How to Find Temperature Distribution in a Rectangular Sheet?

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To find the temperature distribution in a rectangular sheet with specified boundary conditions, the solution involves using a Fourier series approach. The general formula for the temperature distribution is given as U(x, y) = (summation) B [sin((n*pi*x)/a) sinh((n*pi*y)/a)]. A modification of this formula is proposed to account for the boundary conditions, leading to U(x, y) = (summation) B [sin((n*pi*x)/a) sinh((n*pi*(b-y))/a)]. The discussion emphasizes the importance of correctly applying boundary conditions to derive the accurate temperature distribution. This method effectively addresses the problem of temperature distribution in the specified rectangular domain.
Void123
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Homework Statement



Find the temperature distribution in the rectangular sheet given the following boundary conditions:

U(0, y) = U(a, y) = U(x, b) = 0; U(x, 0) = f(x)


Homework Equations



U(x, y) = (summation) B (fourier coefficient) [sin ((n*pi*x)/a) sinh ((n*pi*y)/a)]



The Attempt at a Solution



Wouldn't it just be a simple modification of the above formula:

U(x, y) = (summation) B [sin ((n*pi*x)/a) sinh ((n*pi*(b - y))/a)
 
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Void123 said:

Homework Statement



Find the temperature distribution in the rectangular sheet given the following boundary conditions:

U(0, y) = U(a, y) = U(x, b) = 0; U(x, 0) = f(x)


Homework Equations



U(x, y) = (summation) B (fourier coefficient) [sin ((n*pi*x)/a) sinh ((n*pi*y)/a)]



The Attempt at a Solution



Wouldn't it just be a simple modification of the above formula:

U(x, y) = (summation) B [sin ((n*pi*x)/a) sinh ((n*pi*(b - y))/a)


I can't help you solve it. However, I will bump it by retyping it with LaTeX for you :-)

U\left(x,y\right)=\sum B\sin{\left(\frac{n \pi x}{a}\right)}\sinh{\left(\frac{n\pi(b-y)}{a}\right)}
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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