Answer :
The function f(x, y, z) is given by f(x, y, z) = 10xyz + 5x^2z + 4x + z^2 + g1(x, z) + g2(y, z) + g3(x, y).
The evaluated integral ∫P · dr along the curve C is (5t, 2t^2, 38t) + C, where C is the constant of integration.
To find the function f such that F = ∇f, where F = (5yz, 5xz + 4, 5xy + 2z), we need to find the potential function f(x, y, z) by integrating each component of F with respect to its corresponding variable.
Integrating the first component, we have:
∫(5yz) dy = 5xyz + g1(x, z),
where g1(x, z) is a function of x and z.
Integrating the second component, we have:
∫(5xz + 4) dx = 5x^2z + 4x + g2(y, z),
where g2(y, z) is a function of y and z.
Integrating the third component, we have:
∫(5xy + 2z) dz = 5xyz + z^2 + g3(x, y),
where g3(x, y) is a function of x and y.
Now, we can write the potential function f(x, y, z) as:
f(x, y, z) = 5xyz + g1(x, z) + 5x^2z + 4x + g2(y, z) + 5xyz + z^2 + g3(x, y).
Combining like terms, we get:
f(x, y, z) = 10xyz + 5x^2z + 4x + z^2 + g1(x, z) + g2(y, z) + g3(x, y).
Therefore, the function f(x, y, z) is given by:
f(x, y, z) = 10xyz + 5x^2z + 4x + z^2 + g1(x, z) + g2(y, z) + g3(x, y).
To evaluate ∫P · dr along the curve C, where P = (5, 2, 44 – 6) and C is parameterized by r(t) = (t, t^2 + 5, 2t), we substitute the values of P and r(t) into the dot product:
∫P · dr = ∫(5, 2, 44 – 6) · (dt, d(t^2 + 5), 2dt).
Simplifying, we have:
∫P · dr = ∫(5dt, 2d(t^2 + 5), (44 – 6)dt).
∫P · dr = ∫(5dt, 2(2t dt), 38dt).
∫P · dr = ∫(5dt, 4tdt, 38dt).
Evaluating the integrals, we get:
∫P · dr = (5t, 2t^2, 38t) + C,
where C is the constant of integration.
Therefore, the evaluated integral ∫P · dr along the curve C is given by:
∫P · dr = (5t, 2t^2, 38t) + C.
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