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27 October, 16:40

Suppose tha:

∇f (x, y, z) = 2xyzex^2i + zex^2j + yex^2k.

1. If f (0,0,0) = -4, find f (2,2,4)

Hint: As a first step, define a path from (0,0,0) to (2, 2, 4) and compute a line integral.

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  1. 27 October, 19:51
    0
    f (2,2,4) = 8e^4 - 4

    Step-by-step explanation:

    Note that by integrating each entry of "f" with respect to the proper variable:

    ∇ (yze^ (x^2) + C) = {^ (x^2), ze^ (x^2), ye^ (x^2) }

    Hence, f (x, y, z) = yze^ (x^2) + C

    So, the Fundamental Theorem of Calculus for line integral yields

    ∫c f (x, y, z) · dr, where C is a path from (0,0,0) to (2,2,4)

    = (yze^ (x^2) {from (0,0,0) to (2,2,4) }

    = 8e^4.

    Now, let's solve for C, using f (0,0,0) = - 4:

    -4 = 0 + C ⇒ C = - 4.

    So, f (x, y, z) = yze^ (x^2) - 4

    ⇒ f (2,2,4) = 8e^4 - 4.
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