where F(x, y, z) = (2x+yz)i + (2y+xz)j + (4z+xy)k and C is the path from (1,6,-1) to (4,2,3) given by x(t) = 3t+1, y(t) = 4+2cos(πt/2), z(t) = t^2 - 1, for 0
-
-
Evaluate the curl of the function. If the curl of the vector field function gives a zero vector, then this field is conservative. There exists a function f such that the gradient of f is the F vector field.
This means that:
df/dx = 2x + yz
df/dy = 2y + xz
df/dz = 4z + xy
Where f is the potential function of x y and z.
Solving for f in the first line:
f(x,y,z) = int(2x + yz dx)
f = x^2 + xyz + g(y,z)
Take the first derivative with respect to y:
df/dy = xz + g'(y,z)
But this is equal to 2y + xz:
xz + g'(y,z) = 2y + xz
Solve for g:
g' (y,z) = 2y
g(y,z) = y^2 + h(z)
Plugging this in the term for f:
f = x^2 + xyz + y^2 + h(z)
Take derivative wrt z:
df/dz = xy + h'(z) = 4z + xy
Solve for h'
h'(z) = 4z
h(z) = 2z^2 + c
Plugging into f:
f = x^2 + xyz + y^2 + 2z^2 + c is our potential function.
The fundamental theorem of Calculus is applied by saying that the line integral of the gradient of f *dr = f(x,y,z)) (t=2) - f(x,y,z) when t = 0
Solve for x y and a for t = 2 and t = 0 to evaluate the above.
*edit to add: the above works because we har a conservative vector field. A conservative vector field is path independent: only the final and initial positions determine the value of the function, not the path that it goes through. A few real world examples would be gravity. The gravitational field is conservative and the work done by gravity on an object depends on the change in the potential energy function. Electrostatic charge has a conservative electric field. Spring work is conservative.
The process to find the potential function is equivalent to finding the anti derivative of a scalar function in introductory calculus. This is like the vector counterpart of the scalar method.
This means that:
df/dx = 2x + yz
df/dy = 2y + xz
df/dz = 4z + xy
Where f is the potential function of x y and z.
Solving for f in the first line:
f(x,y,z) = int(2x + yz dx)
f = x^2 + xyz + g(y,z)
Take the first derivative with respect to y:
df/dy = xz + g'(y,z)
But this is equal to 2y + xz:
xz + g'(y,z) = 2y + xz
Solve for g:
g' (y,z) = 2y
g(y,z) = y^2 + h(z)
Plugging this in the term for f:
f = x^2 + xyz + y^2 + h(z)
Take derivative wrt z:
df/dz = xy + h'(z) = 4z + xy
Solve for h'
h'(z) = 4z
h(z) = 2z^2 + c
Plugging into f:
f = x^2 + xyz + y^2 + 2z^2 + c is our potential function.
The fundamental theorem of Calculus is applied by saying that the line integral of the gradient of f *dr = f(x,y,z)) (t=2) - f(x,y,z) when t = 0
Solve for x y and a for t = 2 and t = 0 to evaluate the above.
*edit to add: the above works because we har a conservative vector field. A conservative vector field is path independent: only the final and initial positions determine the value of the function, not the path that it goes through. A few real world examples would be gravity. The gravitational field is conservative and the work done by gravity on an object depends on the change in the potential energy function. Electrostatic charge has a conservative electric field. Spring work is conservative.
The process to find the potential function is equivalent to finding the anti derivative of a scalar function in introductory calculus. This is like the vector counterpart of the scalar method.
12
keywords: Fundamental,of,int,Calculus,Integrals,middot,compute,Use,for,Line,Theorem,to,dr,the,Use the Fundamental Theorem of Calculus for Line Integrals to compute ∫ F · dr