Finding Work Done by a Force Field: A Math Problem

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To find the work done by the force field F = 2xy(ux) + (x^2 - z^2)(uy) - 3xz^2(uz) on a particle moving from A(0,0,0) to B(2,1,3), the integral of the dot product of F and the differential path vector dl must be calculated. The work can be expressed as the integral from A to B of F·dl, where dl is defined as dx(ux) + dy(uy) + dz(uz). A common variable of integration, t, can be introduced to express both the position vector r and the force F as functions of t. The integral can then be evaluated over the limits corresponding to the parameterization of the straight line path from A to B. This approach will yield the total work done by the force field along the specified path.
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i have a force field F = 2xy*(ux)+(x^2-z^2)(uy)-3xz^2(uz) where ux,uy,and uz are unit vectors in the direction indicated. I have to find the work done by the field on a particle that travels from A(0,0,0) to
B(2,1,3) on a straight line. work is regarded as the integral from A to B of the dot product of F and dl = dx(ux)+dy(uy)+dz(uz). I'm not quite sure how to do this.
 
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You can introduce a common vairable of integration to solve this.
Since
\int_{A}^{B} \vec{F}.d\vec{r} = \int_{t1}^{t2} \vec{F}.(\vec{\frac{dr}{dt}}) dt

Where \vec{r} (and F) can be expressed as a function of t.
 
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ok, thanks
 
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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