Compute the work done by the force field.

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To compute the work done by the force field F(xyz) along the path C, one must substitute the parameterized functions x(t), y(t), and z(t) into the force equation. The derivative of r(t) should then be calculated to perform the dot product with the force field. The integral formula ∫F·dr is applicable, specifically expressed as ∫_0^1 F·(dr/dt) dt. Clarifications were made regarding the notation, emphasizing the use of lowercase r for consistency. The discussion highlights the importance of understanding the independent variable t and the limits of integration.
bfusco
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Homework Statement


compute the work done by the force field F(xyz)=[(yze^(xyz)+y^2+1)i+(xze^(xyz)+2xy)j+(xye^(xyz))k], in moving the object along the path C from beginning to end. here C is the path paramterized by r(t)=<t, (t^2)-1, t+2>, 0<_t<_1.

The Attempt at a Solution


Would i be correct if i said that i should use the functions of x(t), y(t), z(t), and substituting them into the original equation? then take the derivative of r(t) and do the dot product of them. then use the formula: ∫F*dr?
 
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bfusco said:

Homework Statement


compute the work done by the force field F(xyz)=[(yze^(xyz)+y^2+1)i+(xze^(xyz)+2xy)j+(xye^(xyz))k], in moving the object along the path C from beginning to end. here C is the path paramterized by r(t)=<t, (t^2)-1, t+2>, 0<_t<_1.

The Attempt at a Solution


Would i be correct if i said that i should use the functions of x(t), y(t), z(t), and substituting them into the original equation? then take the derivative of r(t) and do the dot product of them. then use the formula: ∫F*dr?

If you mean$$
\int_0^1\vec F\cdot \frac{d\vec R}{dt}\, dt$$then yes.
 
bfusco said:
Would i be correct if i said that i should use the functions of x(t), y(t), z(t), and substituting them into the original equation? then take the derivative of r(t) and do the dot product of them. then use the formula: ∫F*dr?
Some of that sounds right, other parts bewildering. To make it clearer, try it and post your working.
 
LCKurtz said:
If you mean$$
\int_0^1\vec F\cdot \frac{d\vec R}{dt}\, dt$$then yes.
Actually, given that \vec{r}(t) is a function of t, "d\vec{r}" is a reasonable way of writing "(d\vec{r}/dt) dt". But I would object to using the capital R when only "r" was given before.
 
HallsofIvy said:
Actually, given that \vec{r}(t) is a function of t, "d\vec{r}" is a reasonable way of writing "(d\vec{r}/dt) dt". But I would object to using the capital R when only "r" was given before.

Yeah, I used the capital letter out of habit. But my point was to make sure the OP understood the independent variable was ##t## and the use of the ##t## limits.
 
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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