# Is this enough to get full marks?

• sa1988
In summary, the conversation discusses a student's concerns about the difficulty level of a past exam question involving Laplace's equation and boundary conditions. The student provides a summary of their working, which involves separating variables and using boundary conditions to find the solution. The expert notes that the student may lose marks for using the same symbol to represent multiple variables.
sa1988

## The Attempt at a Solution

I'd first like to apologise in advance because I may be making a few more threads like this. I have an exam on Tuesday and am looking at past papers which don't have any answers for me to check whether I'm doing the right thing, so I'm turning to the heroes of Physics Forums in my hour need.

For the above question I'm pretty sure I've done it correctly but I can't see how the work I've done is even remotely worth 30 marks.

Here's my working:

a)

##T(x,y) = T(x)T(y)##

Put into Laplace equation:

##\frac{d^2T(x)}{dx^2}T(y) + \frac{d^2T(y)}{dy^2}T(x) = 0##

##\frac{d^2T(x)}{dx^2}\frac{1}{T(x)} + \frac{d^2T(y)}{dy^2}\frac{1}{T(y)} = 0##

##\frac{d^2T(x)}{dx^2}\frac{1}{T(x)} = - \frac{d^2T(y)}{dy^2}\frac{1}{T(y)}##

Then for separation constant ##m^2## :

##\frac{d^2T(x)}{dx^2} = m^2 T(x)##

##\frac{d^2T(y)}{dy^2} = -m^2 T(y)##

Done. Is that really worth 8 marks?

b)

By inspection I can already see the solutions:

##T(x) = Ae^{mx}+Be^{-mx}##

##T(y) = Ce^{imy}+De^{-imy}##

So,

##T(x,y) = T(x)T(y) = (Ae^{mx}+Be^{-mx})(Ce^{imy}+De^{-imy})##

Done. For 10 marks? I suppose I need to show the step by step working to get to those obvious solutions using the the usual ##f(x) = e^{\lambda x}## trial solution route? But even then it seems a bit simple for 10 marks.

c)

Boundary conditions are:
##T(0,0) = 0##
##T(x,0) = 0##
##T(x,L) = 0##
##T(L,y) = sin(\frac{6\pi y}{L})##

These conditions give four equations:

##i. (A+B)(C+D) = 0##
##ii. (Ae^{mx}+Be^{-mx})(C+D) = 0##
##iii. (A+B)(Ce^{imL}+De^{-imL}) = 0##
##iv. (Ae^{mL}+Be^{-mL})(Ce^{imy}+De^{-imy}) = sin(\frac{6\pi y}{L})##

Condition ##ii## leads to ##D = -C##
Condition ##iii## leads to ##B= -A##

These can then be put into condition ##iv##:

##AC(e^{mL}-e^{-mL})(e^{imy}-e^{-imy}) = sin(\frac{6\pi y}{L})##

Then rewrite the exponentials in trig form:

##AC sinh(mL)sin(my) = sin(\frac{6\pi y}{L})##

Now this is clearly only possible for ##m = \frac{6\pi}{L}## and ##AC = \frac{1}{sinh(6\pi)}##,

leading to a full solution of:

##T(x,y) = \frac{sinh(\frac{6\pi x}{L})}{sinh(6\pi)}sin(\frac{6\pi y}{L})##

I've checked this against the Laplace equation and it holds so I'm fairly sure it's correct. But it's a pretty easy 15 marks.

So overall I'm wondering if there's something I've done wrong? Should I be showing more step-by-step working or something? The 10-mark question is the biggest worry for me as it seemed far too easy. I'm convinced I missed something, yet I've come to the correct final answer at the end so what could possibly be penalised?

Thanks for any thoughts/opinions.

Your first boundary condition is redundant, and you're missing a boundary condition. Condition iii doesn't imply A=-B.

vela said:
Your first boundary condition is redundant, and you're missing a boundary condition. Condition iii doesn't imply A=-B.

So I am missing something then.

I know the first condition was left unused, but I can't see how it was needed.

If condition ##iii## doesn't imply ##A=-B##, the only thing I could say is that it 'suggests':

##A=-B##
OR
##m=\frac{n\pi}{L}##
or both.

Similarly condition ##i## can only 'suggest':
##A=-B##
OR
##C=-D##
or both.

Is there some other route I should take to achieve values for A, B, C, D and m with certainty instead of having to use this loose 'trial guess' method?

And you say I'm missing another boundary condition altogether. Which condition is that?

Thanks.

Reread the problem. You'll see you left out the condition for one of the edges.

sa1988
vela said:
Reread the problem. You'll see you left out the condition for one of the edges.

Ha. Christ. I see it now. That changes everything.

##i. T(0,y) = 0##
##ii. T(x,0) = 0##
##iii. T(x,L) = 0##
##iv. T(L,y) = sin(\frac{6\pi y}{L})##

So,

##i. (A+B)(Ce^{imy}+De^{-imy}) = 0##

## \implies B = -A##

##ii. A(e^{mx}-e^{-mx})(C+D) = 0##

## \implies D = -C##

##iii. AC(e^{mx}-e^{-mx})(e^{imL}-e^{-imL}) = 0##

## \implies m = \frac{n\pi}{L}##, ##\Big( n\in \mathbb{N}\Big)##

##iv. AC(e^{n\pi}-e^{-n\pi})(e^{i\frac{n\pi y}{L}}-e^{-i\frac{n\pi y}{L}}) = sin(\frac{6\pi y}{L})##

rewrite ##iv## as trig:

##iv. ACsinh(n\pi)(sin(\frac{n\pi y}{L}) = sin(\frac{6\pi y}{L})##
##\implies AC = \frac{1}{sinh(n\pi)}##
and
##n = 6##

Which gives the same answer as before.

But still, my error was only due to a tiny 'typo' so to speak. It still seems like a surprisingly small amount of steps across parts a, b and c to gain the full 30 marks.

In the last part, I'd say
$$\sum_{n=1}^\infty A_n \sinh n\pi \sin \left(\frac{n \pi y}{L}\right) = \sin\left(\frac{6 \pi y}{L}\right),$$ from which you can conclude ##A_6 = \frac{1}{\sinh 6\pi}## and ##A_n=0## for ##n \ne 6##.

sa1988
sa1988 said:

## The Attempt at a Solution

I'd first like to apologise in advance because I may be making a few more threads like this. I have an exam on Tuesday and am looking at past papers which don't have any answers for me to check whether I'm doing the right thing, so I'm turning to the heroes of Physics Forums in my hour need.

For the above question I'm pretty sure I've done it correctly but I can't see how the work I've done is even remotely worth 30 marks.

Here's my working:

a)

##T(x,y) = T(x)T(y)##

Put into Laplace equation:

##\frac{d^2T(x)}{dx^2}T(y) + \frac{d^2T(y)}{dy^2}T(x) = 0##

##\frac{d^2T(x)}{dx^2}\frac{1}{T(x)} + \frac{d^2T(y)}{dy^2}\frac{1}{T(y)} = 0##

##\frac{d^2T(x)}{dx^2}\frac{1}{T(x)} = - \frac{d^2T(y)}{dy^2}\frac{1}{T(y)}##

Then for separation constant ##m^2## :

##\frac{d^2T(x)}{dx^2} = m^2 T(x)##

##\frac{d^2T(y)}{dy^2} = -m^2 T(y)##

Done. Is that really worth 8 marks?

b)

By inspection I can already see the solutions:

##T(x) = Ae^{mx}+Be^{-mx}##

##T(y) = Ce^{imy}+De^{-imy}##

So,

##T(x,y) = T(x)T(y) = (Ae^{mx}+Be^{-mx})(Ce^{imy}+De^{-imy})##

Done. For 10 marks? I suppose I need to show the step by step working to get to those obvious solutions using the the usual ##f(x) = e^{\lambda x}## trial solution route? But even then it seems a bit simple for 10 marks.

c)

Boundary conditions are:
##T(0,0) = 0##
##T(x,0) = 0##
##T(x,L) = 0##
##T(L,y) = sin(\frac{6\pi y}{L})##

These conditions give four equations:

##i. (A+B)(C+D) = 0##
##ii. (Ae^{mx}+Be^{-mx})(C+D) = 0##
##iii. (A+B)(Ce^{imL}+De^{-imL}) = 0##
##iv. (Ae^{mL}+Be^{-mL})(Ce^{imy}+De^{-imy}) = sin(\frac{6\pi y}{L})##

Condition ##ii## leads to ##D = -C##
Condition ##iii## leads to ##B= -A##

These can then be put into condition ##iv##:

##AC(e^{mL}-e^{-mL})(e^{imy}-e^{-imy}) = sin(\frac{6\pi y}{L})##

Then rewrite the exponentials in trig form:

##AC sinh(mL)sin(my) = sin(\frac{6\pi y}{L})##

Now this is clearly only possible for ##m = \frac{6\pi}{L}## and ##AC = \frac{1}{sinh(6\pi)}##,

leading to a full solution of:

##T(x,y) = \frac{sinh(\frac{6\pi x}{L})}{sinh(6\pi)}sin(\frac{6\pi y}{L})##

I've checked this against the Laplace equation and it holds so I'm fairly sure it's correct. But it's a pretty easy 15 marks.

So overall I'm wondering if there's something I've done wrong? Should I be showing more step-by-step working or something? The 10-mark question is the biggest worry for me as it seemed far too easy. I'm convinced I missed something, yet I've come to the correct final answer at the end so what could possibly be penalised?

Thanks for any thoughts/opinions.

You would likely lose marks for using the exact symbol ##T## to stand for three different things in the same problem. Use better notation, such as ##T_1(x), T_2(y)## or ##F(x), G(y)## or something similar. It is never a good idea to make the marker work extra hard trying to decipher what you are doing.

sa1988
vela said:
In the last part, I'd say
$$\sum_{n=1}^\infty A_n \sinh n\pi \sin \left(\frac{n \pi y}{L}\right) = \sin\left(\frac{6 \pi y}{L}\right),$$ from which you can conclude ##A_6 = \frac{1}{\sinh 6\pi}## and ##A_n=0## for ##n \ne 6##.
Ray Vickson said:
You would likely lose marks for using the exact symbol ##T## to stand for three different things in the same problem. Use better notation, such as ##T_1(x), T_2(y)## or ##F(x), G(y)## or something similar. It is never a good idea to make the marker work extra hard trying to decipher what you are doing.

Thanks, duly noted.

## 1. Is following the instructions enough to get full marks?

Following the instructions is a crucial part of any scientific project or assignment. However, it is not the only factor that determines your grade. You also need to demonstrate a clear understanding of the subject matter and present your findings or results accurately and logically.

## 2. Do I need to include all the details in my report to get full marks?

In most cases, including all the relevant details and information in your report is essential to receive full marks. However, it is important to remember that the quality of your analysis and interpretation of the data is also a significant factor in determining your grade. Therefore, make sure to focus on both the quantity and quality of your report.

## 3. Can I get full marks if my experiment does not yield expected results?

The purpose of conducting an experiment is to test a hypothesis and gather data, regardless of the outcome. As long as you can demonstrate a clear understanding of your experimental design, methodology, and analysis of the data, you can still receive full marks even if your results are unexpected.

## 4. Will my writing style affect my grade for this project?

While writing style is not the primary focus of a scientific project or assignment, it can still impact your grade. Your writing should be clear, concise, and free of grammatical errors to effectively communicate your ideas and findings. Additionally, following the required formatting guidelines for your report is also essential.

## 5. How important is it to cite my sources in a scientific project?

Citing your sources is crucial in any scientific project or report. It not only gives credit to the original authors but also helps to support and strengthen your arguments and findings. Failure to cite your sources can result in plagiarism and significantly impact your grade.

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