Answer:
a)∇f = 2y + 2x + 18z
b) [tex]\int\limits^._C {F} \, dr[/tex] =108
Step-by-step explanation:
Given:
f (x,y,z ) = [tex](2xz+ y^{2})i + (2xy) j +(x^{2} + 9z^{2})k[/tex]
The curve C :
[tex]x=t^{2} ,\\y= t+3\\z= 3t-1[/tex]
where 0 ≤ t ≤ 1
Required:
(a) F = ∇f =? (F is a vector here)
(b) [tex]\int\limits^._C {F} \, dr[/tex] =?
Solution
First we will find the directional derivative F = ∇f
for that , we will use the formula :
∇f = [tex]F_{x}i+ F_{y} j+F_{z}k[/tex]
Fx= δf/δx = δ/δx [tex](2xz+ y^{2})i[/tex] = 2z i
Fy= δf/δy=δ/δy (2xy)j = 2x j
Fz= δf/δz=δ/δz[tex](x^{2} + 9z^{2})k[/tex] = 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:
[tex]\int\limits^._C {F} \, dr[/tex]
to calculate dr, we will need the curve C:
r = x(t)+y(t)+z(t)
r=[tex](t^{2})i + (t+3) j +(3t-1) k[/tex]
[tex]\frac{dr}{dt}=\frac{dx}{dt} +\frac{dy}{dt} + \frac{dz}{dt}[/tex]
[tex]\frac{dx}{dt} = 2t[/tex]
[tex]\frac{dy}{dt} = 1[/tex]
[tex]\frac{dz}{dt} = 3[/tex]
[tex]\int\limits^._C {F} \, dr[/tex] = [tex]\int\limits^1_0 {F_{x} } \, dx+ F_{y} dy +F_{z} dz[/tex]
= [tex]\int\limits^1_0 {(2z(2t) + 2x(1) + 18z (3)} \, )[/tex]
put values of y, x and z
= [tex]\int\limits^1_0 {2(3t-1) + 2(t^{2}) +18 (3) (3t-1)} \,[/tex]
=[tex]{2t^{2} + 6t+162t -54-2}\, |^1_0[/tex]
=[tex]{ 2t^{2}+ 168t - 56} \,|^1_0[/tex] (Note : f(1)-f(0))
=2(1)+162(1)+2(0)+162(0)-56
= 2+162 -56
[tex]\int\limits^._C {F} \, dr[/tex] =108
