Question

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.

Answer 1

Answer:

a)∇f = 2y + 2x + 18z

b) $\int\limits^._C {F} \, dr$ =108

Step-by-step explanation:

Given:

f (x,y,z ) = $(2xz+ y^{2})i + (2xy) j +(x^{2} + 9z^{2})k$

The curve C :

$x=t^{2} ,\\y= t+3\\z= 3t-1$

where 0 ≤ t ≤ 1

Required:

(a) F = ∇f =? (F is a vector here)

(b) $\int\limits^._C {F} \, dr$ =?

Solution

First we will find the directional derivative F = ∇f

for that , we will use the formula :

∇f = $F_{x}i+ F_{y} j+F_{z}k$

Fx= δf/δx = δ/δx $(2xz+ y^{2})i$ = 2z i

Fy= δf/δy=δ/δy (2xy)j = 2x j

Fz= δf/δz=δ/δz$(x^{2} + 9z^{2})k$ = 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:

$\int\limits^._C {F} \, dr$

to calculate dr, we will need the curve C:

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

r=$(t^{2})i + (t+3) j +(3t-1) k$

$\frac{dr}{dt}=\frac{dx}{dt} +\frac{dy}{dt} + \frac{dz}{dt}$

$\frac{dx}{dt} = 2t$

$\frac{dy}{dt} = 1$

$\frac{dz}{dt} = 3$

$\int\limits^._C {F} \, dr$ = $\int\limits^1_0 {F_{x} } \, dx+ F_{y} dy +F_{z} dz$

= $\int\limits^1_0 {(2z(2t) + 2x(1) + 18z (3)} \, )$

put values of y, x and z

= $\int\limits^1_0 {2(3t-1) + 2(t^{2}) +18 (3) (3t-1)} \,$

=${2t^{2} + 6t+162t -54-2}\, |^1_0$

=${ 2t^{2}+ 168t - 56} \,|^1_0$               (Note : f(1)-f(0))

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

= 2+162 -56

$\int\limits^._C {F} \, dr$ =108