(xf(r))= f(r)+ x\frac{\partial f}{/partial x}= f(r)+ \frac{x^2 f'(r)}{r}[/tex].

That first term is what gives you "3f(r)".
 
  • #4
Hey, thanks alot, that helped. Actually what i had problems with was the use of the product rule while evaluating [tex] \partial/ \ {\partial} x (xf(r))[/tex] and i also failed to keep in mind that [tex] r=\sqrt {x^2+y^2+z^2[/tex] but that was sorted out anyway.

thanks alot. I've still got one question regarding the curl operator though...hope that someone could help me with that.
 
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  • #5
In general

[tex] (\nabla\times\vec{f})_{x}=...? [/tex]

Daniel.
 
  • #6
dextercioby said:
In general

[tex] (\nabla\times\vec{f})_{x}=...? [/tex]

Daniel.

For, [tex] \nabla \times \vec f [/tex], it is,

[tex] det \left(
\begin{array}{ccc}

\hat x & \hat y & \hat z\\
{\partial}/{\partial x} & {\partial}/{\partial y} & {\partial}/{\partial z} \\
F_x & F_y & F_z \\

\end{array}
\right)

[/tex]

(the matrix is quite horrible though, i can't seem to type it the right way, but i guess you know what i mean).

hmmm, I've tried doing the curl again, i got,

[tex]\nabla \times f \vec v \vert _x = \frac {\partial}{\partial y} (f V_z)-\frac {\partial}{\partial z} (f V_y)[/tex]

Using the product rule

[tex]( f \frac {\partial{V_z}}{\partial {y}} + {V_z}\frac {\partial {f}}{\partial y})-(f \frac {\partial{V_y}}{\partial {z}} + {V_y}\frac {\partial {f}}{\partial z})[/tex]

Simplifying,

[tex] \frac {\partial}{\partial {y}} (f V_z +V_z f)-\frac {\partial}{\partial {z}} (f V_y+ V_y f)[/tex]

[tex] 2f( \frac {\partial}{\partial {y}} V_z - \frac {\partial}{\partial {z}} V_y) [/tex]

[tex] 2f \nabla \times \vec {v} \vert _x[/tex]

Where do i go from here? Or are my steps wrong?
 
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  • #7
kay,the code is \begin{array}{ccc} (3 columns)...

Okay.

Now solve your problem.

Daniel.
 
  • #8
hmmm, I've tried it in the edited post above.
 
  • #9
Well,use the Leibniz rule
[tex] [\nabla\times (f\vec{V})]_{x} =[(\nabla f\times \vec{V})+f(\nabla\times\vec{V})]_{x} =... [/tex]

Daniel.
 
  • #10
lol, leibniz's rule did everything. But I've arrived at the solution already, thanks alot.
 

1. What is vector calculus?

Vector calculus is a branch of mathematics that deals with the study of vectors and their properties. It is used to solve problems involving quantities that have both magnitude and direction, such as force, velocity, and acceleration.

2. How is vector calculus different from regular calculus?

The main difference between vector calculus and regular calculus is that vector calculus deals with vectors, while regular calculus deals with just scalar quantities. Vector calculus also involves the use of vector operations, such as addition, subtraction, and cross product.

3. What are some real-world applications of vector calculus?

Vector calculus has many applications in physics, engineering, and other sciences. It is used to model and analyze physical systems, such as the motion of objects, fluid flow, and electromagnetic fields. It is also used in computer graphics to create realistic 3D images.

4. What are some important concepts in vector calculus?

Some important concepts in vector calculus include vector fields, line integrals, surface integrals, and the gradient, divergence, and curl operators. These concepts are used to describe and analyze vector quantities in different dimensions.

5. How can I improve my understanding of vector calculus?

To improve your understanding of vector calculus, it is important to have a strong foundation in basic calculus and linear algebra. Practice solving problems and working with vectors and vector operations. You can also seek out additional resources, such as textbooks, online courses, and tutorials, to supplement your learning.

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