Consider f and c below. f(x, y, z) = 2xz + y2 i + 2xy j + x2 + 9z2 k c: x = t2, y = t + 3, z = 3t − 1, 0 ≤ t ≤ 1 (a) find a function f such that f = ∇f. f(x, y, z) = (b) use part (a) to evaluate c ∇f · dr along the given curve c.

a)∇f = 2y + 2x + 18z

b) =108

Step-by-step explanation:

Given:

f (x,y,z ) =

The curve C :

where 0 ≤ t ≤ 1

Required:

(a) F = ∇f =? (F is a vector here)

(b) =?

Solution

First we will find the directional derivative F = ∇f

for that , we will use the formula :

∇f =

Fx= δf/δx = δ/δx = 2z i

Fy= δf/δy=δ/δy (2xy)j = 2x j

Fz= δf/δz=δ/δz = 18z k

∇f = (2z) i .i + (2x) j.j + (18z) k.k

∇f = 2z + 2x + 18z

For part b):

we will use line integral formula:

to calculate dr, we will need the curve C:

r = x(t)+y(t)+z(t)

r=

=

=

put values of y, x and z

=

=

=               (Note : f(1)-f(0))

=2(1)+162(1)+2(0)+162(0)-56

= 2+162 -56

=108

4 + 5 7/9= 9 7/9

step-by-step explanation: