Griffith's E&M Derivation

In summary, the conversation discusses the use of tangents and partial derivatives in understanding waves and the connection between the two. The second question also involves the definition of the derivative and the legitimacy of multiplying both sides of the equation by a small value. The focus is on finding the displacement of a string at any point along its length at any given time.
  • #1
darkchild
155
0

Homework Statement


Please see the attached pdf file, which is the bottom of page 365 from the 3rd edition of the book. This is a lesson about (generalized) waves, and f(z,t) is the vertical displacement of the medium at point z, time t. F is net force and T is the tension on the string in the picture.

I don't understand how we get from the tangents of angles to partial derivatives in this derivation. I've asked a professor about this before, and he said something about tangent lines to curves being connected to derivatives, but it didn't make sense because the tangent line to a curve is a different use of tangent than what I have here, tangent the trig function. Is there some connection between the two?

Also, the last step (from first to second partial derivatives) seems too hand wavy to be legitimate.

Homework Equations



None for my first question.

For my second question, I understand that it involves the definition of the derivative:

[tex]\lim_{\Delta z \to 0} \frac{\frac{\partial f(z+ \Delta z)}{\partial t} -
\frac{\partial f(z)}{\partial t}}{\Delta z} = \frac{\partial^{2} f}{\partial t^{2}}[/tex]

What I'm not sure about is if it's mathematically sound to just multiply both sides of that equation by [tex]\Delta z[/tex], seeing as how one side has a limit.

The Attempt at a Solution



I have no idea.
 

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  • #2
darkchild said:

Homework Statement




I don't understand how we get from the tangents of angles to partial derivatives in this derivation. I've asked a professor about this before, and he said something about tangent lines to curves being connected to derivatives, but it didn't make sense because the tangent line to a curve is a different use of tangent than what I have here, tangent the trig function. Is there some connection between the two?

Also, the last step (from first to second partial derivatives) seems too hand wavy to be legitimate.

Homework Equations



None for my first question.


I'm responding to your first question. Then I'll look at your second question.

Consider the definition of the trig tangent when you are looking at a right triangle. Its tangent equals opposite over adjacent.

Now consider the derivative expressed as

[tex] \frac{dy}{dx} [/tex]

for a curve increasing from left to right. (imagine a parabola) Draw a tangent to the curve at some point P. What is the slope of the tangent? It's the change in y over the change in x, or
[tex] \frac{dy} {dx} [/tex] which is the "opposite", dy, over the "adjacent", dx. (Draw in little lines to represent the change in dy and dx to make this clear, if you need to ).
 
  • #3
darkchild said:
For my second question, I understand that it involves the definition of the derivative:

[tex]\lim_{\Delta z \to 0} \frac{\frac{\partial f(z+ \Delta z)}{\partial t} -
\frac{\partial f(z)}{\partial t}}{\Delta z} = \frac{\partial^{2} f}{\partial t^{2}}[/tex]

What I'm not sure about is if it's mathematically sound to just multiply both sides of that equation by [tex]\Delta z[/tex], seeing as how one side has a limit.

Surely you mean

[tex]\lim_{\Delta z \to 0} \frac{\frac{\partial f(z+ \Delta z)}{\partial z} -
\frac{\partial f(z)}{\partial z}}{\Delta z} = \frac{\partial^{2} f}{\partial z^{2}}[/tex]

...right? :wink:

Anyways, the point is, that for sufficiently small [itex]\Delta z[/itex], you have

[tex]\lim_{\Delta z \to 0} \frac{\frac{\partial f(z+ \Delta z)}{\partial z} -
\frac{\partial f(z)}{\partial z}}{\Delta z} \approx\frac{\frac{\partial f(z+ \Delta z)}{\partial z} -
\frac{\partial f(z)}{\partial z}}{\Delta z}[/tex]

In other words, the slope of the line connecting the point [tex](z,\frac{\partial f(z)}{\partial z}})[/itex] to the point [tex](z+\Delat z,\frac{\partial f(z+\Delta z)}{\partial z}})[/itex] is approximately equal to the tangent to the curve [tex]\frac{\partial f(z)}{\partial z}}[/itex] at the point [tex](z,\frac{\partial f(z)}{\partial z}})[/itex]
 
Last edited:
  • #4
gabbagabbahey said:
Surely you mean

Yes, that's what I meant.

Anyways, the point is, that for sufficiently small [itex]\Delta z[/itex], you have

[tex]\lim_{\Delta z \to 0} \frac{\frac{\partial f(z+ \Delta z)}{\partial z} -
\frac{\partial f(z)}{\partial z}}{\Delta z} \approx\frac{\frac{\partial f(z+ \Delta z)}{\partial z} -
\frac{\partial f(z)}{\partial z}}{\Delta z}[/tex]

I don't understand why that is a legitimate approximation. If [tex]\Delta z[/tex] is small, why wouldn't we say instead that the expression becomes large or approaches infinity?
 
  • #5
darkchild said:
I don't understand why that is a legitimate approximation. If [tex]\Delta z[/tex] is small, why wouldn't we say instead that the expression becomes large or approaches infinity?

If [tex]\Delta z[/tex] is small, then so is the difference [tex]\frac{\partial f(z+ \Delta z)}{\partial z} -
\frac{\partial f(z)}{\partial z}[/tex]...one small number, divided by another small number can easily produce finite result.

Think about the relationship between speed and position in one-dimension...the exact speed of a particle at time [itex]t[/itex] is given by [tex]v(t)=\frac{dx}{dt}=\lim_{\Delta t\to 0}\frac{x(t+\Delta t)-x(t)}{\Delta t}[/itex]...but if you measure the average speed over some very small time interval from [itex]t[/itex] to [itex]t+\Delta t[/itex], [tex]v_{av}=\frac{x(t+\Delta t)-x(t)}{\Delta t}[/tex] would you not expect it to be close to the exact speed at time [itex]t[/itex]?
 
  • #6
gabbagabbahey said:
If [tex]\Delta z[/tex] is small, then so is the difference [tex]\frac{\partial f(z+ \Delta z)}{\partial z} -
\frac{\partial f(z)}{\partial z}[/tex]...one small number, divided by another small number can easily produce finite result.

Oh, I see. In that case, my question is: Why assume or focus on the case in which [tex]\Delta z[/tex] is small?
 
  • #7
darkchild said:
Oh, I see. In that case, my question is: Why assume or focus on the case in which [tex]\Delta z[/tex] is small?


You are ultimately interested in finding the displacement of the string at any point along its length at any given time...looking at the difference in tension (force) over very small intervals [itex]\Delta z[/itex] allows you to determine the average acceleration that small piece of the string experiences at any given time...in the limit that [itex]\Delta z\to 0[/itex], you thus determine the exact acceleration of each point along the string, and hence you can then find the displacement of every point along the string (at any given point in time).
 

What is Griffith's E&M derivation?

Griffith's E&M derivation is a mathematical derivation that explains the behavior of electromagnetic waves in a vacuum. It is based on Maxwell's equations and provides a comprehensive understanding of how electric and magnetic fields interact with each other.

Why is Griffith's E&M derivation important?

Griffith's E&M derivation is important because it provides a fundamental understanding of the principles behind electromagnetic waves, which are essential to many fields of science and technology. It also serves as the basis for advanced theories and applications in electromagnetism.

What are the key equations in Griffith's E&M derivation?

The key equations in Griffith's E&M derivation are Maxwell's equations, which describe the relationship between electric and magnetic fields, and the wave equation, which describes the propagation of electromagnetic waves. These equations are used to derive important concepts such as the speed of light and the properties of electromagnetic waves.

How is Griffith's E&M derivation used in practical applications?

Griffith's E&M derivation is used in many practical applications, such as in the design of electronic devices and communication systems. It is also used in fields like optics, where understanding electromagnetic waves is crucial for understanding the behavior of light.

Are there any limitations to Griffith's E&M derivation?

Like any scientific theory, Griffith's E&M derivation has its limitations. It is based on certain assumptions and simplifications, which may not accurately reflect real-world situations. Additionally, it only applies to electromagnetic waves in a vacuum and may not be applicable in other mediums.

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