Question

Given that ∇f ( X, Y, Z) = 2XYZ^3ˆi +X^2Z^3ˆj + 3X^2YZ^2kˆ and f (1, − 2, 2) = 4 show that
f ( X, Y, Z) =X^2YZ^3 + 20 .

Answer 1

Answer on 39181, Math, Other From

$$\delta f = 2 x y z ^ {3} \vec {i} + x ^ {2} z ^ {3} \vec {j} + 3 x ^ {2} y z ^ {2} \vec {k}$$

we see that

$$\frac {\partial f}{\partial x} = 2 x y z ^ {3}, \quad \frac {\partial f}{\partial y} = x ^ {2} z ^ {3}, \quad \frac {\partial f}{\partial z} = 3 x ^ {2} y z ^ {2}$$

from where we can easily find (by integrating with respect to $x$ first one equation)

$$f = x ^ {2} y z ^ {3} + C$$

where $C$ in integration constant which can be found from condition

$$f (1, - 2, 2) = 4$$

we have

$$1 ^ {2} \cdot (- 2) \cdot 2 ^ {3} + C = 4$$ $$C = 4 + 16 = 20$$

Hence

$$f = x ^ {2} y z ^ {3} + 20$$