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anonymous
 3 years ago
Solve the semiimfinite plate problem if the bottom edge of width 20 is held at
T = 0o for 0 < x < 10, and T = 100o for 10< x<20. And the other sides are at 0o
anonymous
 3 years ago
Solve the semiimfinite plate problem if the bottom edge of width 20 is held at T = 0o for 0 < x < 10, and T = 100o for 10< x<20. And the other sides are at 0o

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Solve the semiimfinite plate problem if the bottom edge of width 20 is held at: \[T = 0^{o} \rightarrow 0 < x < 10\]\[T = 100^{o} \rightarrow 10 < x < 20\] and the other sides are at \(\ 0^{o}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1366518323249:dw @oldrin.bataku hbu mate ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0have idea ?? i'm little bit confused for \(\ 100^{o}\) > 10 < x < 20..,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is this similar to the Fourier heat problem?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0gerry this is the same problem except that boundary condition for bottom edge

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we had (note that 10 turens to 20)\[T = \sum_{n=1}^{\infty} a_ne^{\frac{n \pi}{20} y} \sin (\frac{n \pi}{20} x)\]now for evaluating \(a_n\) using fourier series\[T = 0 \rightarrow 0 < x < 10 \\ T = 100 \rightarrow 10 < x < 20 \\ \ \ @ \ \ y=0\]so\[a_n=\frac{2}{20} (\int_{0}^{10} 0 \times \sin (\frac{n \pi}{20} x) \ \text{d}x+\int_{10}^{20} 100 \times \sin (\frac{n \pi}{20} x) \ \text{d}x)\]\[a_n=\frac{1}{10} \int_{10}^{20} 100 \times \sin (\frac{n \pi}{20} x) \ \text{d}x=...\]makes sense?
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