heat diffusion

[logistic_guy]

no comment

pleae comment as i don't understand where i go wrongi thought in integration we've to deal with constant of integration

[imath]\displaystyle \int_{0}^{L}\sin\left(\frac{n\pi x}{L}\right) \ dx = \frac{L(1 - \cos n\pi)}{n\pi}[/imath]

where is the constant of integration? why you don't write it. i'm confused

Now think about it. What happens if [imath]n[/imath] is odd and what happens if [imath]n[/imath] is even?

\(\displaystyle n = 1,2,3,....\) i understand this go to infinite, so you want me to check 1 trillion odd number and 1 trillion even number?

 

[khansaheb]

where is the constant of integration? why you don't write it. i'm confused

When you evaluate "definite" integral - you would have two expressions (containing the constant of integration) being subtracted from each other, in the "result". Thus those two constants vanish - after subtraction.

 

[mario99]

pleae comment as i don't understand where i go wrongi thought in integration we've to deal with constant of integration


where is the constant of integration? why you don't write it. i'm confused


\(\displaystyle n = 1,2,3,....\) i understand this go to infinite, so you want me to check 1 trillion odd number and 1 trillion even number?

What do you want me to say? I don't understand what you don't understand!

If you forgot how to evaluate definite integrals, open your calculus book again! And I agree with you that the indices go to infinite, but the idea is to check the first few numbers and you stop once you see a pattern!

 

[khansaheb]

.....where is the constant of integration? why you don't write it. i'm confused

Can you explain why/how the process of differentiation does not introduce a "constant of differentiation"?

Can you explain why/how does constant of integration was introduced during the process of integration?

 

[logistic_guy]

Can you explain why/how the process of differentiation does not introduce a "constant of differentiation"?

Can you explain why/how does constant of integration was introduced during the process of integration?

both have constant of integration. you don't give function to test on. i'll create function

\(\displaystyle f(x) = x^2 + cx\)

differentiation, you mean derivative right?

\(\displaystyle f'(x) = 2x + c\)

\(\displaystyle \int f(x) dx = \int (x^2 + cx) dx = \frac{x^3}{3} + c\frac{x^2}{2} + c = \frac{x^3}{3} + c(\frac{x^2}{2} + 1)\)

the constant \(\displaystyle c\) appear in both of them

What do you want me to say? I don't understand what you don't understand!

If you forgot how to evaluate definite integrals, open your calculus book again! And I agree with you that the indices go to infinite, but the idea is to check the first few numbers and you stop once you see a pattern!

i'm not junior student to study calculus again. i'm engineering 3rd year and i fell very good in calculus. maybe i'm just confused by the concept of the constant of integration. i don't see where is my mistake

 

[khansaheb]

3x3+c2x2+c

That c should be d (or something else - but not c)

Now get your old textbook (or Google) and REALLY understand the need for "constant of integration".

Why do we NEED to include constant of integration?

Why do you not encounter constant of differentiation?

Specially since you are student of engineering, you should discover the answers to the questions above.

 

[mario99]

i'm not junior student to study calculus again. i'm engineering 3rd year and i fell very good in calculus. maybe i'm just confused by the concept of the constant of integration. i don't see where is my mistake

I did not mean to offend you in any way. I just don't understand what you don't understand. I am confused too. I will show you the difference between definite and indefinite integrals and I hope that this is what you need to know.

Let me start with indefinite:

[imath]\displaystyle \int \sin x \ dx = -\cos x + C[/imath]

And now focus on the definite:

[imath]\displaystyle \int_{0}^{\pi} \sin x \ dx = -\cos x \bigg|_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = -(-2) = 2[/imath]

I think that you want to know why there is no [imath]C[/imath] in the definite integral. The answer is that you don't need to write it because eventually it will vanish.

Let us write it and see what happens:

[imath]\displaystyle \int_{0}^{\pi} \sin x \ dx = -\cos x + C\bigg|_{0}^{\pi} = -\cos \pi + C - \left(-\cos 0 + C\right)[/imath]

 

[mario99]

Continuation.

[imath]\displaystyle \int_{0}^{\pi} \sin x \ dx = -\cos \pi + C + \cos 0 - C = -\cos \pi + \cos 0 = -(-1) + 1 = 1 + 1 = 2[/imath]

Which is the same as the previous result.

 

[logistic_guy]

I did not mean to offend you in any way. I just don't understand what you don't understand. I am confused too. I will show you the difference between definite and indefinite integrals and I hope that this is what you need to know.

Let me start with indefinite:

[imath]\displaystyle \int \sin x \ dx = -\cos x + C[/imath]

And now focus on the definite:

[imath]\displaystyle \int_{0}^{\pi} \sin x \ dx = -\cos x \bigg|_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = -(-2) = 2[/imath]

I think that you want to know why there is no [imath]C[/imath] in the definite integral. The answer is that you don't need to write it because eventually it will vanish.

Let us write it and see what happens:

[imath]\displaystyle \int_{0}^{\pi} \sin x \ dx = -\cos x + C\bigg|_{0}^{\pi} = -\cos \pi + C - \left(-\cos 0 + C\right)[/imath]

thank mario99 very much. i think i understand my mistake now

[imath]\displaystyle \int_{0}^{L}\sin\left(\frac{n\pi x}{L}\right) \ dx = \frac{L(1 - \cos n\pi)}{n\pi}[/imath]

Now think about it. What happens if [imath]n[/imath] is odd and what happens if [imath]n[/imath] is even?

\(\displaystyle \frac{L(1 - \cos 1\pi)}{1\pi} = \frac{L(1 - -1)}{1\pi} = \frac{L(2)}{1\pi}\)

\(\displaystyle \frac{L(1 - \cos 2\pi)}{2\pi} = \frac{L(1 - 1)}{2\pi} = 0\)

\(\displaystyle \frac{L(1 - \cos 3\pi)}{3\pi} = \frac{L(1 - -1)}{3\pi} = \frac{L(2)}{3\pi}\)

\(\displaystyle \frac{L(1 - \cos 4\pi)}{4\pi} = \frac{L(1 - 1)}{4\pi} = 0\)

i think the pattern is zero when i write even number

 

[mario99]

thank mario99 very much. i think i understand my mistake now


\(\displaystyle \frac{L(1 - \cos 1\pi)}{1\pi} = \frac{L(1 - -1)}{1\pi} = \frac{L(2)}{1\pi}\)

\(\displaystyle \frac{L(1 - \cos 2\pi)}{2\pi} = \frac{L(1 - 1)}{2\pi} = 0\)

\(\displaystyle \frac{L(1 - \cos 3\pi)}{3\pi} = \frac{L(1 - -1)}{3\pi} = \frac{L(2)}{3\pi}\)

\(\displaystyle \frac{L(1 - \cos 4\pi)}{4\pi} = \frac{L(1 - 1)}{4\pi} = 0\)

i think the pattern is zero when i write even number

Yes, or better you may say:

[imath]\cos n\pi = -1 \rightarrow n \ \text{odd}[/imath]
[imath]\cos n\pi = 1 \rightarrow n \ \text{even}[/imath]

Or

[imath]\cos n\pi = (-1)^n, \ \ \ n=1,2,3,\cdots[/imath]

Or

[imath]-\cos n\pi = (-1)^{n+1}, \ \ \ n=1,2,3,\cdots[/imath]

Therefore,

[imath]\displaystyle \frac{L(1 - \cos n\pi)}{n\pi} =L\frac{(-1)^{n+1} + 1}{n\pi}[/imath]

And we got the same form in the original solution. Now you need to collect (and simplify) everything we got so far to get the exact solution in post #1.

 

[logistic_guy]

Yes, or better you may say:

[imath]\cos n\pi = -1 \rightarrow n \ \text{odd}[/imath]
[imath]\cos n\pi = 1 \rightarrow n \ \text{even}[/imath]

Or

[imath]\cos n\pi = (-1)^n, \ \ \ n=1,2,3,\cdots[/imath]

Or

[imath]-\cos n\pi = (-1)^{n+1}, \ \ \ n=1,2,3,\cdots[/imath]

Therefore,

[imath]\displaystyle \frac{L(1 - \cos n\pi)}{n\pi} =L\frac{(-1)^{n+1} + 1}{n\pi}[/imath]

And we got the same form in the original solution. Now you need to collect (and simplify) everything we got so far to get the exact solution in post #1.

thank

i think i understand the pattern

\(\displaystyle \theta_n(x,y) = B_n \sinh(\frac{n\pi y}{L})\sin(\frac{n\pi x}{L})\)

\(\displaystyle B_n = \frac{2}{L\sinh(\frac{n\pi W}{H})}\int_{0}^{L}\sin(\frac{n\pi x}{L}) dx \ \ \ n = 1,2,3,...\)

\(\displaystyle \displaystyle \int_{0}^{L}\sin(\frac{n\pi x}{L})dx = \frac{L(1 - \cos n\pi)}{n\pi}\)

\(\displaystyle \theta_n(x,y) = \frac{2}{L\sinh(\frac{n\pi W}{H})}\frac{L(1 - \cos n\pi)}{n\pi} \sinh(\frac{n\pi y}{L})\sin(\frac{n\pi x}{L})\)

\(\displaystyle \theta(x,y) = \sum_{n=1}^{\infty} \frac{2}{L\sinh(\frac{n\pi W}{H})}\frac{L(1 - \cos n\pi)}{n\pi} \sinh(\frac{n\pi y}{L})\sin(\frac{n\pi x}{L})\)

\(\displaystyle \frac{L(1 - \cos n\pi)}{n\pi} =L\frac{(-1)^{n+1} + 1}{n\pi}\)

\(\displaystyle \theta(x,y) = \sum_{n=1}^{\infty} \frac{2}{L\sinh(\frac{n\pi W}{H})}L\frac{(-1)^{n+1} + 1}{n\pi} \sinh(\frac{n\pi y}{L})\sin(\frac{n\pi x}{L})\)

i think i solve the equation. i want to show the temperature is \(\displaystyle 94.5\) how?

 

[mario99]

Show that the temperature at the midpoint is \(\displaystyle 94.5^{\circ}C\) by considering the first five nonzero terms of the infinite series that must be evaluated.

What do you understand from this?