Question

Consider f and c below. f(x, y, z) = (y2z + 2xz2)i + 2xyzj + (xy2 + 2x2z)k,
c.x = t , y = t + 5, z = t2, 0 ≤ t ≤ 1 (a) find a function f such that f = ∇f.

Answer 1

$\dfrac{\partial f}{\partial x}=y^2z+2xz^2$
$f(x,y,z)=xy^2z+x^2z^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=2xyz+\dfrac{\partial g}{\partial y}=2xyz$
$\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$\dfrac{\partial f}{\partial z}=xy^2+2x^2z+\dfrac{\mathrm dh}{\mathrm dz}=xy^2+2x^2z$
$\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=0$

$\implies f(x,y,z)=xy^2z+x^2z^2+C$

Answer 2

Based on the calculations, the function (f) such that f = ∇f is given by

f(x, y, z) = xy²z + x²z².

What is a function?

A function refers to a mathematical expression that can be used to define and represent the relationship that exists between two or more variable.

How to find a function (f = ∇f).

From the given function, fₓ(x, y, z) = y²z+2xz² simply means that f(x, y, z) = xy²z + x²z² + g(y, z). Thus, we can represent this function as follows:

$f_y(x, y, z) = 2xyz + g_y(y, z)$

Since $f_y(x, y, z)$ = 2xyz,  so $g_y(y, z)$ = 0.

Let's assume g(y, z) = h(z), so the function becomes:

f(x, y, z) = xy²z + x²z² + h(z)

On the z-plane, the function is given by:

fz(x, y, z) = xy²+2x²z + h'(z)

When h(z) = K and h'(z) = 0, the function (f) such that f = ∇f is given by:

f(x, y, z) = xy²z + x²z².

Read more on functions here: https://brainly.com/question/4246058