Solution to II.2.8-II.2.10: Textbook Equations

In summary, the homework equations say that: 1) E changed to e somewhere between II.2.8 and II.2.92) There should be a minus sign in front of II.2.9, otherwise it's correct.3) Remember: d/dr(rn) = nrn-1 etc.4) No way I can determine the correctness of II.2.10 since a new parameter (r0) is introduced.According to the homework equations, E changed to e somewhere between II.2.8 and II.2.9. There should be a minus sign in front of II.2.9, otherwise it's correct. Remember: d/
  • #1
Mechdude
117
1

Homework Statement


please look at equations II.2.8 , II.2.9, II.2.10 in the attached scan,

Homework Equations


none

The Attempt at a Solution


im unable to get from II.2.8 to II.2.9, i wonder where they got the extra r in the denominator of the stuff in the brackets,
here is what i get for II.2.10:
[tex] \frac{1}{18} \left( \frac{ (n+2) n C} { r^{n+3}} - \frac{3 M e^2}{r^4} \right) [/tex]
what I am i doing wrong?
 

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  • #2
Looks like sloppy work to me:

1. in going from II.2.8 to II.2.9 somehow E changed to e.

2. There should be a minus sign in front of II.2.9, otherwise it's correct.

remember: d/dr(rn) = nrn-1 etc.

No way I can determine the correctness of II.2.10 since a new parameter (r0) is introduced.
 
  • #3
rude man said:
Looks like sloppy work to me:

1. in going from II.2.8 to II.2.9 somehow E changed to e.

2. There should be a minus sign in front of II.2.9, otherwise it's correct.

remember: d/dr(rn) = nrn-1 etc.

No way I can determine the correctness of II.2.10 since a new parameter (r0) is introduced.

how did u do that? in II.2.8 there is (for stuff after partial differentiation***{EDIT})
[tex] \frac{1}{6r^2} \left( ME^2 \frac{1}{r} - \frac{nC}{r^n} \right) [/tex]
how does this end up being
[tex] \frac{1}{6} \left( ME^2 \frac{1}{r^4} - \frac{nC}{r^{n+3}} \right) [/tex]
when the stuff just outside the bracket gets taken in? r is only to the inverse second power, why does r in II.2.9 end up to the inverse fourth power? 2+1 =4 ? remember the partial diff hasn't yet been done.

***EDIT
i mean partial differentiation sign not after performing the operation.
 
Last edited:
  • #4
though i think the [itex] r_o [/itex] is the r corresponding to minimum P.E., looking at II.2.7 and the paragraph preceeding it.
 
  • #5
Oops - I hadn't noticed - in II.2.9 there should not be a partial differential anymore. The derivative is taken already in going from II.2.8 to II.2.9.

Like I said - sloppy work! And there still needs to be a minus sign in front of the corrected II.2.9.
 
  • #6
rude man said:
Oops - I hadn't noticed - in II.2.9 there should not be a partial differential anymore. The derivative is taken already in going from II.2.8 to II.2.9.

Like I said - sloppy work! And there still needs to be a minus sign in front of the corrected II.2.9.

Ok, i see that now, i had not noticed that the differentiation had already been done, so that's why you kept on insisting on the negative sign, ( i admit i was clueless why until now).
thanks. Let me work on it and see whether i can better this guy's work.
cheers.
 
  • #7
one quick last one, has anyone seen this question before? is n=8.1 really? is that the correct answer that i need to find when i correct this guy's unholy mess?
 

1. What is the meaning of a solution in the context of textbook equations?

A solution to a textbook equation refers to the set of values that satisfy the equation and make it true. In other words, it is the answer to the equation.

2. How do I know if my solution to a textbook equation is correct?

You can check the validity of your solution by substituting the values into the equation and ensuring that it holds true. Additionally, you can also use a calculator or computer software to verify the solution.

3. Can there be more than one solution to a textbook equation?

Yes, there can be multiple solutions to a textbook equation. For example, a quadratic equation may have two solutions or roots. These solutions may be real or complex numbers.

4. What if my solution to a textbook equation does not make sense in the given context?

If your solution does not make sense in the given context, it is possible that you have made a mistake in solving the equation. Check your work and make sure you have followed all the steps correctly. If you are still unsure, seek help from a teacher or tutor.

5. Can a textbook equation have no solution?

Yes, there are equations that have no solution. For example, an equation with two variables and one equation may not have a unique solution. This is known as an inconsistent system of equations. In other cases, the solution may involve imaginary numbers, which may not be relevant in certain contexts.

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