Calc Work: F<y,x,z^3> from P(1,1,1) to Q(2,3,4)

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In summary, to calculate the work performed by the force field F = <y,x,z^3> from P = (1,1,1) to Q = (2,3,4), you can use the potential difference between the two points, which is obtained by calculating the potential function f = xy + (z^4)/4 and finding the difference between the values at the two points. This yields a work value of 68.75.
  • #1
PsychonautQQ
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Homework Statement


Calculate the work performed by the force field F = <y,x,z^3> from P = (1,1,1) to Q = (2,3,4).




Homework Equations





The Attempt at a Solution


So first I found that F must be a conservative field because ∇(xy + (z^4)/r) = <y,x,z^3>

Since it is a conservative field it is path independent... which means something that my teacher would be mad at me for not knowing (lol)

Can I parameterize the line integral in terms of t?

like
c(t) = <1+t,1+2t,1+3t>
c'(t) = <1,2,3>

∫F dot c'(t) dt = W? then what would be the bounds on my integral?
 
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  • #2
Since you've found the potential function, why not just take the potential difference?
 
  • #3
^what he or she said. Which is, coincidentally, a very useful result of path independence!
 
Last edited:
  • #4
PsychonautQQ said:
c(t) = <1+t,1+2t,1+3t>
c'(t) = <1,2,3>

∫F dot c'(t) dt = W? then what would be the bounds on my integral?

What values of ##t## correspond to the two points?
 
  • #5
0 and 1?
so ∫F dot c'(t) dt = W between 0 and 1 will give me the correct answer?
 
  • #6
how do I put F into terms of t though? like x = cos(t) y = sin(t) and z=z? ;-( I'm confused
 
  • #7
LCKurtz said:
What values of ##t## correspond to the two points?

PsychonautQQ said:
0 and 1?

Why do you answer every hint with another question? Aren't you capable of plugging ##t=0## and ##t=1## into your equation of the line and checking whether you get the two points yourself?

so ∫F dot c'(t) dt = W between 0 and 1 will give me the correct answer?

Yes, if you insist on doing the problem the hard way. Had you heeded the advice in post #2 you would have been done with this thread long ago.
 
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  • #8
I got the work = 68.75.

f(r(Q) - f(r(P)
f = xy + (z^4)/4
from <1,1,1> to <2,3,4>
so
(2)(3) + ((4)^4)/4 - ((1)(1) + ((1)^4)/4
70 - 1.25
68.75

Is this the correct method for doing this problem?
 
  • #9
PsychonautQQ said:
I got the work = 68.75.

f(r(Q) - f(r(P)
f = xy + (z^4)/4
from <1,1,1> to <2,3,4>
so
(2)(3) + ((4)^4)/4 - ((1)(1) + ((1)^4)/4
70 - 1.25
68.75

Is this the correct method for doing this problem?

Yes. (Btw, you wrote /r instead of /4 in the OP. Adjacent keys, I see.)
 
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1. What is the formula for calculating work in a vector field?

The formula for calculating work in a vector field is W = ∫F⋅dr, where W is work, F is the vector field, and dr is the differential displacement vector.

2. How do I find the vector field in this problem?

In this problem, the vector field is given as F. This means that at any given point in the field, the vector F has components of y, x, and z^3 in the y, x, and z directions, respectively.

3. What are the coordinates of the starting and ending points in this problem?

The starting point in this problem is P(1,1,1), which means that the initial coordinates are x = 1, y = 1, and z = 1. The ending point is Q(2,3,4), which means that the final coordinates are x = 2, y = 3, and z = 4.

4. How do I calculate the differential displacement vector in this problem?

The differential displacement vector, dr, is calculated by taking the difference between the final and initial coordinates. In this problem, dr = (2-1, 3-1, 4-1) = (1, 2, 3).

5. What is the final result of the calculation?

The final result of the calculation is the value of W, which represents the work done by the vector field from P to Q. It is important to note that this value may be positive, negative, or zero depending on the direction of the vector field and the displacement vector.

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