Find Vector Field Given The Curl

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To find a vector field A in ℝ3 such that the curl of A equals a specified vector, the user derived a set of partial differential equations from the curl expression. They proposed a solution with components A1, A2, and A3, which were verified to satisfy the curl condition. The user expressed concern about the rigor of their method due to their professor's strictness but received reassurance that trial and error is an acceptable approach for this type of problem. The discussion emphasizes the validity of the proposed solution while acknowledging the subjective nature of academic expectations. Overall, the method used to arrive at the solution is deemed satisfactory.
tazzzdo
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Homework Statement



Find a vector field \vec{A}(\vec{r}) in ℝ3 such that:

\vec{\nabla} \times \vec{A} = y2cos(y)e-y\hat{i} + xsin(x)e-x2\hat{j}

The Attempt at a Solution



I broke it down into a series of PDE's that would be the result of \vec{\nabla} \times \vec{A}:

∂A3/∂y - ∂A2/∂z = y2cos(y)e-y

∂A3/∂x - ∂A1/∂z = -[-xsin(x)e-x2] (since the j component has a negative sign)

∂A2/∂x - ∂A1/∂y = 0

By a little trial and error I came up with:

A1 = zxsin(x)e-x2
A2 = -zy2cos(y)e-y
A3 = z

This can be verified that:

\vec{\nabla} \times \vec{A} = y2cos(y)e-y\hat{i} + xsin(x)e-x2\hat{j}

Is this a valid answer then?
 
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Sure it is. Why would you think it isn't?
 
Lol my Vector Calc professor tends to be picky. I didn't know if there was a more "rigorous" way to do it. But if this works, then I'm fine.
 
tazzzdo said:
Lol my Vector Calc professor tends to be picky. I didn't know if there was a more "rigorous" way to do it. But if this works, then I'm fine.

Trial and error is a perfectly fine way to solve a problem like this.
 
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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