Question

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.

Answer 1

Answer:

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.